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Punjab State Board PSEB 12th Class Physics Book Solutions Chapter 9 Ray Optics and Optical Instruments Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

PSEB 12th Class Physics Guide Ray Optics and Optical Instruments Textbook Questions and Answers

Question 1.
A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?
Answer:
Size of the candle, h = 2.5 cm
Image size = h’
Object distance, u = -27 cm
Radius of curvature of the concave mirror, R = -36 cm
Focal length of the concave mirror, f = \(\frac{R}{2}=\frac{-36}{2}\) = -18 cm
Image distance = v

The image distance can be obtained using the mirror formula

The negative sign shows that the image is formed in front of the mirror i.e., on the side of the object itself. Thus the screen must be placed at a distance of 54 cm in front of the mirror.
The magnification of the image is given as

The height of the candle’s image is 5 cm. The negative sign indicates that the image is inverted and virtual.
If the candle is moved closer to the mirror, then the screen will have to be moved away from the mirror in order to obtain the image.

Question 2.
A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror.
Answer:
Given u = -12 cm, f = +15 cm. (convex mirror)

That is image is formed at a distance of 6.67 cm behind the mirror.
Magnification m = \(-\frac{v}{u}=-\frac{\frac{20}{3}}{-12} \) = \(\frac{5}{9}\)
Size of image I = mO = \(\frac{5}{9}\) x 4.5 = 2.5 cm
The image is erect, virtual and has a size 2.5 cm.

Its position is 6.67 cm behind the mirror when needle is moved farther, the image moves towards the focus and its size goes on decreasing.

Question 3.
A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
Answer:
Case I:
When tank is filled with water Actual depth of the needle in water, h₁ = 12.5cm
Apparent depth of the needle in water, h₂ =9.4cm
Refractive index of water = μ
The value μ can be obtained as follows
μ = \(\frac{\text { Actual depth }}{\text { Apparent depth }}\)
= \(\frac{h_{1}}{h_{2}}=\frac{12.5}{9.4}\) ≈ 1.33
Hence, the refractive index of water is about 1.33

Case II: When tank is filled with liquid
Water is replaced by a liquid of refractive index, μ’ = 1.63
The actual depth of the needle remains the same, but its apparent depth changes.
Let y be the new apparent depth of the needle. Hence, we can write the relation
μ’ = \(\frac{h_{1}}{y}\)
y = \(\frac{h_{1}}{\mu^{\prime}}=\frac{12.5}{1.63}\) = 7.67 cm
Hence, the new apparent depth of the needle is 7.67cm. It is less than h2 Therefore, to focus the needle again, the microscope should be moved up. Distance by which the microscope should be moved up =9.4-7.67 = 1.73 cm.

Question 4.
Figures 9.34 (a) and (b) show refraction of a ray in air incident at 60° with the normal to a glass air and water-air interface, respectively. Predict the angle of refraction in glass when the angle of incidence in water is 45° with the normal to a water-glass interface [Fig. 9.34 (c)]

Answer:
As per the given figure, for the glass-air interface
Angle of incidence, i = 60°
Angle of refraction, r = 35°
The relative refractive index of glass with respect to air is given by Snell’s law as
^aμ_g = \( \frac{\sin i}{\sin r}\)
= \(\frac{\sin 60^{\circ}}{\sin 35^{\circ}}=\frac{0.8660}{0.5736}\) = 1.51 …………………….. (1)
As per the given figure, for the air-water interface
Angle of incidence, j = 600
Angle of refraction, r = 470
The relative refractive index of water with respect to air is given by Snell’s law as
^wμ_w = \( \frac{\sin i}{\sin r}\)
= \(\frac{\sin 60^{\circ}}{\sin 47^{\circ}}=\frac{0.8660}{0.7314}\) = 1.184 …………………………… (2)

Using equations (1) and (2), the relative refractive index of glass with respect to water can be obtained as
^wμ_g = \(\frac{a_{g}}{a_{w_{w}}}\)
= \( \frac{1.51}{1.184} \) = 1.275

The following figure shows the situation involving the glass-water interface

Angle of incidence, i = 45
Angle of reflection = r
From Snell’s law, r can be calculated as, \(\frac{\sin i}{\sin r}\) = ^wμ_g
\(\frac{\sin 45^{\circ}}{\sin r}\) = 1.275
sin r = \(\frac{\frac{1}{\sqrt{2}}}{1.275}=\frac{0.707}{1.275}\) = 0.5546
r = sin^-1(0.5546) = 38.68°
Hence, the angle of refraction at the water-glass interface is 38.68°

Question 5.
A small bulb is placed at the bottom of a tank containirg water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33 (Consider the bulb to be a point source.)
Answer:
Actual depth of the bulb in water, d₁ = 80 cm = 0.8 m
Refractive index of water, μ = 1.33
The given situation is shown in the following figure

where,
i = Angle of Incidence
r = Angle of Refraction = 90°
Since the bulb is a point source, the emergent light can be considered as a circle of radius,
R = \(\frac{A C}{2}\) = AO = OC
Using Snell’s law, we can write the relation for the refractive index of water as
μ = \(\frac{\sin r}{\sin i}\)
1.33 = \(\frac{\sin 90^{\circ}}{\sin i}\)
i = sin^-1\(\left(\frac{1}{1.33}\right)\) = 48.75°

Using the given figure, we have the relation
tan i = \(\frac{O C}{O B}=\frac{R}{d_{1}}\)
∴R = tan 48.75° x 0.8 = 0.91 m
∴ Area of the surface of water = πR²
= π(0.91)²
= 2.61 m²
Hence, the area of the surface of water through which the light from the bulb can emerge is approximately 2.61 m².

Question 6.
A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. The angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? The refracting angle of the prism is 60°. If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light.
Answer:
Angle of minimum deviation, δ_m = 40 °
Refracting angle of the prism, A = 60°
Refractive index of water, μ = 1.33
Let μ’ be the refractive index of the material of the prism.
The angle of deviation and refracting angle of the prism are related to refractive index (μ’) as
μ’ = \(\frac{\sin \left(\frac{A+\delta_{m}}{2}\right)}{\sin \left(\frac{A}{2}\right)} \)
= \(\frac{\sin \left(\frac{60^{\circ}+40^{\circ}}{2}\right)}{\sin \left(\frac{60^{\circ}}{2}\right)}=\frac{\sin 50^{\circ}}{\sin 30^{\circ}}=\frac{0.766}{0.5}\)
= 1.532
Hence, the refractive index of the material of the prism is 1.532.
Since the prism is placed in water, let 8 ^ be the new angle of minimum deviation for the same prism.

The refractive index of glass with respect to water is given by the relation


Hence, the new minimum angle of deviation is 10.32°.

Question 7.
Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?
Answer:
Lens maker formula is
\(\frac{1}{f}=(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\) …………………………………… (1)
If R is radius of curvature of double convex lens, then,

∴ R = 2(n-1)f
Here, n =1.55, f = +20 cm
∴ R = 2 (1.55 -1) x 20 = 22 cm

Question 8.
A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, and (b) a concave lens of focal length 16 cm?
Answer:
In the given situation, the object is virtual and the image formed is real.
Object distance, u = +12cm
(a) Focal length of the convex lens, f = 20 cm
Image distance = v
According to the lens formula, we have the relation

∴ v = \(\frac{60}{8}\) = 7.5cm
Hence, the image is formed 7.5cm away from the lens, toward its right.

(b) Focal length of the concave lens, f = -16 cm
Image distance = v
According to the lens formula, we have the relation

∴ v = 48 cm
Hence, the image is formed 48 cm away from the lens, toward its right.

Question 9.
An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved further away from the lens?
Answer:
Size of object O = 3.0 cm
u = -14 cm, f = -21 cm (concave lens)
∴ Formula \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
⇒ \(\frac{1}{v}=\frac{1}{f}+\frac{1}{u}\)
or \(\frac{1}{v}=\frac{1}{-21}+\frac{1}{-14}=-\frac{2+3}{42}\)
or v = \(-\frac{42}{5}\) = -8.4 cm
Size of image I = \(\frac{v}{u}\) O
= \(\frac{-8.4}{-14}\) x 3.0 cm = 1.8 cm

That is, image is formed at a distance of 8.4 cm in front of lens. The image is virtual, erect and of size 1.8 cm. As the object is moved farther from the lens, the image goes on shifting towards focus and its size goes on decreasing. The image is never formed beyond the focus of the concave lens.

Question 10.
What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system
a converging or a diverging lens? Ignore thickness of the lenses.
Answer:
Given f₁ = +30 cm, f₂ = -20 cm
The focal length (F) of combination is given by
\(\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\)
⇒ F = \(\frac{f_{1} f_{2}}{f_{1}+f_{2}}\)
= \(\frac{30 \times(-20)}{30-20}\) = -60 cm
That is, the focal length of combination is 60 cm and it acts like a diverging lens.

Question 11.
A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25cm), and (b) at infinity? What is the magnifying power of the microscope in each case?
Answer:
Focal length of the objective lens, f₀ = 2.0 cm
Focal length of the eyepiece, f_e = 6.25cm
Distance between the objective lens and the eyepiece, d = 15cm
(a) Least distance of distinct vision, d’ = 25cm
∴ Image distance for the eyepiece, v_e = -25cm
Object distance for the eyepiece = u_e
According to the lens formula, we have the relation
\(\frac{1}{v_{e}}-\frac{1}{u_{e}}=\frac{1}{f_{e}}\)
or \(\frac{1}{u_{e}}=\frac{1}{v_{e}}-\frac{1}{f_{e}}\)
= \(\frac{1}{-25}-\frac{1}{6.25}=\frac{-1-4}{25}=\frac{-5}{25}\)
∴ u_e = -5cm
Image distance for the objective lens, v₀ = d + ue =15-5 = 10 cm
Object distance for the objective lens = u₀
According to the lens formula, we have the relation
\(\frac{1}{v_{o}}-\frac{1}{u_{o}}=\frac{1}{f_{o}}\)
\(\frac{1}{u_{0}}=\frac{1}{v_{0}}-\frac{1}{f_{0}}=\frac{1}{10}-\frac{1}{2}=\frac{1-5}{10}=\frac{-4}{10}\)
∴ u₀=-2.5cm
Magnitude of the object distance, |u₀| = 2.5 cm
The magnifying power of a compound microscope is given by the relation
m = \(\frac{v_{o}}{\left|u_{o}\right|}\left(1+\frac{d^{\prime}}{f_{e}}\right)\)
= \(\frac{10}{2.5}\left(1+\frac{25}{6.25}\right)\) = 4(1+4) = 20
Hence, the magnifying power of the microscope is 20.

(b) The final image is formed at infinity.
∴ Image distance for the eyepiece, v_e = ∞
Object distance for the eyepiece = u_e
According to the lens formula, we have the relation
\(\frac{1}{v_{o}}-\frac{1}{u_{o}}=\frac{1}{f_{o}}\)
\(\frac{1}{u_{o}}=\frac{1}{v_{o}}-\frac{1}{f_{o}}=\frac{1}{8.75}-\frac{1}{2.0}=\frac{2-8.75}{17.5}\)
∴ u₀ = \(\frac{17.5}{6.75}\) = -2.59 cm
Magnitude of the object distance, |u₀| = 2.59 cm
The magnifying power of a compound microscope is given by the relation
m = \(\frac{v_{o}}{\left|u_{o}\right|}\left(1+\frac{d^{\prime}}{f_{e}}\right)\)
= \(\frac{8.75}{2.59} \times\left(1+\frac{25}{6.25}\right)\) = 13.51
Hence, the magnifying power of the microscope is 13.51.

Question 12.
A person with a normal near point (25cm) using a compound microscope with objective of focal length 8.0 mm and an eyepiece of focal length 2.5 cm can bring an object placed at 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? Calculate the magnifying power of the microscope.
Answer:
Focal length of the objective lens, f₀= 8 mm = 0.8cm
Focal length of the eyepiece, f_e = 2.5 cm
Object distance for the objective lens, u₀ = -9.0 mm = -0.9 cm
Least distance of distant vision, d = 25 cm
Image distance for the eyepiece, v_e = -d = -25 cm
Object distance for the eyepiece = u_e

Using the lens formula, we can obtain the value of u_e as

∴ u_e = \(-\frac{25}{11}\) = -2.27 cm
We can also obtain the value of the image distance for the objective lens (v₀) using the lens formula.

∴ v₀ = 7.2 cm
The distance between the objective lens and the eyepiece = |u_e|+v₀
= 2.27+ 7.2 = 9.47cm
The magnifying power of the microscope is calculated as \(\frac{v_{o}}{\left|u_{o}\right|}\left(1+\frac{d}{f_{e}}\right)\)
= \(\frac{7.2}{0.9}\left(1+\frac{25}{2.5}\right)\)
= 8(1 +10) = 88
Hence, the magnifying power of the microscope is 88.

Question 13.
A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece?
Answer:
Focal length of the objective lens, f₀ = 144 cm
Focal length of the eyepiece, f_e = 6.0 cm
The magnifying power of the telescope is given as, m = \(\frac{f_{o}}{f_{e}}=\frac{144}{6}\) = 24
The separation between the objective lens and the eyepiece is calculated as
= f_o + f_e
= 144 + 6 = 150 cm
Hence, the magnifying power of the telescope is 24 and the separation between the objective lens and the eyepiece is 150 cm.

Question 14.
(a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope?
(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 x 10⁶ m, and the radius of lunar orbit is 3.8 x 10⁸ m.
Answer:
(a) Given f₀ = 15 m,
f_e = 1.0 cm = 1.0 x 10^-2 m
Angular magnification of telescope,
m = \(-\frac{f_{o}}{f_{e}}=-\frac{15}{1.0 \times 10^{-2}}\) = -1500
Negative sign shows that the final image is inverted.
(b) Let D be diameter of moon, d diameter of image of moon formed by objective and r be the distance of moon from objective lens, then

Question 15.
Use the mirror equation to deduce that:
(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
(b) a convex mirror always produces a virtual image independent of the location of the object.
(c) the virtual image produced by a convex mirror is always diminished In size and is located between the focus and the
pole.
(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image.
[Note: This exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]
Answer:
(a) For a concave mirror, the focal length (f) is negative
∴ f<o
When the object is placed on the left side of the mirror, the object distance (u) is negative.
∴ u<O
For image distance v, we can write the mirror formula
\(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\) …………………………………… (1)
The object lies between f and 2f.
∴ 2f < u < f (∵ u and f are negative) ∴ \(\frac{1}{2 f}>\frac{1}{u}>\frac{1}{f}\)
\(-\frac{1}{2 f}<-\frac{1}{u}<-\frac{1}{f}\)
\(\frac{1}{f}-\frac{1}{2 f}<\frac{1}{f}-\frac{1}{u}<0\) ………………………………… (2)
Using equation (1), we get
\(\frac{1}{2 f}<\frac{1}{v}<0\)

∴ \(\frac{1}{v}\) is negative, i.e., v is negative.
\(\frac{1}{2 f}<\frac{1}{v}\) 2f > v
-v > -2 f
Therefore, the image lies beyond 2f.

(b) For a convex mirror, the focal length (f) is positive.
∴ f>o
When the object is placed on the left side of the mirror, the object distance (u) is negative.
∴ u<O
For image distance y, we have the mirror formula
\(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\)
Using equation (2), we can conclude that
\(\frac{1}{\nu}\) < 0 v v> 0
Thus, the image is formed on the back side of the mirror.
Hence, a convex mirror always produces a virtual image, regardless of the object distance.

(c) For a convex mirror, the focal length (f) is positive.
∴ f> 0
When the object is placed on the left side of the mirror, the object distance (u) is negative.
∴ u< 0
For image distance v, we have the mirror formula
\(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\)
But we have u < 0 ∴ \(\frac{1}{v}>\frac{1}{f}\)
v < f
Hence, the image formed is diminished and is located between the focus (f) and the pole.

(d) For a concave mirror, the focal length (f) is negative.
∴ f< 0
When the object is placed on the left side of the mirror, the object distance (u) is negative.
∴ u< 0 It is placed between the focus (f) and the pole. ∴f > u > 0
\(\frac{1}{f}<\frac{1}{u}\) < 0 \(\frac{1}{f}-\frac{1}{u}\) > 0
For image distance v, we have the mirror formula

The image is formed on the right side of the mirror. Hence, it is a virtual image.
For u < 0 and v > 0, we can write
\(\frac{1}{u}>\frac{1}{v}\)
v > u
Magnification, m = \(\frac{v}{u}\) > 1 u
Hence, the formed image is enlarged.

Question 16.
A small pin fixed on a tabletop is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab?
Answer:
Actual depth of the pin, d = 15cm
Apparent depth of the pin = d’
Refractive index of glass, µ = 1.5

Ratio of actual depth to the apparent depth is equal to the refractive index of glass, i.e.
µ = \(\frac{d}{d^{\prime}}\)
∴ d’ = \(\frac{d}{\mu}\)
= \(\frac{15}{1.5}\) = 10 cm
The distance at which the pin appears to be raised = d-d’=15-10 = 5 cm
For a small angle of incidence, this distance does not depend upon the location of the slab.

Question 17.
(a) Figure 9.35 shows a cross-section of a ‘light pipe’ made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure.
(b) What is the answer if there is no outer covering of the pipe?

Answer:
(a) Refractive index of the glass fibre, µ₂ = 1.68
Refractive index of the outer covering of the pipe, µ₁ =1.44
Angle of incidence = i
Angle of refraction = r
Angle of incidence at the interface = i’

The refractive index (µ) of the inner core-outer core interface is given as
µ = \(\frac{\mu_{2}}{\mu_{1}}=\frac{1}{\sin i^{\prime}}\)
sin i’ = \(\frac{\mu_{1}}{\mu_{2}}=\frac{1.44}{1.68}\) = 0.8571
∴ i’ = 59°

For the critical angle, total internal reflection (TIR) takes place only when i > i’. i.e., i > 59°
Maximum angle of reflection, r_max = 90°-i’ = 90°-59°= 31°
Let, i_max be the maximum angle of incidence.
The refractive index at the air – glass interface, µ₂ =1.68
µ₂ = \(\frac{\sin i_{\max }}{\sin r_{\max }}\)
sin i_max = µ₂ sin r_max = 1.68 sin31°
= 1.68 x 0.5150
= 0.8652
∴i_max = sin^-1 (0.8652) ≈ 60°
Thus, all the rays incident at angles lying in the range 0 < i < 60° will suffer total internal reflection.

(b) If the outer covering of the pipe is not present, then
Refractive index of the outer pipe, µ₁ = Refractive index of air = 1
For the angle of incidence i =90°, we can write Snell’s law at the air-pipe interface as
\(\frac{\sin i}{\sin r}\) = µ₂ = 1.68
sin r = \(\frac{\sin 90^{\circ}}{1.68}=\frac{1}{1.68}\)
r = sin^-1(0.5952)
∴ i’ = 90°-36.5°= 53.5°
Since i’ > r, all incident rays will suffer total internal reflection.

Question 18.
Answer the following questions:
(a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.
(b) A virtual image, we always say, cannot be caught on a screen. Yet when we ‘see’ a virtual image, we are obviously bringing it onto the ‘screen’(i.e., the retina) of our eye. Is there a contradiction?
(c) A diver underwater, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter to the diver than what he actually is?
(d) Does the apparent depth of a tank of water change if viewed obliquely? If so, does the apparent depth increase or decrease?
(e) The refractive index of diamond is much greater than that – of ordinary glass. Is this fact of some use to a diamond cutter?
Answer:
(a) Yes, they produce real images under some circumstances. If the object is virtual, i.e., if the light rays converging at a point behind a plane mirror (or a convex mirror) are reflected to a point on a screen placed in front of the mirror, then a real image will be formed.

(b) No, there is no contradiction. A virtual image is formed when light rays diverge. The convex lens of the eye causes these divergent rays to converge at the retina. In this case, the virtual image serves as an object for the lens to produce a real image.

(c) The diver is in the water and the fisherman is on land (i.e., in the air). Water is a denser medium than air. It is given that the diver is viewing the fisherman. This indicates that the light rays are traveling from a denser medium to a rarer medium. Hence, the refracted rays will move away from the normal. As a result, the fisherman will appear to be taller.

(d) Yes, the apparent depth of a tank of water changes when viewed obliquely. This is because light bends on traveling from one medium to another. The apparent depth of the tank, when viewed obliquely, is less than the near-normal viewing.

(e) Yes, the refractive index of diamond (2.42) is more than that of ordinary glass (1.5). The critical angle for diamond is less than that for glass. A diamond cutter uses a large angle of incidence to ensure that the light entering the diamond is totally reflected from its faces. This is the reason for the sparkling effect of a diamond.

Question 19.
The image of a small electric bulb on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose? ’’
Answer:
Here, u + v = 3 m, :.v = 3 -u
From lens formula,

or u = \(\frac{3 \pm \sqrt{3^{2}-4.3 f}}{2}\)
For real solution, 9 -12, f should be positive.
It., 9 -12f > 0
or 9 >12f.
or f < \(\frac{9}{12}\) < \(\frac{3}{4}\) m
∴ The maximum focal length of the lens required for the purpose is \(\frac{3}{4}\) m
i.e, f_max = 0.7 m

Question 20.
A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.
Answer:
Here, O is a position of object and I is position of image (screen).
Distance OI = 90 cm
L₁ and L₂ are the two positions of the lens.
∴ Distance between L₁ and L₂ = O₁ O₂ = 20 cm
For Position L₁ of the Lens: Let x be the distance of the object from the lens.
∴ u₁ = -x
∴ Distance of the image from the lens, v₁ = +(90 – x)

If f be the focal length of the lens, then using lens formula,
\(-\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\) we get
\(-\frac{1}{-x}+\frac{1}{90-x}=\frac{1}{f}\)
or \(\frac{1}{f}=\frac{1}{x}+\frac{1}{90-x}\) ……………………………….. (1)
For Position L₂ of the Lens : Let u₂ and v₂ be the distances of the object and image from the lens in this position.
∴ u₂=-(X + 20),
v₂ = +[90-(x+20)] = +(70-x)
∴ Using lens formula,

Question 21.
(a) Determine the ‘effective focal length’ of the combination of the two lenses in Exercise 9.10, if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side of the combination a beam of parallel light is incident? Is the notion of effective focal length of this system useful at all?

(b) An object 1.5 cm in size is placed on the side of the convex lens in the arrangement (a) above. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system and the size of the image.
Answer:
Focal length of the convex lens, f₁ = 30 cm
Focal length of the concave lens,f₂ = -20 cm
Distance between the two lenses, d = 8.0 cm

(a)
(i) When the parallel beam of light is incident on the convex lens first.
According to the lens formula, we have
\(\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}\)
where, μ₁ = Object distance = ∞, v₁ = Image distance = ?
\(\frac{1}{v_{1}}=\frac{1}{30}-\frac{1}{\infty}=\frac{1}{30}\)
∴ v₁ = 30 cm
The image will act as a virtual object for the concave lens. Applying lens formula to the concave lens, we have
\(\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}\)
where, u₂ = Object distance = (30 – d) = 30 – 8 = 22 cm,
v₂ = Image distance=?
\(\frac{1}{v_{2}}=\frac{1}{22}-\frac{1}{20}=\frac{10-11}{220}=\frac{-1}{220}\)
∴ v₂ = -220 cm
The parallel incident beam appears to diverge from a point that is \(\left(220-\frac{d}{2}=220-\frac{8}{2}=220-4=216 \mathrm{~cm}\right)\) from the centre of the combination of the two lenses.

(ii) When the parallel beam of light is incident, on the concave lens first. According to the lens formula, we have
\(\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}\)
\(\frac{1}{v_{2}}=\frac{1}{f_{2}}+\frac{1}{u_{2}}\)
where, u₂ = Object distance = -∞, v₂ = Image distance = ?
\(\frac{1}{v_{2}}=\frac{1}{-20}+\frac{1}{-\infty}=-\frac{1}{20}\)
∴ v₂ = -20 cm
The image will act as a real object for the .convex lens.
Applying lens formula to the convex lens, we have
\(\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}\)
where, u₁ = Object distance = -(20 + d) = -(20 + 8) = -28 cm v₁ = Image distance = ?
\(\frac{1}{v_{1}}=\frac{1}{30}+\frac{1}{-28}=\frac{14-15}{420}=\frac{-1}{420}\)
∴ v₁ = -420 cm
Hence, the parallel incident beam appear to diverge from a point that is (420 – 4 = 416 cm) from the left of the centre of the combination of the two lenses. The answer depends on the side of the combination at which the parallel beam of light is incident. The notion of effective focal length does not seem to be useful for this combination.

(b) Height of the object, h₁ =1.5 cm
Object distance from the side of the convex lens, u₁ = -40 cm
|u_i| = 40 cm

According to the lens formula
\(\frac{1}{v_{1}}-\frac{1}{u_{1}}=\frac{1}{f_{1}}\)
where, v₁ = Image distance =?
\(\frac{1}{v_{1}}=\frac{1}{30}+\frac{1}{-40}=\frac{4-3}{120}=\frac{1}{120}\)
∴ v₁ = 120 cm
Magnification, m= \(\frac{v_{1}}{\left|u_{1}\right|}=\frac{120}{40}\) = 3

Hence, the magnification due to the convex lens is 3.
The image formed by the convex lens acts as an object for the concave lens.
According to the lens formula
\(\frac{1}{v_{2}}-\frac{1}{u_{2}}=\frac{1}{f_{2}}\)
where,
u₂ = Object distance = +(120 —8)=112 cm
v₂= Image distance =?

Magnification, m’ = \(\left|\frac{v_{2}}{u_{2}}\right|=\frac{2240}{92} \times \frac{1}{112}=\frac{20}{92}\)
Hence, the magnification due to the concave lens is \(\frac{20}{92}\)
The magnification produced by the combination of the two lenses is calculated as m x m’ = \(3 \times \frac{20}{92}=\frac{60}{92}\) = 0.652
The magnification of the combination is given as
\(\frac{h_{2}}{h_{1}}\) = 0.652
h₂ = 0.652 x h₁
where, h₁ = Object size = 1.5 cm,
h₂ = Size of the image
∴ h₂ = 0.652 x 1.5 = 0.98 cm
Hence, the height of the image is 0.98 cm.

Question 22.
At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524.
Answer:
The incident, refracted, and emergent rays associated with a glass prism ABC are shown in the given figure

Angle of prism, A = 60°
Refractive index of the prism, μ = 1.524
i₁ = Incident angle
r₂ = Refracted angle
r₂ = Angle of incidence at the face
AC = e = Emergent angle = 90°

According to Snell’s law, for face AC, we can have sine
\(\frac{\sin e}{\sin r_{2}}\) = μ

It is clear from the figure that angle A = r₁ + r₂
According to Snell’s law, we have the relation
μ = \(\frac{\sin i_{1}}{\sin r_{1}} \)
sin i₁ = μ sin r₁
= 1.524 x sin19°= 0.496
∴ i₁= 29.75°
Hence, the angle of incidence is 29.75°.

Question 23.
You are given prisms made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms which will
(a) deviate a pencil of white light without much dispersion,
(b) disperse (and displace) a pencil of white light without much deviation.
Answer:
(a) Place the two prisms beside each other. Make sure that their bases are on the opposite sides of the incident white light, with their faces touching each other. When the white light is incident on the first prism, it will get dispersed. When this dispersed light is incident on the second prism, it will recombine and white light will emerge from the combination of the two prisms.

(b) Take the system of the two prisms as suggested in answer (a). Adjust (increase) the angle of the flint-glass-prism so that the deviations due to the combination of the prisms become equal. This combination will disperse the pencil of white light without much deviation.

Question 24.
For a normal eye, the far point is at infinity and the near point of distinct vision is about 25 cm in front of the eye. The cornea of . the eye provides a converging power of about 40 dioptres, and the least converging power of the eye-lens behind the cornea is about 20 dioptres. From this rough data estimate the range of accommodation (i.e., the range of converging power of the eye-lens) of a normal eye.
Answer:
Least distance of distinct vision, d = 25 cm
Far point of a normal eye, d’ = ∞
Converging power of the cornea, P_c = 40 D
Least converging power of the eye- lens, P_e = 20 D
To see the objects at infinity, the eye uses its least converging power.
Power of the eye-lens, P = P_c + P_e =40+20 = 60 D
Power of the eye-lens is given as
P = \(\frac{1}{\text { Focal length of the eye lens }(f)} \)
f = \(=\frac{1}{P}=\frac{1}{60 D}=\frac{100}{60}=\frac{5}{3}\) cm

To focus an object at the near point, object distance (u) = -d = -25 cm
Focal length of the eye-lens = Distance between the cornea and the retina = Image distance
Hence, image distance, v = \( \frac{5}{3}\) cm
According to the lens formula, we can write
\(\frac{1}{f^{\prime}}=\frac{1}{v}-\frac{1}{u}\)
Where f’ = Focal length

Power of the eye-lens = 64-40 = 24 D
Hence, the range of accommodation of the eye-lens is from 20 D to24D.

Question 25.
Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accommodation? If not, what might cause these defects of vision?
Answer:
A myopic or hypermetropic person can also possess the normal ability of accommodation of the eye-lens. Myopia occurs when the eyeballs get elongated from front to back. Hypermetropia occurs when the eye-balls get shortened.
When the eye-lens loses its ability of accommodation, the defect is called presbyopia.

Question 26.
A myopic person has been using spectacles of power -1.0 dioptre for distant vision. During old age he also needs to use separate reading glass of power +2.0 dioptres. Explain what may have happened.
Answer:
The power of the spectacles used by the myopic person, P = -1.0 D
Focal length of the spectacles, f = \(\frac{1}{P}=\frac{1}{-1 \times 10^{-2}}\) = -100 cm
Hence, the far point of the person is 100 cm. He might have a normal near point of 25 cm. When he uses the spectacles, the objects placed at infinity produce virtual images at 100 cm.
He uses the ability of accommodation of the eye-lens to see the objects placed between 100 cm and 25 cm.
During old age, the person uses reading glasses of power, P’ = +2D The ability of accommodation is lost in old age.
This defect is called presbyopia. As a result, he is unable to see clearly the objects placed at 25 cm.

Question 27.
A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected?
Answer:
In the given case, the person is able to see vertical lines more distinctly than horizontal lines. This means that the refracting system (cornea and eye-lens) of the eye is not working in the same way in different planes. This defect is tailed astigmatism. The person’s eye has enough curvature in the vertical plane. However, the curvature in the horizontal plane is insufficient. Hence, sharp images of the vertical lines are formed on the retina, but horizontal lines appear blurred. This defect can be corrected by using cylindrical lenses.

Question 28.
A man with normal near point (25cm) reads a book with small print using a magnifying glass: a thin convex lens of focal length 5 cm.
(a) What is the closest and the farthest distance at which he should keep the lens from the page so that he can read the book when viewing through the magnifying glass?
(b) What is the maximum and the minimum angular magnification (magnifying power) possible using the above simple microscope?
Answer:
(a) Focal length of the magnifying glass, f = 5 cm
Least distance of distinct vision, d = 25 cm
Closest object distance = u
Image distance, v = -d = -25 cm
According to the lens formula, we have


Hence, the closest distance at which the person can read the book is 4.167 cm.
For the object at the farthest distance (u’), the image distance (v’) = ∞

According to the lens formula, we have

∴ u’ = -5 cm
Hence, the farthest distance at which the person can read the book is 5 cm.
(b) Maximum angular magnification is given by the relation
α_max= \(\frac{d}{|u|}=\frac{25}{\frac{25}{6}} \) = 6
Minimum angular magnification is given by the relation
α_min = \(\frac{d}{\left|u^{\prime}\right|}=\frac{25}{5} \) = 5.

Question 29.
A card sheet divided into squares each of size 1 mm² is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 10 cm) held close to the eye.
(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?
(b) What is the angular magnification (magnifying power) of the lens?
(c) Is the magnification in (a) equal to the magnifying power in (b)?
Explain.
Answer:
(a) Area of each square, A = 1 mm²
Object distance, u = -9 cm
Focal length of the converging lens, f = 10 cm
For image distance v, the lens formula can be written as

∴ v = -90 cm
Magnification, m = \(=\frac{v}{u}=\frac{-90}{-9}\) =10
∴ Area of each square in the virtual image = (10)²A
= 10² x 1 =100 mm² = 1 cm²
(b) Magnifying power of the lens = \(\frac{d}{|u|}=\frac{25}{9}\) = 2.8
(c) The magnification in (a) is not the same as the magnifying power in(b).
The magnification magnitude is \(\left(\left|\frac{v}{u}\right|\right)\) and the magnifying power is \(\left(\frac{d}{|u|}\right) \) .
The two quantities will be equal when the image is formed at the near point (25 cm).

Question 30.
(a) At what distance should the lens be held from the figure in Exercise 9.29 in order to view the squares distinctly with the maximum possible magnifying power?
(b) What is the magnification in this case?
(c) Is the magnification equal to the magnifying power in this case? Explain.
Answer:
(a) The maximum possible magnification is obtained when the image is formed at the near point (d = 25cm).
Image distance, v = -d = -25 cm
Focal length, f = 10 cm
Object distance = u
According to the lens formula, we have

∴ u = \(-\frac{50}{7}\) = -7.14 cm
Hence, to view the squares distinctly, the, lens should be kept 7.14 cm away from them. .
(b) Magnifying = \(\left|\frac{v}{u}\right|=\frac{25}{50}\) =3.5
(c) Magnifying power = \(\frac{d}{u}=\frac{25}{\frac{50}{7}}\) = 3.5
Since the image is formed at the near point (25 cm), the magnifying power is equal to the magnitude of magnification.

Question 31.
What should be the distance between the object in Exercise 9.30 and the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm². Would you be able to see the squares distinctly with your eyes very close to the magnifier? [Note: Exercises 9.29 to 9.31 will help you clearly understand the difference between magnification in absolute size and the angular magnification (or magnifying power) of an instrument.]
Answer:
Area of the virtual image of each square, A = 6.25 mm
Area of each square, A₀ = 1 mm²
Hence, the linear magnification of the object can be calculated as
m = \(\sqrt{\frac{A}{A_{0}}}=\sqrt{\frac{6.25}{1}} \) = 2.5
But m = \(\frac{\text { Image distance }(v)}{\text { Object distance }(u)} \)
∴ v = mu = 2.5 u
Focal length of the magnifying glass, f = 10 cm

According to the lens formula, we have the relation

∴ u = \(-\frac{1.5 \times 10}{2.5}\) = -6 cm
and v = 2.5 u = 2.5 x 6 = -15 cm
The virtual image is formed at a distance of 15 cm, which is less than the near point (i.e., 25 cm) of a normal eye. Hence, it cannot be seen by the eyes distinctly.

Question 32.
Answer the following questions:
(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?
(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?
(d) Why must both the objective and the eyepiece of a compound microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?
Answer:
(a) Though the image size is bigger than the object, the angular size of the image is equal to the angular size of the object. A magnifying glass helps one see the objects placed closer than the least distance of distinct vision (i.e., 25 cm). A closer object causes a larger angular size. A magnifying glass provides angular magnification. Without magnification, the object cannot be placed closer to the eye. With magnification, the object can be placed much closer to the eye.

(b) Yes, the angular magnification changes when the distance between the eye and a magnifying glass is increased, the angular magnification decreases a little. This is because the angle subtended at the eye is slightly less than the angle subtended at the lens. Image distance does not have any effect on angular magnification.

(c) The focal length of a convex lens cannot be decreased by a greater amount. This is because making lenses having very small focal lengths is not easy. Spherical and chromatic aberrations are produced by a convex lens having a very small focal length.

(d) The ang lar magificarin produced by’the eyepiece of a compound microscope is \(\left[\left(\frac{25}{f_{e}}\right)+1\right]\)
Where f_e = Focal length of the eyepiece
It can be inferred that f_e is small, then angular magnification of the eyepiece will be large.
The angular magnification of the objective lens of a compound microscope is given as
\(\frac{1}{\left(\left|u_{o}\right| f_{o}\right)}\)
Where, u₀ = Object distance for the objective lens, f₀ = Focal length of the objective
The magnification is large when u₀> f₀ . In the case of a microscope, the object is kept close to the objective lens. Hence, the object distance is very little.
Since u₀ is small, f₀ will be even smaller. Therefore, f_e and f₀ are both small in the given condition.

(e) When we place our eyes too close to the eyepiece of a compound microscope, we are unable to collect much-refracted light. As a result, the field of view decreases substantially. Hence, the clarity of the image gets blurred.
The best position of the eye for viewing through a compound microscope is at the eye-ring attached to the eyepiece. The precise location of the eye depends on the separation between the objective lens and the eyepiece.

Question 33.
An angular magnification (magnifying power) of 30X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm. How will you set up the compound microscope?
Answer:
Focal length of the objective lens, f₀ = 1.25 cm
Focal length of the eyepiece, f_e = 5 cm
Least distance of distinct vision, d = 25 cm

Angular magnification of the compound microscope = 30X
Total magnifying power of the compound microscope, m = 30
The angular magnification of the eyepiece is given by the relation
m_e = \(\left(1+\frac{d}{f_{e}}\right)=\left(1+\frac{25}{5}\right)\) = 1+5 = 6
The angular magnification of the objective lens (m₀) is related to me as
m_om_e=m
or m₀ = \(\frac{m}{m_{e}}=\frac{30}{6}\) = 5

We also have the relation
m = \( \frac{\text { Image distance for the objective lens }\left(v_{o}\right)}{\text { Object distance for the objective lens }\left(-u_{0}\right)}\)
5 = \(\frac{v_{o}}{-u_{o}}\)
∴ v₀ = -5u₀ …………………………….. (1)
Applying the lens formula for the objective lens

and v₀ = -5u₀
= -5 x (-1.5) = 7.5 cm
The object should be placed 1.5 cm away from the objective lens to obtain the desired magnification.
Applying the lens formula for the eyepiece
\(\frac{1}{v_{e}}-\frac{1}{u_{e}}=\frac{1}{f_{e}}\)
where,
v_e = Image distance for the eyepiece = -d = -25 cm
u_e = Object distance for the eyepiece
\(\frac{1}{u_{e}}=\frac{1}{v_{e}}-\frac{1}{f_{e}}=\frac{-1}{25}-\frac{1}{5}=-\frac{6}{25}\)
u_e =-4.17 cm
Separation between the objective lens and the eyepiece = \(\left|u_{e}\right|+\left|v_{o}\right|\)
= 4.17 + 7.5 = 11.67 cm
Therefore, the separation between the objective lens and the eyepiece should be 11.67 cm.

Question 34.
A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when
(a) the telescope is in normal adjustment (i.e., when the final image is at infinity)?
(b) the final image is formed at the least distance of distinct vision (25 cm)?
Answer:
Focal length of the objective lens, f₀ =140 cm
Focal length of the eyepiece, f_e = 5 cm
Least distance of distinct vision, d = 25 cm
(a) When the telescope is in normal adjustment, its magnifying power is given as
m = \(\frac{f_{o}}{f_{e}}=\frac{140}{5} \) = 28
(b) When the final image is formed at d, the magnifying power of the telescope is given as
\(\frac{f_{o}}{f_{e}}\left[1+\frac{f_{e}}{d}\right]=\frac{140}{5}\left[1+\frac{5}{25}\right]\)
= 28[1 +0.2] = 28×1.2 = 33.6

Question 35.
(a) For the telescope described in Exercise 9.34 (a), what is the separation between the objective lens and the eyepiece?
(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?
(c) What is the height of the final image of the tower if it is formed at 25 cm?
Answer:
Focal length of the objective lens, f₀ =140 cm
Focal length of the eyepiece, f_e= 5 cm
(a) In normal adjustment, the separation between the objective lens and the eyepiece = f₀ + f_e = 140 + 5 = 145 cm
(b) Height of the tower, h₁ = 100 m
Distance of the tower (object) from the telescope, u = 3 km = 3000 m
The angle subtended by the tower at the telescope is given as
θ’ = \(\frac{h_{2}}{f_{o}}=\frac{h_{2}}{140}\) rad
where,
h₂ = Height of the image of the tower formed by the objective lens
\(\frac{1}{30}=\frac{h_{2}}{140}\) (∵θ=θ’)
∴ h₂ = \(\frac{140}{30}\) = 4.7 cm
Therefore, the objective lens forms a 4.7 cm tall image of the tower.

(c) Image is formed at a distance, d = 25 cm
The magnification of the eyepiece is given by the relation
m = 1 + \(\frac{d}{f_{e}}\)
= 1+ \(\frac{25}{5}\) =1 + 5 = 6
Height of the final image = mh₂ = 6 x 4.7 = 28.2 cm
Hence, the height of the final image of the tower is 28.2 cm.

Question 36.
A Cassegrain telescope uses two mirrors as shown in Fig. 9.33. Such a telescope is built with the mirrors 20 mm apart.
If the radius of curvature of the large mirror is 220 mm and the small mirror is 140 mm, where will the final image of an object at infinity be?
Answer:
Given, r₁ = 220 mm, f₁ = \(\frac{r_{1}}{2}\) = 110 mm = 11 cm
r₂ = 140 mm, f₂ = \(\frac{r_{2}}{2}\) = 70 mm = 7.0 cm
Distance between mirrors, d = 20 mm = 2.0 cm
The parallel incident rays coming from distant objects fall on the concave mirror and try to be focused at the principal focus of concave lens, i. e., v₁ = -f₁ = -11 cm
But in the path of rays reflected from concave mirror, a convex mirror is placed. Therefore the image formed by the concave mirror acts as a virtual object for convex mirror.

For convex mirror f₂ = -7.0 cm, u₂ = -(11 -2) = -9 cm

v₂ = \(-\frac{63}{2}\) cm = -31.5 cm
This is the distance of the final image formed by the convex mirror. Thus, the final image is formed at a distance of 31.5 cm from the smaller (convex) mirror behind the bigger mirror.

Question 37.
Light incident normally on a plane mirror attached to a galvanometer coil retraces backward as shown in Fig. 9.36. A current in the coil produces a deflection of 3.5° of the mirror. What is the displacement of the reflected spot of light on a screen placed 1.5 m away?

Answer:
Angle of deflection, θ = 3.5°
Distance of the screen from the mirror, D = 1.5 m
The reflected rays get deflected by an amount twice the angle of deflection i.e., 2θ = 2 x 3.5 = 7.0°
The displacement (d) of the reflected spot of light on the screen is given as
tan 2θ = \(\frac{d}{1.5}\) d =1.5 x tan7°= 0.184 m = 18.4 cm
Hence, the displacement of the reflected spot of light is 18.4 cm.

Question 38.
Figure 9.37 shows an biconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror.
A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to he 30.0 cm. What is the refractive index of the liquid?

Answer:
Focal length of the convex lens, f₁ = 30 cm
The liquid acts as a mirror. Focal length of the liquid = f₂
Focal length of the system (convex lens + liquid), f = 45 cm
For a pair of optical systems placed in contact, the equivalent focal length is given as
\(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\)
\(\frac{1}{f_{2}}=\frac{1}{f}-\frac{1}{f_{1}}\)
= \(\frac{1}{45}-\frac{1}{30}=-\frac{1}{90}\)
∴ f₂ = -90 cm
Let the refractive index of the lens be μ₁ and the radius of curvature of one surface be R. Hence, the radius of curvature of the other surface is R.
R can be obtained using the relation \(\frac{1}{f_{1}}=\left(\mu_{1}-1\right)\left(\frac{1}{R}+\frac{1}{-R}\right)\)
\(\frac{1}{30}=(1.5-1)\left(\frac{2}{R}\right)\)
∴ R = \(\frac{30}{0.5 \times 2}\) = 30 cm

Let μ₂ be the refractive index of the liquid.
Radius of curvature of the liquid on the side of the plane minor = ∞
Radius of curvature of the liquid on the side of the lens, R = -30 cm
The value of μ₂, can be calculated using the relation
\(\frac{1}{f_{2}}=\left(\mu_{2}-1\right)\left[\frac{1}{-R}-\frac{1}{\infty}\right]\)
\(\frac{-1}{90}=\left(\mu_{2}-1\right)\left[\frac{1}{+30}-0\right]\)
μ₂ – 1 = \(\frac{1}{3} \)
∴ μ₂ = \(\frac{4}{3} \) = 133
Hence, the refractive index of the liquid is 1.33.

Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 9 Coordination Compounds Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds

PSEB 12th Class Chemistry Guide Coordination Compounds InText Questions and Answers

Question 1.
Explain the bonding in coordination compounds in terms of Werner’s postulates.
Answer:
(i) The primary valencies are satisfied by negative ions and equal-to the oxidation state of the metal.

(ii) The secondary valencies can be satisfied by neutral or negative ions. It is equal to the coordination number of the central metal atom and is fixed.

(iii) The ions bound to the central metal ion to secondary linkages have definite spatial arrangements and give geometry to the complex. While primary valency is non-directional.

Question 2.
FeSO₄ solution mixed with (NH₄)₂SO₄ solution in 1 : 1 molar ratio gives the test of Fe²⁺ ion but CuSO₄ solution mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu²⁺ ion. Explain why?
Answer:
FeSO⁴ solution mixed with (NH⁴)²SO⁴ solution in 1 : 1 molar ratio forms double salt, FeSO⁴∙(NH⁴)²SO⁴∙6H²O which ionises in the solution to give Fe2+ ions. Hence, it gives the test of Fe²⁺ ions.

CuSO₄ solution mixed with aqueous ammonia in 1 : 4 molar ratio forms a complex, with the formula [Cu(NH₃)₄]SO₄. The complex ion, [CU(NH₃)]²⁺ does not ionise to give Cu²⁺ ions. Hence, it does not give the test of Cu²⁺ ion.

Question 3.
Explain with two examples each of the following: Coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.
Answer:
Coordination entity: A coordination entity constitutes usually a central metal atom or ion, to which a fixed number of other atoms or ions or groups are attached by coordinate bonds. A coordination entity may be neutral, positively or negatively charged. For examples : [Ni(CO)₄], [CoCl₃(NH₃)₃], [Co(NH₃)₆]^3+.

Ligand : A ligand is an ion or a small molecule having at least one lone pair of electrons and capable of forming a coordinate bond with central atom or ion in the coordination entity. For example: Cl^–, OH^–, CN^–, CO, NH₃, H₂O etc.

Coordination number : The coordination number of the central atom or ion is determined by the number of a bonds between the ligands and the central atom or ion. n bonds are not consider for the determination of coordination number. The a bonding electrons may be indicated by a pair of dots (:). For example, [Co(:NH₃)₆]³⁺ and [Fe(:CN)₆]^3-.

Coordination polyhedron : The spatial arrangement of the ligands which are directly attached to the central atom or ion called coordination polyhedron.
For example: [Co(NH₃)₆]³⁺ is octahedral, [Ni(CO)₄] is tetrahedral and [PtCl₄ ]^2- is square planar.

Homoleptic and heteroleptic : Complexes in which a metal is bound to only one type of donor groups are known as homoleptic.
For example : [Co(NH₃)₆]³⁺, [PtCl₆]^2- .
Complexes in which a metal is bound to more than one kind of donor groups are known as heteroleptic. ‘
For example : [Co(NH₃)₄Cl₂]⁺, [PdI₂(ONO)₂ (H₂O)₂],

Question 4.
What is meant by unidentate, didentate and ambidentate ligands? Give two examples for each.
Answer:
A molecule or an ion which has only one donor atom to form one coordinate bond with the central metal atoms is called unidentate ligand, e.g., Cl^– and NH₃.

A molecule or an ion which contains two donor atoms and hence forms two coordinate bonds with the central metal atoms is called a didentate ligand, e.g., NH₂—CH₂—CH₂—NH₂ and ^–OOC — COO^–.
A molecule or an ion which contains two donor atoms but only one of them forms a coordinate bond at a time with the central metal atom is called ambidentate ligand, e.g., CN^–or NC^– and \(\) or : ONO^–.

Question 5.
Specify the oxidation numbers of the metals in the following coordination entities:
(i) [Co(H₂O)(CN)(en)₂]²⁺
(ii) [CoBr₂(en)₂]⁺
(iii) [PtCl₂]^2-
(iv) K₃Fe(CN)₆]
(v) [Cr(NH₃)₂Cl₃]
Answer:
(i) x + (-1) + (0) + (0) = + 2 so x = +3 (III)
(ii) x + 2(-1) + 0 = +1 so x = +3 (III)
(iii) x + 4(-1) = -2 so x = +2(11)
(iv) x + 6(-1) = -3 so x = +3 (III)
(v) x + 3(-1) + 0 = 0 so x = +3 (III)

Question 6.
Using IUPAC norms write the formulas for the following:
(i) Tetrahydroxozincate(II)
(ii) Potassium tetrachloridopalladate(II)
(iii) Diamminedichloridoplatinum(II)
(iv) Potassium tetracyanonickelate(II)
(v) Pentaamminenitrito-O-cobalt(III)
(vi) Hexaamminecobalt(III) sulphate
(vii) Potassium tri(oxalato)chromate(III)
(viii) Hexaammineplatinum(IV)
(ix) Tetrabromidocuprate(Il)
(x) Pentaamminenitrito-N-cobalt(lll)
Answer:
(i) [Zn(OH)₄]^2-
(ii) K₂[PdCl₄]
(iii) pt(NH₃)₂Cl₂]
(iv) K₂[Ni(CN)₄]
(v) [Co(ONO) (NH₃)₅]²⁺
(vi) [CO(NH₃)₆]₂ (SO₄)₃
(vii) K₃[Cr(C₂O₄)₃]
(viii) [Pt(NH₃)₆]⁴⁺
(ix) [Cu(Br)₄]^2-
(x) [Co (NO₂) (NH₃)₅]²⁺

Question 7.
Using IUPAC norms write the systematic names of the following:
(i) [CO(NH₃)₆]Cl₃
(ii) [Pt(NH₃)₂Cl(NH₂CH₃)]Cl
(iii) [Ti(H₂O)₆]³⁺
(iv) [CO(NH₃)₄Cl(NO₂)]CI
(v) [Mn(H₂O)₆]²⁺
(vi) [NiCl₄]^2-
(vii) [Ni(NH₃)₆]Cl₂
(viii) [Co(en)₃]³⁺
(ix) [Ni(CO)₄]
Answer:
(i) Hexaamminecobalt(III) chloride
(ii) Diamminechlorido(methylamine) platinum(II) chloride
(iii) Hexaquatitanium(III) ion
(iv) Tetraamminechloridonitrito-N-Cobalt(III) chloride
(v) Hexaquamanganese(II) ion
(vi) Tetrachloridonickelate(II) ion
(vii) Hexamminenickel(II) chloride
(viii) Tris(ethane-1, 2-diamine) cobalt(III) ion
(ix) Tetracarbonylnickel(O)

Question 8.
List various types of isomerism possible for coordination compounds giving an example of each.
Answer:
Two principal types of isomerism are known among coordination compounds :
(A) Sterioisomerism,
(B) Structural isomerism.
Each of which can be further sub-divided as :
(A) Stereoisomerism
(i) Geometrical isomerism : It arises in heteroleptic complexes due to different possible geometric arrangements of the ligands.
Example: Pt[(NH₃)₂Cl₂]

(ii) Optical isomerism : It is common in octahedral complexes involving didentate ligands.
Example : [Pt Cl₂(en) ₂]²⁺

Optical isomers (d and l) of cis-[PtCl₂(en)₂]²⁺

(B) Structural isomerism
(i) Linkage isomerism.
Example: [Co(NH₃)₅ (NO₂)]Cl₂
(ii) Coordination isomerism.
Example: [Co(NH₃)₆] [Cr(CN)₆]
(iii) Ionisation isomerism.
Example: [Co(NH₃)₅SO₄]Br and [CO(NH₃)₅ Br]SO₄
(iv) Solvate isomerism.
Example : [Cr(H₂O)₆] Cl₃ (violet) its solvate isomer
[Cr(H₂O)₅Cl]Cl₂. H₂O (grey-green)

Question 9.
How many geometrical isomers are possible in the following coordination entities? ’
(i) [Cr(C₂O₄)₃]^3-
(ii) [Co(NH₃)₃Cl₃]
Answer:
(i) [Cr(C₂O₄)₃]^3-,
No geometric isomer is possible as it is a bidentate ligand.
(ii) [CO(NH₃)₃Cl₃] .
Two geometrical isomers are possible.

Question 10.
Draw the structures of optical isomers of:
(i) [Cr(C₂O₄)₃]^3-
(ii) [PtCl₂(en)₂]²⁺
(iii) [Cr(NH₃)₂ Cl₂ (en)]⁺
Answer:
(i) [Cr(C₂O₄)₃]^3-

Question 11.
Draw all the isomers (geometrical and optical) of:
(i) [CoCl₂ (en)₂]⁺
(ii) [Co(NH₃)Cl(en)₂]²⁺
(iii) [Co(NH₃)₂Cl₂(en)]⁺
Answer:
(i) [CoCl₂ (en)₂]⁺

Question 12.
Write all the geometrical isomers of [Pt(NH₃)(Br)(Cl)(py)] and how many of these will exhibit optical isomers?
Answer:
Three isomers are possible as follows :

Isomers of this type do not show any optical isomerism. Optical isomerism rarely occurs in square planar or tetrahedral complexes and that too when they contain unsymmetrical chelating ligand.

Question 13.
Aqueous copper sulphate solution (blue in colour) gives :
(i) a green precipitate with aqueous potassium fluoride, and
(ii) a bright green solution with aqueous potassium chloride. Explain these experimental results.
Answer:
Aqueous copper sulphate exists as [Cu(H₂O)₄]SO₄. It is a labile complex. The blue colour of the solution is due to [Cu(H₂O)₄]²⁺ ions,

(i) When KF is added, the weak H₂O ligands are replaced by F^– ligands forming [CuF₄]^2- ions, which is a green precipitate.

(ii) When KCl is added, Cl^– ligands replace the weak H₂O ligands forming [CuCl₄]^2- ion, which has bright green colour.

Question 14.
What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H₂S (g) is passed through this solution?
Answer:
K₂[Cu(CN)₄] is formed when excess of aqueous KCN is added to an aqueous solution of CuSO₄.

As CN^–ions are strong ligands the complex is very stable. It is not replaced by S^2- ions when H₂S gas is passed through the solution and thus no precipitate of CuS is obtained.

Question 15.
Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:
(i) [Fe(CN)₆]^4-
(ii) [FeFe₆]^3-
(iii) [Co(C₂O₄)₃]^3-
(iv) [CoF₆]^3-
Answer:
(i) [Fe(CN)₆]^4-
In the above coordination complex, iron exists in the +2 oxidation state.
Fe = [Ar] 3d⁶ 4s²
Outer configuration of Fe²⁺ = 3d⁶ 4s⁰
Orbitals of Fe²⁺ ion:

As CN^– is a strong field ligand, it causes the pairing of the unpaired 3d electrons.

Since, there are six ligands around the central metal ion, the most feasible hybridisation is d²sp³. d²sp³ hybridised orbitals of Fe²⁺ are :

6 electron pairs from CN ions occupy the six hybrid d²sp³ orbitals.
Then,

Hence, the geometry of the complex is octahedral and the complex is diamagnetic (as there are no unpaired electrons).

(ii) [FeF₆]^3-
In this complex, the oxidation state of Fe is + 3.
Fe³⁺ = 3d⁵ 4s⁰
Orbitals of Fe³⁺ ion:

There are 6F^– ions. Thus, it will undergo d²sp³ or sp³d² hybridisation. As F^– is a weak field ligand, it does not cause the pairing of the electrons in the 3d orbital. Hence, the most feasible hybridisation is sp³d². sp³d² hybridised orbitals of Fe are:

Hence, the geometry of the complex is found to be octahedral.

(iii) [Co(C₂O₄)₃]^3-
Cobalt exists in the + 3 oxidation state in the given complex.
Outer configuration of Co = 3d₇ 4s₂
Co₃₊ = 3d₆4s₀
Orbitals of Co₃₊ ion:

Oxalate is a weak field ligand. Therefore, it cannot cause the pairing of the 3d electrons. As there are 6 ligands, hybridisation has to be either sp³d² or d²sp³ hybridisation. sp³d² hybridisation of Co³⁺.

The 6 electron pairs from the 3 oxalate ions (oxalate anion is a bidentate ligand) occupy these sp³d² orbitals.

Hence, the geometry of the complex is found to be octahedral.

(iv) [CoF₂]^3-
Cobalt exists in the + 3 oxidation state.
Orbitals of Co³⁺ ion:

Again, fluoride ion is a weak field ligand. It cannot cause the pairing of the 3d electrons. As a result, the Co³⁺ ion will undergo sp³d² hybridisation.
sp³d² hybridised orbitals of Co³⁺ ion are :

Hence, the geometry of complex is octahedral, 6 electron pants.

Question 16.
Draw figure to show the splitting of d-orbitals in an octahedral crystal field.
Answer:

Question 17.
What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
Answer:
The arrangement of ligands in order of their increasing field strengths, i.e., increasing crystal field splitting energy (CFSE) values is called spectrochemical series.

The ligands with a small value of CFSE (△₀) are called weak field ligands whereas those with a large value of CFSE are called strong field ligands.

Question 18.
What is crystal field splitting energy? How does the magnitude of △₀ decide the actual configuration of d-orbitals in a coordi-nation entity?
Answer:
When ligands approach a transition metal ion, the d-orbitals split into two sets, one with lower energy and the other with higher energy. The difference of energy between the two sets of orbitals is called crystal field splitting energy (△₀) in case of octahedral field.

If △₀ < P, (pairing energy), the 4th electron enters one of the e_g orbitals giving the configuration \(t_{2 g}^{3} e_{g}^{1}\), thereby forming high spin complexes.

Such ligands for which A 0 < P are called weak field ligands.
If △₀ > P, the 4th electron pairs up in one of the t2g orbitals giving the configuration \(t_{2 g}^{4} e_{g}^{0}\), thus forming low spin complexes. Such ligands for which △₀ > P are called strong field ligands.

Question 19.
[Cr(NH₃)₆]³⁺ is paramagnetic while [Ni(CN)₄]^2- is diamagnetic. Explain why?
Answer:
Cr is in the +3 oxidation state i.e., d³ configuration. Also, NH₃ is a weak field ligand that does not cause the pairing of the electrons in the orbital.

Therefore, it undergoes d²sp³ hybridisation and the electrons in the 3d orbitals remain unpaired. Hence, it is paramagnetic in nature.
In [Ni(CN)₄]^2-, Ni exists in the + 2 oxidation state i. e., d⁸ configuration.

CN^– is a strong field ligand. It causes the pairing of the 3d electrons. Then, Ni²⁺ undergoes dsp² hybridisation.

As there are no unpaired electrons, it is diamagnetic.

Question 20.
A solution of [Ni(H₂O)₆]²⁺ is green but a solution of [Ni(CN)₄]^2- is colourless. Explain.
Answer:
In [Ni(H₂O)6]²⁺, \(\mathrm{H}_{2} \ddot{\mathrm{O}}\) is a weak field ligand. Therefore, there are unpaired electrons in Ni²⁺. In this complex, the d electrons from the lower energy level can be excited to the higher energy level i. e., the possibility of d-d transition is present. Hence, [Ni(H₂O)₆]²⁺ is coloured.

In [Ni(CN)₄]²⁺, the electrons are all paired as CN^– is a strong field ligand. Therefore, d-d transition is not possible in [Ni(CN)₄]^2-. Hence, it is colourless.

Question 21.
[Fe(CN)₆]^4- and [Fe(H₂O)₆]²⁺ are of different colours in dilute solutions. Why?
Answer:
In both the complex compounds, Fe is in +2 oxidation state with configuration 3d⁶, i.e., it has four unpaired electrons. In the presence of weak H₂O ligands, the unpaired electrons do not pair up. But in the presence of strong ligand CN^– they get paired up. Then no unpaired electron is left. Due to this, difference in the number of unpaired electrons, both complex ions have different colours.

Question 22.
Discuss the nature of bonding in metal carbonyls.
Answer:
The metal carbon in metal carbonyls possesses both CT and π character. The ligand to metal is CT bond and metal to ligand is π bond. The effect of CT bond strengthens the rcbond and vice-versa. This is called synergic effect. This unique synergic provides stability to metal carbonyls.

Question 23.
Give the oxidation state, d-orbital occupation and coordination number of the central metal ion in the following complexes:
(i) K₃[CO(C₂O₄)₃]
(ii) cis-[Cr(en)₂Cl₂]Cl
(iii) (NH₄)₂[CoF₄]
(iv) [Mn(H₂O)₆]S0₄
Solution:
(i) K₃[CO(C₂O₄)₃]
The central metal ion is Co.
The oxidation state can be given as :
(+1) × 3 + × + (- 2) × 3 = 0
x – 6 = -3 ⇒ x = + 3
The d orbital occupation for Co³⁺ is \(t_{2 g}^{6} e g^{0}\).
(as \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) is strong field ligand)
Coordination number of Co = 3 × denticity of C₂O₄
= 3 × 2 (as \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) is a bidentate ligand) = 6

(ii) cis-[Cr(en)₂Cl₂]Cl
The central metal ion is Cr.
The oxidation state can be given as:
x + 2(0) + 2(-1) + (-1) = 0
x – 2 – 1 = 0
x = + 3
The d orbital occupation for Cr³⁺ is \(t_{2 g}^{3}\).
Coordination number of Cr
= 2 × denticity of en + 2
= 2 × 2 + 2 = 6

(iii) (NH₄)₂[CoF₄]
The central metal ion is Co.
The oxidation state can be given as:
(+1) × 2 + × + (-1) × 4 = 0
x – 4 = -2
x = + 2
The d orbital occupation for Co²⁺ is d⁷ or \(t_{2 g}^{5} e_{g}^{2}\). (as F^– is a weak ligand)
Coordination number of Co = 4

(iv) [Mn(H₂O)₆]S0₄
The central metal ion is Mn.
The oxidation state can be given as:
x + (0) × 6 + (- 2) = 0
x = + 2
The d orbital occupation for Mn is d⁵ or [latext_{2 g}^{3} e_{g}^{2}][/latex].
Coordination number of Mn = 6

Question 24.
Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex:
(i) K[Cr(H₂O)₂(C₂O₄)₂] 3H₂O
(ii) [Co(NH₃)₅Cl]Cl₂
(iii) CrCl₃(py)₃
(iv) Cs[FeCl₄]
(v) K₄[Mn(CN)₆]
Answer:
(i) K[Cr(H₂O)₂ (C₂O₄)2 ] ∙ 3H₂O
IUPAC name : Potassium diaquadioxalatochromate (III) hydrate.
Oxidation state of chromium
+1 + x + (0) × 2 + (- 2) × 2 + 3(0) = 0
+ 1 + x – 4 = 0
x = + 3
Electronic configuration of Cr⁺³= 3d³ = (\(t_{2 g}^{3} e_{g}^{0}\))
Coordination number = 6
Shape : Octahedral


Magnetic moment (μ) = \(\sqrt{n(n+2)}\)
= \(\sqrt{3(3+2)}\)
= \(\sqrt{15}\) = 3.87 BM

(ii) [Co(NH₃)₅Cl]Cl₂
IUPAC name : Pentaammine chlorido cobalt(III) chloride
Oxidation state of Co
x + (0)5 + (-1) + (-1) × 2 = 0
x – 3 =0
x = + 3
Coordination number = 6
Shape: Octahedral.
Electronic configuration of Co³⁺ = 3d⁶ = \(t_{2 g}^{6} e_{g}^{0}\)
The complex does not exhibit geometrical as well as optical isomerism.
Magnetic Moment (μ) = \(\sqrt{n(n+2)}\)BM = \(\sqrt{0(0+2)}\) BM = 0 BM

(iii) CrCl₃(py)₃
IUPAC name : Trichlorido tripyridine chromium (III) Oxidation state of Cr
x + (-1) × 3 + (0)3 = 0
x = + 3
Electronic configuration of Cr = 3d³ = (\(t_{2 g}^{3} e_{g}^{0}\))
Coordination number = 6
Shape : Octahedral
Stereochemistry

Both isomers are optically active. Therefore, a total of 4 isomers exist.
Magnetic moment (μ) = \(\sqrt{n(n+2)}\) = \(\sqrt{3(3+2)}\)
= \(\sqrt{15}\) = 3.87 BM

(iv) Cs[FeCl₄]
IUPAC name : Caesium tetrachlorido ferrate (III)
Oxidation state of Fe
+ 1 + x + (-1) × 4= 0
x – 3 = 0
x = + 3
Electronic configuration of Fe = 3d⁵(\(t_{2 g}^{3} e_{g}^{2}\))
Coordination number = 4
Shape : Tetrahedral
The complex does not exhibit geometrical or optical isomerism, (stereo isomerism).
Magnetic moment (μ) = \(\sqrt{n(n+2)}\)
= \(\sqrt{5(5+2)}\)
= \(\sqrt{35}\) = 5.92 BM

(v) K₄[Mn(CN)₆]
IUPAC name : Potassium hexacyanomanganate(II)
Oxidation state of Mn
(+1) × 4 + x + (-1) × 6 = 0
x – 2 = 0
x = + 2
Electronic configuration of Mn = 3d⁵ (\(t_{2 g}^{5} e_{g}^{0}\))
Coordination number = 6
Shape : Octahedral.
The complex does not exhibit stereo isomerism.
Magnetic moment (μ) = \(\sqrt{n(n+2)}\)
= \(\sqrt{1(1+2)}\)
= \(\sqrt{3}\)
= 1.732 BM

Question 25.
What is meant by stability of a coordination compound in solution? State the factors which govern the stability of complexes.
Answer:
The stability of a coordination compound in solution refers to the degree of association between the two species involved in the state of equilibrium. The stability of the coordination compound is measured in term of magnitude of stability or formation of equilibrium constant.
M + 4L → ML₄
K = \(\frac{\left[\mathrm{ML}_{4}\right]}{[\mathrm{M}][\mathrm{L}]^{4}}\)
Larger the stability constant, the higher is the proportion of ML₄ that exists in solution.

Factors on which stability of the complex depends are as follows :

1. Charge on the central metal ion : Greater the charge on the central metal ion, greater is the stability of the complex.
2. Nature of the metal ion : Groups 3 to 6 and inner transition element form stable complexes when donor atoms of the ligands are N, O and F. The element after group 6 of the transition metals which have d-orbitals (e.g., Rh, Pd, Ag, Au, Hg, etc.) form stable complexes when the donor atoms of the ligands are heavier members of N, O and F family.
3. Basic nature of the ligand : Greater the basic strength of the ligand, greater is the stability of the complex.
4. Chelate effect: Presence of chelate rings in the complex increases its stability. It is called chelate effect. It is maximum for the 5- and 6- membered rings.
5. Effect of multidentate cyclic ligands : If the ligands happen to be multidentate and cyclic without any steric effect, the stability of the complex is further increased.

Question 26.
What is meant by chelate effect? Give an example.
Answer:
When a didentate or a polydentate ligand contains donor atoms positioned in such a way that when they coordinate with the central metal ion, a five or a six-membered ring is formed, the effect is called chelate effect. Example, [PtCl₂(en)].

Question 27.
Discuss briefly giving an example in each case the role of coordination compounds in:
(i) biological systems
(ii) medicinal chemistry
(iii) analytical chemistry
(iv) extraction/metallurgy of metals
Answer:
(i) Role of coordination compounds in biological systems :

• Haemoglobin, the oxygen carrier in blood, is a complex of Fe²⁺ with porphyrin.
• The pigment chlorophyll in plants, responsible for photosynthesis, is a complex of Mg²⁺ with porphyrin.
• Vitamin B₁₂ (cyanocobalamine) the antipemicious anaemia factor, is a complex of cobalt.

(ii) Role of coordination compounds in medicinal chemistry :

• The platinum complex cis-[Pt(NH₃)₂Cl₂] (cis-platin) is used in the treatment of cancer.
• EDTA complex of calcium is used in the treatment of lead poisoning. Ca-EDTA is a weak complex; when it is administered, calcium in the complex is replaced by the lead present in the body and is eliminated in the urine.
• The excess of copper and iron present in animal system are removed by the chelating ligands D-penicillamine and desferroxime B via the formation of complexes.

(iii) Role of coordination compounds in analytical chemistry :
Complex formation is frequently encountered in qualitative and quantitative chemical analysis.
(a) Qualitative analysis
I. Detection of Cu²⁺ is based on the formation of a blue tetraammine copper (II) ion.

II. Ni²⁺ is detected by the formation of a red complex with dimethyl glyoxime (DMG).

III. The separation of Ag⁺ and Hg²⁺ in group I is based on the fact that while AgCl dissolves in NH₃, forming a soluble complex, Hg₂Cl₂ forms an insoluble black substance.

(b) Quantitative analysis : Gravimetric estimation of Ni²⁺ is carried out by precipitating Ni²⁺ as red nickel dimethyl glyoxime complex in the presence of ammonia.

EDTA is used in the complexometric determination of several metal ions such as Ca²⁺, Zn²⁺, Fe²⁺, Co²⁺, Ni²⁺ etc.

(iv) Role of coordination compounds in extraction/metallurgy of metals : Extraction of various metals from their ore involves complex formation. For example, silver and gold are extracted from their ore by forming cyanide complex.

Purification of some metals can be achieved through complex formation. For example in Mond process, impure nickel is converted into [Ni(CO)₄] which is decomposed to yield pure nickel.

Question 28.
How many ions are produced from the complex Co(NH₃)₆ Cl₂ in solution?
(i) 6
(ii) 4
(iii) 3
(iv) 2
Answer:
The correct option is (iii)
Coordination number of cobalt = 6. It ionises in the solution as

Hence, 3 ion are produced.

Question 29.
Amongst the following ions, which one has the highest magnetic moment value?
(i)[Cr(H₂O)₆]³⁺
(ii)[Fe(H₂O)₆]²⁺
(iii) [Zn(H₂O)₆]²⁺
Answer:
The oxidation state are: Cr (III), Fe (II) and Zn (II).
Electronic configuration of Cr³⁺ = 3d³, unpaired electrons = 3
Electronic configuration of Fe²⁺ = 3d⁶, unpaired electrons = 4
Electronic configuration of Z²⁺ = 3d¹⁰, unpaired electrons = 0
As μ = \(\sqrt{n(n+2)}\), therefore, (ii) has the highest magnetic moment.

Question 30.
The oxidation number of cobalt in K[Co(CO)₄] is
(i) +1
(ii) +3
(iii) -1
(iv) -3
Solution:
Oxidation number of Co : K[Co(CO)₄]
x+ (4 × 0) = -1; x = -1
Thus, correct answer is (iii).

Question 31.
Amongst the following, the most stable complex is
(i) [Fe(H₂O)₆]³⁺
(ii) [Fe(NH₃)₆]³⁺
(iii) [Fe(C₂O₄)₃]^3-
(iv) [FeCl₆]^3-
Answer:
In all these complexes, Fe is in +3 oxidation state. However, the complex (iii) is a chelate because three \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) ions acts as the chelating ligands. Thus, the most stable complex is [Fe(C₂O₄)₃]^3-. Thus, correct answer is (iii).

Question 32.
What will be the correct order for the wavelengths of absorption in the visible region of the following:
[Ni(NO₂)₆]^4-, [Ni(NH₃)₆]²⁺, [Ni(H₂O)₆]²⁺
Answer:
As metal ion is fixed, the increasing CFSE values of the ligands from the spectrochemical series are in the order :
H₂O < NH₃ < \(\mathrm{NO}_{2}^{-}\)
Hence, the energies absorbed for excitation will be in the order :
[Ni(H₂O)₆]²⁺ < [Ni(NH₃)₆]²⁺ < [Ni(NO₂)₆]^4-
As E = \(\frac{h c}{\lambda}\), therefore, the wavelengths absorbed will be in the opposite order,
[Ni(NO₂)₆]^4- < [Ni(NH₃)₆]²⁺ < [Ni(H₂O)₆]²⁺

Chemistry Guide for Class 12 PSEB Coordination Compounds Textbook Questions and Answers

Question 1.
Write the formulas for the following coordination compounds :
(i) Tetraamminediaquacobalt (III) chloride
(ii) Potassiumtetracyanidonickelate(II)
(iii) Tris(ethane-l,2-diammine)chromium(III) chloride
(iv) Amminebromidochloridonitrito-N-platinate(II)
(v) Dichloridobis(ethane-l,2-diammine) platinum (IV) nitrate
(vi) Iron(III)hexacyanidoferrate(II).
Answer:
(i) [Co(NH₃)₄(H₂O)₂]Cl₃
(ii) K₂[Ni(CN)₄
(iii) (Cr(en)₃]Cl₃
(iv) [Pt(NH₃)BrCl(NO₂)]^–
(v) [PtCl₂(en)₂] (NO₃)₂
(vi) Fe₄[Fe(CN)₆]₃

Question 2.
Write the IUPAC names of the following coordination compounds:
(i) [CO(NH₃)₆]Cl₃
(ii) [CO(NH₃)₅Cl]Cl₂
(iii) K₃[Fe(CN)₆]
(iv) K₃[Fe(C₂O₄)₃]
(v) K₂[PdCl₄]
(vi) [Pt(NH₃)₂Cl(NH₂CH₃)]Cl
Answer:
(i) Hexaamminecobalt(III)chloride
(ii) Pentaamminechloridocobalt(III)chloride
(iii) Potassiumhexacyanoferrate(III)
(iv) Potassiumtrioxalatoferrate (III)
(v) Potassiumtetrachloridopalladate (II)
(vi) Diamminechloridomethylamine platinum(II) chloride.

Question 3.
Indicate the types of isomerism exhibited by the following complexes and draw the structures of these isomers :
(i) K[Cr(H₂O)₂](C₂O₄)₂]
(ii) [Co(en)₃]Cl₃
(iii) [CO(NH₃)₅(NO₂)](NO₃)₂
(iv) [Pt(NH₃)(H₂O)Cl₂]
Answer:
(i) (a) Both geometrical isomer (cis and traits):

(b) Cis-isomer of this compound can exist as pair of optical is :

(ii) Complex will exist as optical isomers:

The compound will show ionisation as well as linkage isomerism.

(iii) Ionisation isomer :
[Co(NH₃)₅(NO₂)](NO₃)₂,
[Co(NH₃)₅ (NO)₃] (NO₂) (NO₃)
Linkage isomers :
[Co(NH₃)₅ (NO₂)](NO₃)₂;
[CO(NH₃)₅ (ONO)](NO₃)₂

(iv) Geometrical isomerism (cis and trans) :

Question 4.
Give evidence that [Co(NH₃)₅Cl]SO₄ and [Co(NH₃)₅SO₄]Cl are ionisation isomers.
Answer:
When they are dissolved in water, they give different ions in the solution which can be tested by adding AgNO₃ solution and BaCl₂ solution. If Cl^– dons are the counter ions, a white precipitate will be obtained with AgNO₃ solution. If \(\mathrm{SO}_{4}^{2-}\) ions are the counter ions, a white precipitate will be obtained with BaCl₂ solution.

Question 5.
Explain on the basis of valence bond theory that [Ni(CN)₄]^2- ion with square planar structure is diamagnetic and the [Ni(Cl)₂₄]^2- ion with tetrahedral geometry is paramagnetic.
Answer:
Nickel in [Ni(CN)₄]^2- is in the +2 oxidation state. The formation of [[Ni(CN)₄]^2- may be explained through hybridisation as follows :
Ni atom in the ground state

Since no unpaired electrons is present, the square planar complex is diamagnetic. In [Ni(CN)₄]^2-, Cl^– is a weak field ligand. It is, therefore, unable to pair up the unpaired electrons of the 3d orbital. Hence, the hybridisation involved is sp3 and the shape is tetrahedral. Since all the electrons are unpaired, it is paramagnetic

Question 6.
[Ni(CN)₄]^2- is paramagnetic while [Ni(CO)₄] is diamagnetic though both are tetrahedral. Why?
Answer:
In [Ni(CO)₄] Ni is in zero oxidation state whereas in [NiCl₄]^2-, it is in
+ 2 oxidation state. In the presence of strong ligand, CO ligand, the unpaired d electrons of Ni pair up but Cl^– being a weak ligand is unable to pair up the unpaired electrons.

Question 7.
[Fe(H₂O)₆]³⁺ is strongly paramagnetic whereas [Fe(CN)₆]^3- is weakly paramagnetic. Explain.
Answer:
In presence of CN^– (a strong ligand), the 3d⁵ electrons pair up leaving only one unpaired electron. The hybridisation is d²sp³ forming an inner orbital complex. In the presence of H₂O (a weak ligand), 3d electrons do not pair up. The hybridisation is sp³d² forming an outer orbital complex containing five unpaired electrons. Hence, it is strongly paramagnetic.

Question 8.
Explain [Co(NH₃)₆]³⁺ is an inner orbital complex whereas [Ni(NH₃)₆]²⁺ is an outer orbital complex.
Answer:
In [CO(NH₃)₆]³⁺, CO is in +3 oxidation state and has d⁶ electrons. In the presence of NH₃, the 3d electrons pair up leaving two d-orbitals empty to be involved in d²sp³ hybridisation forming inner orbital complex. In [Ni(NH₃)₆]²⁺, Ni is in +2 oxidation state and has d⁸ configuration. The hybridisation involved is sp³d², forming the outer orbital complex.

Question 9.
Predict the number of unpaired electrons in the square planar [Pt(CN)₄]^2- ion.
Answer:
₇₈Pt lies in group 10 with the configuration 5d⁹6s¹. Thus Pt²⁺ has the configuration :

For square planar shape, the hybridisation is dsp². Hence, the unpaired electrons in 5d orbital pair up to make one d orbital empty for dsp² hybridisation.
Thus there is no unpaired electron.

Question 10.
The hexaquomanganese(II) ion contains five impaired electrons, while the hexacyano ion contains only one unpaired electron. Explain using crystal field theory.
Answer:
Mn in the + 2 oxidation state has the configuration 3d⁵. In the presence of H₂O a weak ligand, the distribution of these five electrons is \(t_{2 g}^{3} e_{g}^{2}\)
i.e., all the electrons remain unpaired

However, in the presence of CN^– the distribution of these electrons is \(\), i.e., two t_2g orbitals contain paired electrons while the third t_2g orbital contains one unpaired electron.

Question 11.
Calculate the overall complex dissociation equilibrium constant for the \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) ion, given that β₄ for this complex is 2. 1 × 10¹³.
Solution:
The overall complex dissociation equilibrium constant is the reciprocal of the overall stability constant, β₄.
∴ \(\frac{1}{\beta_{4}}\) = \(\frac{1}{2.1 \times 10^{13}}\)
∴ = 4.7 × 10^-14

Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 10 Haloalkanes and Haloarenes Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 10 Haloalkanes and Haloarenes

PSEB 12th Class Chemistry Guide Haloalkanes and Haloarenes InText Questions and Answers

Question 1.
Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:
(i) (CH₃)₂CHCH(Cl)CH₃
(ii) CH₃CH₂CH(CH₃)CH(C₂H₅)Cl
(iii) CH₃CH₂C(CH₃)₂CH₂I
(iv) (CH₃)₃CCH₂CH(Br)C₆H₅
(v) CH₃CH(CH₃)CH(Br)CH₃
(vi) CH₃C(C₂H₅)₂CH₂Br
(vii) CH₃C(Cl)(C₂H₅)CH₂CH₃
(viii) CH₃CH＝C(Cl)CH₂CH(CH₃)₂
(ix) CH₃CH＝CHC(Br)(CH₃)₂
(x) p-ClC₆H₄CH₂CH(CH₃)₂
(xi) m-ClCH₂C₆H₄CH₂C(CH₃)₃
(xii) o-Br-C₆H₄CH(CH₃)CH₂CH₃
Answer:

Question 2.
Give the IUPAC names of the following compounds:
(i) CH₃CH(Cl)CH(Br)CH₃
(ii) CHF₂CBrClF
(iii) ClCH₂C ☰ CCH₂Br
(iv) (CCl₃)₃CCl
(v) CH₃C(p-ClC₆H₄)₂CH(Br)CH₃
(vi) (CH₃)₃CCH＝CClC₆H₄I-p
Answer:

Question 3.
Write the structures of the following organic halogen compounds.
(i) 2-Chloro-3-methylpentane
(ii) p -Bromochlorobenzene
(iii) 1-Chloro-4-ethylcyclohexane
(iv) 2-(2-Chlorophenyl)-l-iodooctane
(v) 2 -Bromobutane
(vi) 4-terf-Butyl-3-iodoheptane
(vii) l-Bromo-4-sec-butyl-2-methylbenzene
(viii) 1,4-Dibromobut-2 -ene
Answer:

Question 4.
Which one of the following has the highest dipole moment?
(i) CH₂Cl₂
(ii) CHCl₃
(iii) CCl₄
Answer:

1. CCl₄ is a symmetrical molecule. Therefore, the dipole moments of all four C—Cl bonds cancel each other. Hence, its resultant dipole moment is zero.

2. As shown in the above figure, in CHCl₃, the resultant of dipole moments of two C—Cl bonds is opposed by the resultant of dipole moments of one C—H bond and one C—Cl bond. Since the resultant of one C—H bond and one C—Cl bond dipole moments is smaller than two C—Cl bonds, the opposition is to a small extent. As a result, CHC13 has a small dipole moment of 1.08 D.

3. On the other hand, in case of CH₂Cl₂, the resultant of the dipole moments of two C—Cl bonds is strengthened by the resultant of the dipole moments of two C—H bonds. As a result, CH2C12 has a higher dipole moment of 1.60 D than CHCl₃ i.e., CH₂Cl₂ has the highest dipole moment.
Hence, the given compounds can be arranged in the increasing order of their dipole moments as:
CCl₄ < CHCl₃ < CH₂Cl₂

Question 5.
A hydrocarbon C₅H₁₀ does not react with chlorine in dark but gives a single monochloro compound C₅H₉Cl in bright sunlight. Identify the hydrocarbon.
Answer:
CO The hydrocarbon with molecular formula C₅H₁₀ can be either a cycloalkane or an alkene.

Since, the hydrocarbon does not react with Cl₂ in the dark, it cannot be an alkene but must be a cycloalkane.
As the cycloalkane reacts with Cl₂ in the presence of bright sunlight, to give a single monochloro compound, C₅H₉Cl, therefore all the ten hydrogen atoms of the cycloalkane must be equivalent. Therefore, the cycloalkane is cyclopentane.

Question 6.
Write the isomers of the compound having formula C₄H₉Br.
Answer:
There are four isomers of the compound having the formula C₄H₉Br.
These isomers are given below:

Question 7.
Write the equations for the preparation of 1-iodobutane from
(i) 1-butanol
(ii) 1-chlorobutane
(iii) but-l-ene.
Answer:

Question 8.
What are ambident nucleophiles? Explain with an example.
Answer:
Ambident nucleophiles are nucleophiles having two nucleophilic sites. Thus, ambident nucleophiles have two sites through which they can attack.
For example, nitrite ion is an ambident nucleophile.

Nitrite ion can attack through oxygen resulting in the formation of alkyl nitrites. Also, it can attack through nitrogen resulting in the formation of nitroalkanes.

Question 9.
Which compound in each of the following pairs will react faster in Sn₂ reaction with OH^–?
(i) CH₃Br or CH₃I
(ii) (CH₃)₃CCl or CH₃Cl
Answer:
(i) Since I^– ion is a better leaving group than Br^– ion, hence CH₃I reacts faster than CH₃Br in SN₂ reaction with OH^– ion.

(ii) On steric grounds, 1° alkyl halides are more reactive than tert-alkyl halides in S_N 2 reactions. Hence, CH₃Cl will react at a faster rate than (CH₃)₃ CCl in a S_N2 reaction with OH^– ion.

Question 10.
Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene:
(i) 1 -Bromo-1-methylcyclohexane
(ii) 2-Chloro-2-methylbutane
(iii) 2, 2, 3-Trimethyl-3-hromopentane.
Answer:
(i) In 1 -bromo-1 -methylcyclohexane, the β-hydrogens on either side of the Br atom are equivalent, therefore, only 1-alkene is formed.

(ii) All p-hydrogens in 2-chloro-2-methylbutane are not equivalent, hence on treatment with C₂H₅ONa/C₂H₅OH, it gives two alkenes.

(iii) 2, 2, 3-Trimethyl-3-bromopentane has two different sets of p-hydrogen and therefore, in principle, can give two alkenes (I and II). But according to Saytzeff rule, more highly substituted alkene (II), being more stable is the major product.

Question 11.
How will you bring about the following conversions?
(i) Ethanol to but-1-yne
(ii) Ethane to bromoethene
(iii) Propene to 1 -nitropropane
(iv) Toluene to benzyl alcohol
(v) Propene to propynt
(vi) Ethanol to ethyl fluoride
(vii) Bromomethane to propanone
(viii) But-l-ene to but-2-ene
(ix) 1-Chlorobutane to n-octane
(x) Benzene to biphenyl.
Answer:

Question 12.
Explain why
(i) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride ?
(ii) Alkyl halides, though polar are immiscible with water ?
(iii) Grignard reagents should be prepared under anhydrous conditions ?
Answer:
(i) Because of greater s-character, an sp²-hybrid carbon is more electronegative than an sp³-hybrid carbon. Thus, the sp²-hybrid carbon of C—Cl bond in chlorobenzene has less tendency to release electrons to Cl than an sp³-hybrid carbon of cyclohexyl chloride.

Hence, the C—Cl bond in chlorobenzene is less polar than that in cyclohexyl chloride. In other words, the magnitude of negative charge is less on Cl atom of chlorobenzene than in cyclohexyl chloride. Now, due to delocalisation of lone pairs of electrons of the Cl atom over the benzene ring, C—Cl bond in chlorobenzene acquires some double character while the C—Cl bond in cyclohexyl chloride is a pure single bond. Thus, C—Cl bond in chlorobenzene is shorter than in cyclohexyl chloride.

As dipole moment is a product of charge and distance, chlorobenzene has lower dipole moment than cyclohexyl chloride due to lower magnitude of negative charge on the Cl atom and shorter C—Cl distance.

(ii) Alkyl halides, though polar, are immiscible in water because they are unable to form hydrogen bonds with water molecules.

(iii) Grignard reagents are very reactive. They react with moisture present in the apparatus or the starting materials to give hydrocarbons.

Hence, Grignard reagent must be prepared under anhydrous conditions.

Question 13.
Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.
Answer:
Uses of Freon-12(CCl₂F₂)

1. It is used as a refrigerant in refrigerators and air conditioners.
2. It is also used in aerosol spray propellants such as body sprays, hair sprays.

Uses of DDT (p, p’-dichlorodiphenyltrichloroethane)

1. It is very effective against mosquitoes and lice.
2. It is also used in many countries as insecticide for sugarcane and fodder crops. (But due to its harmful effects, its use has been banned in many contries including U.S.A.

Uses of Carbontetrachloride (CCl₄)

1. It is used for manufacturing refrigerants and propellants for aerosol cAnswer:
2. It is used as feedstock in the synthesis of chlorofluorocarbons and other chemicals.
3. It is used as a solvent in the manufacture of pharmaceutical products. Until the mid 1960’s, carbon tetrachloride was widely used as a cleaning fluid, a degreasing agent in industries, a spot reamer in homes, and a fire extinguisher.

Uses of Iodoform (CHI₃)
Iodoform was used earlier as an antiseptic, but now it has been replaced by other formulations-containing iodine-due to its objectionable smell. The antiseptic property of iodoform is only due to the liberation of free iodine when it comes in contact with the skin.

Question 14.
Write the structure of the major organic product in each of the following reactions


(v) C₆H₅ONa + C₂H₆Cl →
(vi) CH₃CH₂CH₂OH + SOCl₂ →

(viii) CH₃CH = C(CH₃)₂ + HBr →
Answer:

Question 15.
Write the mechanism of the following reaction

Answer:
The given reaction is

The given reaction is an S_N2 reaction. In this reaction, CN^– acts as the nucleophile and attacks the carbon atom to which Br is attached. CN^– ion is an ambident nucleophile and can attack through both C and N. In this case, it attacks through the C-atom.

Question 16.
Arrange the compounds of each set in order of reactivity towards S_N2 displacement
(i) 2-Bromo-2-methylbutane, 1 -Bromopentane, 2-Bromopentane
(ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 3-Bromo- 2-methylbutane
(iii) 1-Bromobutane, l-Bromo-2, 2-dimethylpropane, 1-Bromo -2-methylbutane, 1-Bromo-3-methylbutane.
Answer:

Question 17.
Out of C₆H₅CH₂Cl and C₆H₅CHClC₆H₅, which is more easily hydrolysed by aqueous KOH?
Answer:

In S_N1 reaction, reactivity depends upon the stability of carbocations. carbocation is more stable as compared to . Therefore, C₆H₅CHClC₆H₅ gets hydrolysed more easily than C₆H₅CHCl.

Question 18.
p-Dichlorobenzene has higher m.p. and lower solubility than those of o- and m-isomers. Discuss.
Answer:

p-Dichlorobenzene is more symmetrical than o-and m-isomers. For this reason, it fits more closely than o-and m-isomers in the crystal lattice. Therefore, more energy is required to break the crystal lattice of p-dichlorobenzene. As a result, p-dichlorobenzene has a higher melting point and lower solubility than o-and m-isomers.

Question 19.
How the following conversions can be carried out?
(i) Propene to propan-l-ol
(ii) Ethanol to but-l-yne
(iii) 1-Bromopropane to 2-bromopropane
(iv) Toluene to benzyl alcohol
(v) Benzene to 4-bromonitrobenzene
(vi) Benzyl alcohol to 2-phenylethanoic acid
(vii) Ethanol to propanenitrile
(viii) Aniline to chlorobenzene
(ix) 2-Chlorobutane to 3, 4-dimethylhexane
(x) 2-Methyl- 1-propene to 2-chloro-2-methylpropane
(xi) Ethyl chloride to propanoic acid
(xii) But-l-ene to n-butyliodide
(xiii) 2-Chloropropane to 1-propanol
(xiv) Isopropyl alcohol to iodoform
(xv) Chlorobenzene to p-nitrophenol
(xvi) 2-Bromopropane to 1-bromopropane
(xvii) Chloroethane to butane
(xviii) Benzene to diphenyl
(xix) tert-Butyl bromide to isobutyl bromide
(xx) Aniline to phenylisocyanide
Answer:

Question 20.
The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.
Answer:
In an aqueous solution, KOH almost completely ionises to give OH^– ions. OH^– ion is a strong nucleophile, which leads the alkyl chloride to undergo a substitution reaction to form alcohol.

On the other hand, an alcoholic solution of KOH contains alkoxide (RO^–) ion, which is a strong base. Thus, it can abstract a hydrogen from the p carbon of the alkyl chloride and form an alkene by eliminating a molecule of HCl.

OH^– ion is a much weaker base than RO^– ion. Also, OH^– ion is highly solvated in an aqueous solution and as a result, the basic character of OH^– ion decreases. Therefore, it cannot abstract a hydrogen from the β carbon.

Question 21.
Primary alkyl halide C₄H₉Br (A) reacted with alcoholic KOH to give compound (B).Compound (B) is reacted with HBr to give (C) which is an isomer of (A). When (A) is reacted with sodium metal it gives compound (D), C₈H₁₈ which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (A) and write the equations for all the reactions.
Answer:
There are two primary alkyl halides having the formula, C₄H₉Br. They are n-butyl bromide and isobutyl bromide.

Therefore, compound (A) is either n-butyl bromide or isobutyl bromide. Now, compound (A) reacts with Na metal to give compound (B) of molecular formula, C₈H₁₈ which is different from the compound formed when n-butyl bromide reacts with Na metal. Hence, compound (A) must be isobutyl bromide.

Question 22.
What happens when
(i) n-butyl chloride is treated with alcoholic KOH,
(ii) bromobenzene is treated with Mg in the presence of dry ether,
(iii) chlorobenzene is subjected to hydrolysis,
(iv) ethyl chloride is treated with aqueous KOH,
(v) methyl bromide is treated with sodium in the presence of dry ether
(vi) methyl chloride is treated with KCN.
Answer:
(i) When n-butyl chloride is treated with alcoholic KOH, the formation of but-l-ene takes place. This reaction is a dehydrohalogenation reaction.

(ii) When bromobenzene is treated with Mg in the presence of dry ether, phenylmagnesium bromide is formed.

(iii) Chlorobenzene does not undergo hydrolysis under normal conditions. However, it undergoes hydrolysis when heated in an aqueous sodium hydroxide solution at a temperature of 623 K and a pressure of 300 atm to form phenol.

(iv) When ethyl chloride is treated with aqueous KOH, it undergoes hydrolysis to form ethanol.

(v) When methyl bromide is treated with sodium in the presence of dry ether, ethane is formed. This reaction is known as the Wurtz reaction.

(vi) When methyl chloride is treated with KCN, it undergoes a substitution reaction to give methyl cyanide.

Chemistry Guide for Class 12 PSEB Haloalkanes and Haloarenes Textbook Questions and Answers

Question 1.
Write structures of the following compounds :
(i) 2-Chloro-3-methylpentane
(ii) 1-Chloro-4-ethylcyclohexane
(iii) 4-tert-butyl-3-iodoheptane
(iv) 1-4-Dibromobut-2-ene
(v) 1-Bromo-4-sec-butyl-2-methylbenzene
Answer:

Question 2.
Why is sulphuric acid not used during the reaction of alcohols with KI?
Answer:
H₂SO₄ cannot be used along with KI in the conversion of an alcohol to an alkyl iodide as it converts KI to corresponding HI and then oxidises it to I₂.

Question 3.
Write structures of different dihalogen derivatives of propane.
Answer:
(i) ClCH₂CH₂CH₂Cl
(ii) ClCH₂CHClCH₃
(iii) Cl₂CHCH₂CH₃
(iv) CH₃CCl₂CH₃

Question 4.
Among the isomeric alkanes of molecular formula C₅H₁₂ identify the one that on photochemical chlorination yields :
(i) A single monochloride
(ii) Three isomeric monochlorides
(iii) Four isomeric monochlorides
Answer:

All the hydrogen atoms are equivalent and replacement of any hydrogen will give the same product.

(ii)
The equivalent hydrogens are grouped as a, b and c. The replacement of equivalent hydrogens will give the same product. Thus, three isomeric products are possible.

(iii)
The equivalent hydrogens are grouped as a, b, c and d. Thus, four isomeric products are possible.

Question 5.
Draw the structures of major monohalo products in each of the following reactions:

Answer:

Question 6.
Arrange each set of compounds in the order of increasing boiling points.
(i) Bromomethane, Bromoform, Chloromethane, Dibromomethane.
(ii) 1-Chloropropane, Isopropyl chloride, 1 -Chlorobutane.
Answer:
(i) Chloromethane < Bromomethane < Dibromomethane < Bromoform. Boiling point increases with increase in molecular mass.

(ii) Isopropyl chloride < 1-Chloropropane < 1-Chlorobutane. Isopropyl chloride being branched has lower boiling point than 1-Chloropropane.

Question 7.
Which alkyl halide from the following pairs would you expect to react more rapidly by an S_N2 mechanism ? Explain your answer.

Answer:
(i) CH₃CH₂CH₂CH₂Br
Being primary halide, there won’t be any steric hindrance.

(ii)
Being a secondary halide, there will be less crowding around α-carbon than tertiary halide.

(iii)
The presence of methyl group closer to the halide group will increase the steric hindrance and decrease the rate.

Question 8.
In the following pairs of halogen compounds, which compound undergoes faster S_N1 reaction ?

Answer:
(i)
2-Chloro-2-methylpropane as the tertiary carbocation is more stable than secondary carbocation.

(ii)
2-Chloroheptane as the secondary carbocation is more stable than primary carbocation.

Question 9.
Identify A, B, C, D, E, R and R’ in the following:

Answer:

Punjab State Board PSEB 12th Class Physics Book Solutions Chapter 8 Electromagnetic Waves Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Physics Chapter 8 Electromagnetic Waves

PSEB 12th Class Physics Guide Electromagnetic Waves Textbook Questions and Answers

Question 1.
Figure 8.6 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A.

(a) Calculate the capacitance and the rate of charge of potential difference between the plates.
(b) Obtain the displacement current across the plates.
(c) Is Kirchhoffs first rule (junction rule) valid at each plate of the capacitor? Explain.
Answer:
(a) Capacitance of capacitor is given by the relation
C = \(\frac{\varepsilon_{0} A}{d}\) = \(\frac{8.854 \times 10^{-12} \times \pi \times(0.12)^{2}}{5 \times 10^{-2}}\)
= 8.01F
Also \(\frac{d Q}{d t}\) = \(\frac{d V}{d t}\)
∴ \(\frac{d V}{d t}\) = \(\frac{0.15}{8.01 \times 10^{-12}}\)
= 1.87 × 10¹⁰V /s

(b) Displacement current I_d = ε₀ × \(\frac{d}{d t}\) (Φ_E)
Again Φ_E – EA across Hence,(negative end constant).
Hence, I_d = ε₀ A\(\frac{d E}{d t}\)
Again, E = \(\frac{Q}{\varepsilon_{0} A}\)
So, \(\frac{d E}{d t}=\frac{i}{\varepsilon_{0} A}\)
which corresponds i_d = i = 1.5A

(c) Yes, Kirchhoffs law is valid provided by current, we mean the sum of condition and displacement current.

Question 2.
A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s^-1.

(a) What is the rms value of the conduction current?
(b) Is the conduction current equal to the displacement current?
(c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates.
Answer:
(a) I_rms = V_rms × Cω
= 230 × 100 × 10¹² × 300
= 6.9 × 10^-6 A = 6.9 μ A

(b) Yes, we know that the deviation is correct even if I is steady DC or AC (oscillating in time) can be proved as
I_d = ε₀\(\frac{d}{d t}\) (σ) = ε₀\(\frac{d}{d t}\) (EA) (> σ = EA)
ε₀A \(\frac{d E}{d t}\) = ε₀A \(\frac{d}{d t}\) (\(\frac{\sigma}{\varepsilon_{0}}\))
ε₀A \(\frac{d}{d t}\) (\(\frac{\sigma}{\varepsilon_{0} A}\)) (> σ = \(\frac{q}{A}\))
ε₀A × \(\frac{1}{\varepsilon_{0} A} \cdot \frac{d q}{d t}\) = I
which is the required proof.

(c) The region formula for magnetic field
B = \(\frac{\mu_{0} r}{2 \pi R^{2}}\)i_d
even if I_d is oscillating (and so magnetic field B): The formula is valid. I_D oscillates in phase as i₀ = i (peak value of current). Now, we have
B₀ = \(\frac{\mu_{0} r}{2 \pi R^{2}}\)i₀
where B₀ and i₀ are the amplitude of magnetic field and current respectively.
So, i₀ = √2I_rms = 6.96 × 1.414 μA = 9.76μA
Given, r = 3 cm, R = 6cm
B₀ = \(\frac{\mu_{0} r i_{0}}{2 \pi R^{2}}\)
= \(\frac{10^{-7} \times 2 \times 3 \times 10^{-2} \times 9.76 \times 10^{-6}}{(6)^{2} \times\left(10^{-2}\right)^{2}}\)
= 1.633 × 10^-11 T

Question 3.
What physical quantity is the same for X-rays of wavelength 10^-10 m, red light of wavelength 6800 Å and radiowaves of wavelength 500 m?
Answer:
X-rays, red light and radiowaves all are the electromagnetic waves. They have different wavelengths and frequencies. But the physical quantity which is same for all of these is the velocity of light in vacuum which is denoted by c and is equal to 3 × 10⁸ ms^-1 W

Question 4.
A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?
Answer:
In an electromagnetic wave’s propagation vector \(\vec{K}\), electric field vector \(\vec{E}\) and magnetic field vector \(\vec{K}\) form a right handed system. As the propagation vector is along Z-direction, electric field vector will be along X-direction and magnetic field vector will be along Y-direction.
Frequency v = 30 MHz = 30 × 10⁶Hz
Speed of light c = 3 × 10⁸ ms^-1
Wavelength, λ = \(\frac{c}{v}\) = \(\frac{3 \times 10^{8}}{30 \times 10^{6}}\) = 10 m

Question 5.
A radio can tune in to any station in the 7.5 MHz to 12 MHz hand. What is the corresponding wavelength band?
Answer:
Speed of wave c = 3 × 10⁸ ms^-1
When frequency, V₁ = 7.5MHz = 7.5 × 10⁶ Hz
Wavelength, λ₁ = \(\frac{c}{v_{1}}\) = \(\frac{3 \times 10^{8}}{7.5 \times 10^{6}}\) = 40m
When frequency, V₂ 12 MHZ = 12 × 10⁶HZ
Wavelength, λ₂ = \(\frac{c}{v_{2}}\) = \(\frac{3 \times 10^{8}}{12 \times 10^{6}}\) = 25m
Wavelength band is from 25 m to 40 m.

Question 6.
A charged particle oscillates about its mean equilibrium position with a frequency of 10⁹ Hz. What is the frequency of the electromagnetic waves produced by the oscillator?
Answer:
According to Maxwell’s theory, an oscillating charged particle with a frequency v radiates electromagnetic waves of frequency v.
So, the frequency of electromagnetic waves produced by the oscillator is v = 10⁹ Hz.

Question 7.
The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B₀ =510 nT. What is the amplitude of the electric field part of the wave?
The relation between magnitudes of magnetic and electric field vectors in vacuum is
\(\frac{E_{0}}{B_{0}}\) = c
⇒ E₀ = B₀C
Here, B₀ = 510 × 10^-9T, c = 3 × 10⁸ ms^-1
E₀ = 510 × 10^-9 × 3 × 10⁸ = 153N/C

Question 8.
Suppose that the electric field amplitude of an electromagnetic wave is E₀ = 120 N/C and that its frequency is v = 50.0 MHz. (a) Determine, B₀, ω, k and λ. (b) Find expressions for E and B.
Answer:
Electric field amplitude, E₀ = 120 N/C
Frequency of source, v = 50.0 MHz = 50 × 10⁶ Hz
Speed of light, c = 3 × 10⁸ m/s

(a) Magnitude of magnetic field strength is given as
B₀ \(\frac{E_{0}}{\mathcal{C}}\) = \(\frac{120}{3 \times 10^{8}}\)
40 × 10^-8T
= 400 × 10^-9 T
= 400 nT
Angular frequency of source is given as
ω = 2πv = 2π × 50 × 10⁶
= 3.14 × 10⁸ rad/s
Propagation constant is given as
k = \(\frac{\omega}{c}\) = \(\frac{3.14 \times 10^{8}}{3 \times 10^{8}}\) = 1.05 rad /m
Wavelength of wave is given us
λ = \(\frac{c}{v}\) = \(\frac{3 \times 10^{8}}{50 \times 10^{6}}\) = 6.0m

(b) Suppose the wave is propagating in the positive x direction. Then, the electric field vector will be in the positive y direction and the magnetic field vector will be in the positive z direction. This is because all three vectors are mutually perpendicular.
Equation of electric field vector is given as
\(\vec{E}\) = E₀sin (kx – ωt) ĵ
= 120 sin [1.05 x – 3.14 × 10⁸t] ĵ
And, magnetic field vector is given as
\(\vec{B}\) = B₀ sin (kx – ωt)k̂
\(\vec{B}\) = (4 × 10^-7)sin[1.05 x – 3.14 × 10⁸t]k̂

Question 9.
The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E – hv (for energy of a quantum of radiation : photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?
Answer:
Energy of a photon is given as
E = hv = \(\frac{h c}{\lambda}\)
where,
h = Planck’s constant = 6.6 × 10^-34 Js
c = Speed of light = 3 × 10⁸ m/s
λ = Wavelength of radiation
∴ E = \(\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{\lambda}\) = \(\frac{19.8 \times 10^{-26}}{\lambda}\) = J
= \(\frac{19.8 \times 10^{-26}}{\lambda \times 1.6 \times 10^{-19}}\) = \(\frac{12.375 \times 10^{-7}}{\lambda}\) = eV
The given table lists the photon energies for different parts of an electromagnetic spectrum for different λ.

The photon energies for the different parts of the spectrum of a source indicate the spacing of the relevant energy levels of the source.

Question 10.
In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 10¹⁰Hz and amplitude 48 Vm^-1.
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the E field equals the average energy density of the B field, [c = 3 × 10⁸ ms^-1]
Answer:
Frequency of the electromagnetic wave, v = 2.0 × 10¹⁰ Hz
Electric field amplitude, E₀ = 48 V m^-1
Speed of light, c = 3 × 10⁸ m/s

(a) Wavelength of the wave is given as
λ = \(\frac{\mathcal{C}}{\mathrm{v}}\) = \(\frac{3 \times 10^{8}}{2 \times 10^{10}}\) 0.015 m

(b) Magnetic field strength is given as
B₀ = \(\frac{E_{0}}{c}\)
= \(\frac{48}{3 \times 10^{8}}\) = 1.6 × 10^-7 T

(c) Let U_E and U_B be the energy density of \(\) field and \(\) field respectively. Energy density of the electric field is given as
U_E = \(\frac{1}{2}\) ε₀E²
And, energy density of the magnetic field is given as
U_B = \(\frac{1}{2 \mu_{0}}\)²
We have the relation connecting E and B as
E = cB ………….. (1)
where,
c = \(\frac{1}{\sqrt{\varepsilon_{0} \mu_{0}}}\) ……………. (2)
Putting equation (2) in equation (1), we get
E = \(\frac{1}{\sqrt{\varepsilon_{0} \mu_{0}}}\)B
Squaring both sides, we get
E² = \(\frac{1}{\varepsilon_{0} \mu_{0}}\) B²
ε₀E² = \(\frac{B^{2}}{\mu_{0}}\)
\(\frac{1}{2}\)ε₀E² = \(\frac{1}{2} \frac{B^{2}}{\mu_{0}}\)
⇒ U_E = E_B

Question 11.
Suppose that the electric field part of an electromagnetic wave in vacuum is
E = {(3.1 N/C) cos [(1.8 rad/m) y + (5.4 × 10⁶ rad/s) t]}î
(a) What is the direction of propagation?
(b) What is the wavelength λ ?
(c) What is the frequency v?
(d) What is the amplitude of the magnetic field part of the wave?
(e) Write an expression for the magnetic field part of the wave.
Answer:
(a) Wave is propagating along negative y-axis.

(b) Standard equation of wave is \(\vec{E}\) = E₀ cos(ky + cot)î
Comparing the given equation with standard equation, we have
E₀ = 3.1 N/C, k = 1.8 rad/m, ω = 5.4 × 10⁶ rad/s
Propagation constant k = \(\frac{2 \pi}{\lambda}\)
∴ λ = \(\frac{2 \pi}{k}\) = \(\frac{2 \times 3.14}{1.8}\) m = 3.49 m

(c) We have ω = 5.4 × 10⁶ rad/s
Frequency, v = \(\frac{\omega}{2 \pi}\) = \(\frac{5.4 \times 10^{6}}{2 \times 3.14}\) Hz
= 8.6 × 10⁵ Hz

(d) Amplitude of magnetic field,
B₀ = \(\frac{E_{0}}{c}\) = \(\frac{3.1}{3 \times 10^{8}}\) = 1.03 × 10^-8 T

(e) The magnetic field is vibrating along Z-axis because \(\vec{K}\),\(\vec{E}\),\(\vec{B}\) form a right handed system -ĵ × î = k̂
> Expression for magnetic field is
\(\vec{B}\) = B₀ cos(ky+ ωt)k̂
= [1.03 × 10^-8Tcos{(1.8rad / m) y +(5.4 × ⁶ rad/s)t}]k̂

Question 12.
About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation
(a) at a distance of 1 m from the bulb?
(b) at a distance of 10 m?
Assume that the radiation is emitted isotropically and neglect reflection.
Answer:
Power in visible radiation, P = \(\frac{5}{100}\) × 100 = 5W
For a point source, intensity I = \(\frac{P}{4 \pi r^{2}}\), where r is distance from the source.

(a) When distance r = 1 m,
I = \(\frac{5}{4 \pi(1)^{2}}=\frac{5}{4 \times 3.14}\) = 0.4 W/m²

(b) When distance r = 10 m,
I = \(\frac{5}{4 \pi(10)^{2}}=\frac{5}{4 \times 3.14 \times 100}\)
= 0.004 W/m²

Question 13.
Use the formula λ_m T = 0.29 cm K to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you?
Answer:
A body at a particular temperature produces a continuous spectrum of wavelengths. In case of a black body, the wavelength corresponding to maximum intensity of radiation is given according to Planck’s law. It can be given by the relation,
λ_m = \(\frac{0.29}{T}\) cm K
where, λ_m = maximum wavelength
T = temperature
Thus, the temperature for different wavelengths can be obtained as
For λ_m = 10^-4 cm; T = \(\frac{0.29}{10^{-4}}\) = 2900°K
For λ_m = 5 × 10^-5 cm; T = \(\frac{0.29}{5 \times 10^{-5}}\) = 5800°K
For λ_m = 10^-6 cm; T = \(\frac{0.29}{10^{-6}}\) = 290000 °K and so on.

The numbers obtained tell us that temperature ranges are required for obtaining radiations in different parts of an electromagnetic spectrum. As the wavelength decreases, the corresponding temperature increases.

Question 14.
Given below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the part of the electromagnetic spectrum to which each belongs.
(a) 21 cm (wavelength emitted by atomic hydrogen in interstellar space).

(b) 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen; known as Lamb shift).

(c) 2.7 K (temperature associated with the isotropic radiation filling all space-thought to be a relic of the ‘big-bang’ origin of the universe).

(d) 5890 Å – 5896 Å (double lines of sodium).

(e) 14.4 keV [energy of a particular transition in ⁵⁷ Fe nucleus associated with a famous high resolution spectroscopic method (Mossbauer spectroscopy)].
Answer:
(a) 21 cm belongs to short wavelength end of radiowaves (or Hertizan waves).

(b) Wavelength, λ = \(\frac{c}{v}\) = \(\frac{3 \times 10^{8}}{1057 \times 10^{6}}\) = 0.28 m = 28 cm.
This also belongs to short wavelength end of radiowaves.

(c) From relation λ_mT = 0.29 × 10^-2 K,
λ_m = \(\frac{0.29 \times 10^{-2} \mathrm{~K}}{T}=\frac{0.29 \times 10^{2}}{2.7}\)
= 0.107 × 10^-2m= 0.107 cm.
This corresponds to microwaves.

(d) Wavelength doublet 5890Å – 5896Å belongs to the visible region. These are emitted by sodium vapour lamp.

(e) From relation, E = \(\frac{h c}{\lambda}\)
we have λ = \(\frac{h c}{E}\)
λ = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{14.4 \times 10^{3} \times 1.6 \times 10^{-19}} \mathrm{~m}\)
= 0.86 × 10^-10 m = 0.86 Å
It belongs to the X-ray region of electromagnetic spectrum.

Question 15.
Answer the following questions :
(a) Long distance radio broadcasts use short-wave bands. Why?

(b) It is necessary to use satellites for long distance TV transmission. Why?

(c) Optical and radiotelescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth Why?

(d) The small ozone layer on top of the stratosphere is crucial for human survival. Why?

(e) If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now?

(f) Some scientists have predicted that a global nuclear war on the earth would be followed by a severe ‘nuclear winter’ with a devastating effect on life on earth. What might be the basis of this prediction?
Answer:
(a) Long distance radio broadcasts use short-wave bands because only these bands can be refracted by the ionosphere.

(b) Yes, it is necessary to use satellites for long distance TV transmissions because television signals are of high frequencies and high energies. Thus, these signals are not reflected by the ionosphere. Hence, satellites are helpful in reflecting TV signals. Also, they help in long distance TV transmissions.

(c) With reference to X-ray astronomy, X-rays are absorbed by the atmosphere. However, visible and radiowaves can penetrate it. Hence, optical and radiotelescopes are built on the ground, while X-ray astronomy is possible only with the help of satellites orbiting the Earth.

(d) The small ozone layer on the top of the stratosphere is crucial for human survival because it absorbs harmful ultraviolet radiations present in sunlight and prevents it from reaching the Earth’s surface.

(e) In the absence of an atmosphere, there would be no greenhouse effect on the surface of the Earth. As a result, the temperature of the Earth would decrease rapidly, making it chilly and difficult for human survival.

(f) A global nuclear war on the surface of the Earth would have disastrous consequences. Post nuclear war, the Earth will experience severe winter as the war will produce clouds of smoke -that would cover maximum parts of the sky, thereby preventing solar light form reaching the atmosphere. Also, it will lead to the depletion of the ozone layer.

Punjab State Board PSEB 12th Class Physics Book Solutions Chapter 5 Magnetism and Matter Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Physics Chapter 5 Magnetism and Matter

PSEB 12th Class Physics Guide Magnetism and Matter Textbook Questions and Answers

Question 1.
Answer the following questions regarding earth’s magnetism:
(a) A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth’s magnetic field.

(b) The angle of dip at a location in southern India is about 18°
Would you expect a greater or smaller dip angle in Britain?

(c) If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground?

(d) In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or south pole?

(e) The earth’s field, it is claimed, roughly approximates the field due to a dipole of magnetic moment 8 × 10²² JT^-1 located at its centre. Check the order of magnitude of this number in some way.

(f) Geologists claim that besides the main magnetic N-S poles, there are several local poles on the earth’s surface oriented in different directions. How is such a thing possible at all?
Answer:
(a) The three independent quantities conventionally used for specifying earth’s magnetic field are magnetic declination, angle of dip and horizontal component of earth’s magnetic field.

(b) The angle of dip at a point depends on how far the point is located with respect to the North Pole or the South Pole. The angle of dip would be greater in Britain (it is about 70°) than in southern India because the location of Britain on the globe is closer to the magnetic North Pole.

(c) It is hypothetically considered that a huge bar magnet is dipped inside earth with its North Pole near the geographic South Pole and its South Pole near the geographic North Pole.

Magnetic field lines emanate from a magnetic North Pole and terminate at a magnetic South Pole. Hence, in a map depicting earth’s magnetic field lines, the field lines at Melbourne, Australia would seem to come out of the ground.

(d) If a compass is located on the geomagnetic North Pole or South Pole, then the compass will be free to move in the horizontal plane while earth’s field is exactly vertical to the magnetic poles. In such a case, the compass can point in any direction.

(e) Magnetic moment, M = 8 × 10²² JT^-1
Radius of earth, r = 6.4 × 10⁶ m
Magnetic field strength, B = \(\frac{\mu_{0} M}{4 \pi r^{3}}\)
∴ B = \(\frac{4 \pi \times 10^{-7} \times 8 \times 10^{22}}{4 \pi \times\left(6.4 \times 10^{6}\right)^{3}}\) = 0.3G
This quantity is of the order of magnitude of the observed field on earth.

(f) Yes, there are several local poles on earth’s surface oriented in different directions. A magnetised mineral deposit is an example of a local N-S pole.

Question 2.
Answer the following questions :
(a) The earth’s magnetic field varies from point to point in space. Does it also change with time? If so, on what time scale does it change appreciably?

(b) The earth’s core is known to contain iron. Yet geologists do not regard this as a source of the earth’s magnetism. Why?

(c) The charged currents in the outer conducting regions of the earth’s core are thought to be responsible for earth’s magnetism. What might be the ’battery’ (i.e., the source of energy) to sustain these currents?

(d) The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists know about the earth’s field in such distant past?

(e) The earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km). What agencies may be responsible for this distortion?

(f) Interstellar space has an extremely weak magnetic field of the order of 10^-12 T. Can such a weak field he of any significant consequence? Explain.
[Note: Exercise 5.2 is meant mainly to arouse your curiosity. Answers to some questions above are tentative or unknown. Brief answers wherever possible are given at the end. For details, you should consult a good text on geomagnetism.]
Answer:
(a) Earth’s magnetic field changes with time. It takes a few hundred years to change by an appreciable amount. The variation in earth’s magnetic field with the time cannot be neglected.

(b) Earth’s core contains molten iron. This form of iron is not ferromagnetic. Hence, this is not considered as a source of earth’s magnetism.

(c) The radioactivity in earth’s interior is the source of energy that sustains the currents in the outer conducting regions of earth’s core. These charged currents are considered to be responsible for earth’s magnetism.

(d) Earth reversed the direction of its field several times during its history of 4 to 5 billion years. These magnetic fields got weakly recorded in rocks during their solidification. One can get clues about the geomagnetic history from the analysis of this rock magnetism.

(e) Earth’s field departs from its dipole shape substantially at large distances (greater than about 30,000 km) because of the presence of the ionosphere. In this region, earth’s field gets modified because of the field of single ions. While in motion, these ions produce the magnetic field associated with them.

(f) An extremely weak magnetic field can bend charged particles moving in a circle. This may not be noticeable for a large radius path. With reference to the gigantic interstellar space, the deflection can affect the passage of charged particles.

Question 3.
A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10^-2 J. What is the magnitude of magnetic moment of the magnet?
Answer:
Magnetic field strength, B = 0.25 T
Torque on the bar magnet, τ = 4.5 × 10^-2J
Angle between the bar magnet and the external magnetic field, θ = 30°
Torque is related to magnetic moment (M) as
τ = MB sinθ
∴ M = \(\frac{\tau}{B \sin \theta}\)
= \(\frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^{\circ}}\)
\(\frac{4.5 \times 10^{-2} \times 2}{0.25 \times 1}\)
(∵ sin30° = \(\frac{1}{2}\))
= 0.36 JT^-1
Hence, the magnetic moment of the magnet is 0.36 JT^-1.

Question 4.
A short bar magnet of magnetic moment m = 0.32 JT^-1 is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?
Answer:
Given, M = 0.32 JT^-1, B = 0.15T,U = ?
(a) Stable Equilibrium: The magnetic moment should be parallel to the magnetic field. In this position, the potential energy is
U = -MB cos θ =0.32 × 0.15 × 1
= -0.048 J or-4.8 × 10^-2 J\

(b) Unstable Equilibrium: The magnetic moment should be antiparallel to the magnetic field. In this position, the potential energy is
U = -MBcosθ = 0.32 × 0.15 × (-1)
= +0.048 J or + 4.8 × 10^-2 J

Question 5.
A closely wound solenoid of 800 turns and area of cross-section 2.5 × 10^-4 m² carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?
Answer:
Solenoid acts as a bar magnet, its magnetic moment is along the axis of the solenoid, the direction determined by the sense of flow of current. The magnetic moment of a current carrying loop having N turns
= NIA = 800 × 3 × 2.5 × 10^-4
= 6 × 10^-1
= 0.60 A-m²
= 0.60 JT^-1

Question 6.
If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?
Answer:
Magnetic field strength, B = 0.25 T
Magnetic moment, M = 0.6 JT^-1
The angle θ, between the axis of the solenoid and the direction of the applied field is 30°.
Therefore, the torque acting on the solenoid is given as
τ = MB sinθ
= 0.6 × 0.25 × sin30°
= 0.6 × 0.25 × \(\frac{1}{2}\)
= 0.075 N-m

Question 7.
A bar magnet of magnetic moment 1.5 J T^-1 lies aligned with the direction of a uniform magnetic field of 0.22 T.
(a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment:
(i) normal to the field direction,
(ii) opposite to the field direction?

(b) What is the torque on the magnet in cases (i) and (ii)?
Answer:
Given, M = 1.5 JT^-1,B = 0.22 T,θ₁ =0°

(a) To align the dipole normal to the field direction θ₂ = 90°. Therefore,
W = MB(cosθ₁ – cosθ₂)
W = 1.5 × 0.22(cos0° – cos90°) = 0.33 J
Also, τ = MB sinθ₂
or τ = 1.5 × 0.22sin90° = 0.33 Nm

(b) To align the dipole opposite to the field direction θ₂ = 180°. Therefore,
W =MB(cosθ₁ – cosθ₂)
W = 1.5 × 0.22(cos0° – cos180°) = 0.66 J
Also, τ = MB sinθ₂
or τ = 1.5 × 0.22sinl80° = 0 Nm

Question 8.
A closely wound solenoid of2000 turns and area of cross-section 1.6 × 10^-4 m², carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.
(a) What is the magnetic moment associated with the solenoid?
(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10^-2 T is set up at an angle of 30° with the axis of the solenoid?
Answer:
Number of turns on the solenoid, N = 2000
Area of cross-section of the solenoid, A = 1.6 × 10^-4 m^{2</sup
Current in the solenoid, I = 4 A}

(a) Let M = magnetic moment of the solenoid.
∴ Using the relation M = NIA, we get
M = 2000 × 4.0 × 1.6 × 10^{-4</sup
= 1.28 JT-1</sup}

The direction of \(\vec{M}\) is along the axis of the solenoid in the direction related to the sense of current according to right-handed screw rule.

(b) Here θ = 30°
\(\vec{B}\) = 7.5 × 10^-2 T
Let F = force on the solenoid = ?
τ = torque on the solenoid = ?
The solenoid behaves as a bar magnet placed in a uniform magnetic field, so the force is
F = m \(\vec{B}\) + (-m \(\vec{B}\)) = 0
where m = pole strength of the magnet.

Using the relation, τ = MB sinθ, we get
τ = 1.28 × 7.5 × 10^-2 × sin30°
= 1.28 × 7.5 × 10^-2 × \(\frac{1}{2}\) = 0.048 J
The direction of the torque is such that it tends to align the axis of the solenoid (i. e., magnetic moment vector \(\vec{M}\)) along \(\vec{B}\).

Question 9.
A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude 5.0 × 10^-2 T. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 2.0 s^-1. What is the moment of inertia of the coil about its axis of rotation?
Answer:
Number of turns in the circular coil, N = 16
Radius of the coil, r = 10cm = 0.1m
Cross-section of the coil, A = πr² = π × (0.1)² m²
Current in the coil, I = 0.75 A
Magnetic field strength, B = 5.0 × 10^-2 T
Frequency of oscillations of the coil, v = 2.0 s^-1
∴ Magnetic moment, M = NIA = 16 × 0.75 × π × (0.1)² = 0.377 JT^-1
Frequency is given by the relation
v = \(\frac{1}{2 \pi} \sqrt{\frac{M B}{I}}\)
where, I = Moment of inertia of the coil
I = \(\frac{M B}{4 \pi^{2} v^{2}}\) = \(\frac{0.377 \times 5 \times 10^{-2}}{4 \pi^{2} \times(2)^{2}}\)
= 1.19 × 10^-4 kg m²
Hence, the moment of inertia of the coil about its axis of rotation is 1.19 × 10^-4 kg m².

Question 10.
A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at 22° with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be 0.35 G. Determine the magnitude of the earth’s magnetic field at the place.
Answer:
Horizontal component of earth’s magnetic field, B_H = 0.35 G
Angle made by the needle with the horizontal plane
= Angle of dip = δ = 22°
Earth’s magnetic field strength = B
We can relate B and B_H as
B_H = B cosδ
∴ B = \(\frac{B_{H}}{\cos \delta}=\frac{0.35}{\cos 22^{\circ}}=\frac{0.35}{0.9272}\) = 0.377 G
Hence, the strength of earth’s magnetic field at the given location is 0.377 G.

Question 11.
At a certain location in Africa, a compass points 12° west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points 60° above the horizontal. The horizontal component of the earth’s field is measured to he 0.16 G. Specify the direction and magnitude of the earth’s field at the location.
Answer:
Angle of declination, θ = 12°
Angle of dip, δ = 60°
Horizontal component of earth’s magnetic field, B_H = 0.16 G
Earth’s magnetic field at the given location = B
We can relate B and B_H as
B_H = B cosδ
B= \(\frac{B_{H}}{\cos \delta}\) = \(=\frac{0.16}{\cos 60^{\circ}}\) = \(\frac{0.16}{\left(\frac{1}{2}\right)}\) = 0.16 × 2 = 0.32 G
Earth’s magnetic field lies in the vertical plane, 12° west of the geographic meridian, making an angle of 60° (Upward) with the horizontal direction. Its magnitude is 0.32 G.

Question 12.
A short bar magnet has a magnetic moment of 0.48 JT^-1. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.
Answer:
Magnetic moment of the bar magnet, M = 0.48 JT^-1
Distance, d = 10cm = 0.1m

(a) The magnetic field at distance d, from the centre of the magnet on the axis is given by the relation,
B = \(\frac{\mu_{0}}{4 \pi} \frac{2 M}{d^{3}}\)
∴ B = \(\frac{4 \pi \times 10^{-7} \times 2 \times 0.48}{4 \pi \times(0.1)^{3}}\)
= 0.96 × 10^-4 T = 0.96 G
The magnetic field is along the S-N direction.

(b) The magnetic field at a distance of 10 cm (i.e., d = 0.1 m) on the equatorial line of the magnet is given as,
B = \(\frac{\mu_{0} \times M}{4 \pi \times d^{3}}\)
∴ B = \(\frac{4 \pi \times 10^{-7} \times 0.48}{4 \pi \times(0.1)^{3}}\) = 0.48G
The magnetic field is along the N-S direction.

Question 13.
A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at 14 cm from the centre of the magnet. The earth’s magnetic field at the place is 0.36 G and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null-point (i. e., 14 cm) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field.)
Answer:
Distance of the null point from the centre of magnet
d = 14 cm = 0.14 m
The earth’s magnetic field where the angle of dip is zero, is the horizontal component of earth’s magnetic field. i.e., H = 0.36 G
Initially, the null points are on the axis of the magnet. We use the formula of magnetic field on axial line (consider that the magnet is short in length).
B₁ = \(\frac{\mu_{0}}{4 \pi} \cdot \frac{2 m}{d^{3}}\)
This magnetic field is equal to the horizontal component of earth’s magnetic field.
i.e., B₁ = \(\frac{\mu_{0}}{4 \pi} \cdot \frac{2 m}{d^{3}}\) = H ……….(1)
On the equitorial line of magnet at same distance (d) magnetic field due to the magnet
B₂ = \(\frac{\mu_{0}}{4 \pi} \cdot \frac{m}{d^{3}}=\frac{B_{1}}{2}=\frac{H}{2}\) …………….. (2)
The total magnetic field on equitorial line at this point (as given in question)
B = B₂ + H = \(\frac{H}{2}\) + H = \(\frac{3}{2}\)H = \(\frac{3}{2}\) × 0.36 = 0.54G
The direction of magnetic field is in the direction of earth’s field.

Question 14.
If the bar magnet in exercise 5.13 is turned around by 180°, where will the new null points be located?
Answer:
When the bar magnet is turned by 180°, then the null points are obtained on the equitorial line.
So, magnetic field on the equitorial line at distance d’ is
B’ = \(\frac{\mu_{0}}{4 \pi} \cdot \frac{m}{d^{3}}\)
This magnetic field is equal to the horizontal component of earth’s magnetic field
B’ = \(\frac{\mu_{0}}{4 \pi} \cdot \frac{m}{d^{3}}\) = H ………… (1)
From Q.No. 13 MISS
Magnetic field B₁ = \(\frac{\mu_{0}}{4 \pi} \cdot \frac{2 m}{d^{3}}\) = H ………….. (2)
From eqs. (1) and (2), we get

Thus, the null points are located on the equitorial line at a distance of 11.1 cm.

Question 15.
A short bar magnet of magnetic moment 5.25 × 10^-2 JT^-1 is placed with its axis perpendicular to the earth’s field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45° with earth’s field on (a) its normal bisector and (b) its axis. Magnitude of the earth’s field at the place is given to be 0.42 G. Ignore the length of the magnet in comparison to the distances involved.
Answer:
Given, magnetic moment m = 5.25 × 10^-2 J/T
Let the resultant magnetic field is B_net. It makes an angle of 45° with B_e.
∴ B_e = 0.42G =0.42 × 10^-4 T
(a) At normal bisector
Let r is the distance between axial line and point P.
The magnetic field at point P, due to a short magnet
B = \(\frac{\mu_{0}}{4 \pi} \cdot \frac{m}{r^{3}}\) …………. (1)

The direction of B is along PB, i.e., along N pole to S pole.
According to the vector analysis,

r = 0.05 m
r = 5 cm

(b) When point lies on axial line
Let the resultant magnetic field B_net makes an angle 45° from B_e. The magnetic field on the axial line of the magnet at a distance of r from the centre of magnet
B’ = \(\frac{\mu_{0}}{4 \pi} \cdot \frac{2 m}{r^{3}}\) (S to N)

Direction of magnetic field is from S to N.
According to the vector analysis,

Question 16.
Answer the following questions :
(a) Why does a paramagnetic sample display greater magnetisation (for the same magnetising field) when cooled?

(b) Why is diamagnetism, in contrast, almost independent of temperature?

(c) If a toroid uses bismuth for its core, will the field in the core be (slightly) greater or (slightly) less than when the core is empty?

(d) Is the permeability of a ferromagnetic material independent of the magnetic field? If not, is it more for lower or higher fields?

(e) Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. (This fact is analogous to the static electric field lines being normal to the surface of a conductor at every point). Why?

(f) Would the maximum possible magnetisation of a paramagnetic sample be of the same order of magnitude as the magnetisation of a ferromagnet?
Answer:
(a) Owing to the random thermal motion of molecules, the alignments of dipoles get disrupted at high temperatures. On cooling, this disruption is reduced.’Hence, a paramagnetic sample displays greater magnetisation when cooled.

(b) The induced dipole moment in a diamagnetic substance is always opposite to the magnetising field. Hence, the internal motion of the atoms (which is related to the temperature) does not affect the diamagnetism of a material.

(c) Bismuth is a diamagnetic substance. Hence, a toroid with a bismuth core has a magnetic field slightly greater than a toroid whose core is empty.

(d) The permeability of ferromagnetic materials is not independent of the applied magnetic field. It is greater for a lower field and vice versa.

(e) The permeability of ferromagnetic materials is not less than one. It is always greater than one. Hence, magnetic field lines are always nearly normal to the surface of such materials at every point.

(f) The maximum possible magnetisation of a paramagnetic sample can be of the same order of magnitude as the magnetisation of a ferromagnet. This requires high magnetising fields for saturation.

Question 17.
Answer the following questions:
(a) Explain qualitatively on the basis of domain picture the irreversibility in the magnetisation curve of a ferromagnet.

(b) The hysteresis loop of a soft iron piece has a much smaller area than that of a carbon steel piece. If the material is to go through repeated cycles of magnetisation, which piece will dissipate greater heat energy?

(c) ‘A system displaying a hysteresis loop such as a ferromagnet, is a device for storing memory?’ Explain the meaning of this statement.

(d) What kind of ferromagnetic material is used for coating magnetic tapes in a cassette player,’ or for building ‘memory stores’ in a modern computer?

(e) A certain region of space is to be shielded from magnetic fields. Suggest a method.
Answer:
(a) The hysteresis curve (B-H curve) of a ferromagnetic material is shown in the following figure

It can be observed from the given curve that magnetisation persists even when the external field is removed. This reflects the irreversibility of a ferromagnet.

(b) The dissipated heat energy is directly proportional to the area of a hysteresis loop. A carbon steel piede has a greater hysteresis curve area. Hence, it dissipates greater heat energy.

(c) The value of magnetisation is memory or record of hysteresis loop cycles of magnetisation. These bits of information correspond to the cycle of magnetisation. Hysteresis loops can be used for storing information.

(d) Ceramic is used for coating magnetic tapes in cassette players and for building memory stores in modern computers.

(e) A certain region of space can be shielded from magnetic fields if it is – surrounded by soft iron rings. In such arrangements, the magnetic lines are drawn out of the region.

Question 18.
A long straight horizontal cable carries a current of 2.5 A in the direction 10° south of west to 10° north of east. The magnetic meridian of the place happens to be 10° west of the geographic meridian. The earth’s magnetic field at the location is 0.33 G, and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable). (At neutral points, magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth’s magnetic field.)
Answer:
Given, current in the cable
I = 2.5 A

Magnetic meridian M_NM_S is 10° west of geographical meridian G_NG_S earth’s magnetic field R = 0.33 G
= 0.33 × 10^-4 T ……….. (1)
Angle of dip δ = _S0°
The neutral point is the point where the magnetic field due to the current carrying cable is equal to the horizontal component of earth’s magnetic field.
Horizontal component of earth’s magnetic field
H = Rcosθ = 0.33 × 10-4 cos0°
= 0.33 × 10^-4 T
Using the formula of magnetic field at distance r due to an infinite long current carrying conductor
B = \(\frac{\mu_{0}}{4 \pi} \cdot \frac{2 I}{r}\)
At neutral points,
H = B
0.33 × 10^-4 = \(\frac{\mu_{0}}{4 \pi} \cdot \frac{2 I}{r}\)
0.33 × 10^-4 = \(\frac{10^{-7} \times 2 \times 2.5}{r}\)
or r = \(\frac{5 \times 10^{-7}}{0.33 \times 10^{-4}}\)
or r = 1.5 × 10^-2 m = 1.5cm
Thus, the line of neutral points is at a distance of 1.5 cm from the cable.

Question 19.
A telephone cable at a place has four long straight horizontal wires carrying a current of 1.0 A in the same direction east to west. The earth’s magnetic field at the place is 0.39 G, and the angle of dip is 35°. The magnetic declination is nearly zero. What are the resultant magnetic fields at points 4.0 cm above and below the cable?
Answer:
Number of horizontal wires in the telephone cable, n = 4
Current in each wire, I = 1.0 A
Earth’s magnetic field at the location, H = 0.39 G = 0.39 × 10^-4 T
Angle of dip at the location, δ = 35°
Angle of declination, θ = 0°
For a point 4 cm below the cable
Distance, r = 4 cm = 0.04 m
The horizontal component of earth’s magnetic field can be written as
H_h, = Hcosδ – B
where,
B = Magnetic field at 4 cm due to current I in the four wires
= 4 × \(\frac{\mu_{0} I}{2 \pi r}\)
μ₀ = 4π × 10^-7 TmA^-1
∴ B = 4 × \(\frac{4 \pi \times 10^{-7} \times 1}{2 \pi \times 0.04}\)
= 0.2 × 10^-4 T = 0.2G
∴ H_h = 0.39 cos35°- 0.2
= 0.39 × 0.819 – 0.2 ≈ 0.12 G

The vertical component of earth’s magnetic field is given as
H_v = H sinδ
= 0.39 sin35°= 0.22 G
The angle made by the field with its horizontal component is given as
θ = tan^-1 \(\frac{H_{v}}{H_{b}}\)
= tan^-1 \(\frac{0.22}{0.12}\) = 61.39°
The resultant field at the point is given as
H₁ = \(\sqrt{\left(H_{v}\right)^{2}+\left(H_{h}\right)^{2}}\)
= \(\sqrt{(0.22)^{2}+(0.12)^{2}}\) = 0.25 G

For a point 4 cm above the cable
Horizontal component of earth’s magnetic field
H_h = Hcosδ +B = 0.39 cos35° + 0.2 = 0.52 G
Vertical component of earth’s magnetic field
H_v = H sinδ
= 0.39 sin35° = 0.22 G
Angle, θ = tan^-1 \(\frac{H_{v}}{H_{h}}\) = tan^-1\(\frac{0.22}{0.52}\) = 22.90
And resultant field
H₂ = \(\sqrt{\left(H_{v}\right)^{2}+\left(H_{h}\right)^{2}}\)
= \(\sqrt{(0.22)^{2}+(0.52)^{2}}\) = 0.56 G

Question 20.
A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of 45° with the magnetic meridian. When the current in the coil is 0.35 A, the needle points west to east.
(a) Determine the horizontal component of the earth’s magnetic field at the location.
(b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of 90° in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.
Answer:
Number of turns in the circular coil, N = 30
Radius of the circular coil, r = 12cm = 0.12m
Current in the coil, I = 0.35 A
Angle of dip, δ = 45°

(a) The magnetic field due to current I, at a distance r, is given as
B = \(\frac{\mu_{0} 2 \pi N I}{4 \pi r}\)
∴ B = \(\frac{4 \pi \times 10^{-7} \times 2 \pi \times 30 \times 0.35}{4 \pi \times 0.12}\)
= 5.49 × 10^-5 T
The compass needle points from west to east. Hence, the horizontal component of earth’s magnetic field is given as
B_H = B sinδ
= 5.49 × 10^-5 sin 45°
= 3.88 × 10^-5 T = 0.388G

(b) When the current in the coil is reversed and the coil is rotated about its vertical axis by an angle of 90°, the needle will reverse its original direction. In this case, the needle will point from east to west.

Question 21.
A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is 60°, and one of the fields has a magnitude of 1.2 × 10^-2 T. If the dipole comes to stable equilibrium at an angle of 15° with this field, what is the magnitude of the other field?
Answer:
The two fields \(\vec{B}\)₁ and \(\vec{B}\)₂ are shown in the figure

here in which a magnet is placed s.t.
∠NOB = 15°
∠B₁OB₂ = 60°
∴ ∠NOB₂ = 60 – 15 = 45°
B₁ = 1.2 × 10^-2 T
B₂ = ?
Let θ₁ and θ₂ he the inclination of the dipole
with \(\vec{B}\)₁ and \(\vec{B}\)₂ respectively.
∴ θ₁ = 15°,θ₂ = 45°
If τ₁ and τ₂ be the torques on the dipole due to \(\vec{B}\)₁ and \(\vec{B}\)₂ respectively, then
Using the relation,
τ = MB sin θ, we get
τ₁ = MB₁ sinθ₁
and τ₂ = MB₂ sinθ₂
As the dipole is in equilibrium, the torques on the dipole due to \(\vec{B}\)₁ and \(\vec{B}\)₂ are equal and opposite, i. e.,

Question 22.
A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of 0.40 G normal to the initial direction. Estimate the up or down deflection of the beam over a distance of 30 cm (me = 9.11 × 10^-31 kg). [Note : Data in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.]
Answer:
Here, energy = E = 18 KeV = 18 × 1.6 × 10^-16 J
(∵ 1 KeV=10³eV = 10³ × 1.6 × 10^-19 J)
B = horizontal magnetic field = 0.40 G = 0.40 × 10^-4 J
m = 9.11 × 10^-31 kg, e = 1.6 × 10^-19 C
x = 30 cm = 0.30 m
As the magnetic field is normal to the velocity, the charged particle follows circular path in magnetic field. The centrepetal force \(\frac{m v^{2}}{r}\) required for this purpose is provided by force on electron due to magnetic field i. e., BeV

Question 23.
A sample of paramagnetic salt contains 2.0 × 10²⁴ atomic dipoles each of dipole moment 1.5 × 10^-23 JT^-1. The sample is placed under a homogeneous magnetic field of 0.64 T, and cooled to a temperature of 4.2 K. The degree of magnetic saturation achieved is equal to 15%. What is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K? (Assume Curie’s law)
Answer:
Number of atomic dipoles, n = 2.0 × 10²⁴
Dipole moment of each atomic dipole, M = 1.5 × 10^-23 JT^-1
When the magnetic field, B₁ = 0.64 T
The sample is cooled to a temperature, T₁ = 4.2 K
Total dipole moment of the atomic dipole, M_tot = n × M
= 2 × 10²⁴ × 1.5 × 10^-23 = 30 JT^-1
Magnetic saturation is achieved at 15%.
Hence, effective dipole moment, M₁ = \(\frac{15}{100}\) × 30 = 4.5 JT^-1
When the magnetic field, B₂ = 0.98 T
Temperature, T₂ = 2.8 K
Its total dipole moment = M₂
According to Curie’s law, we have the ratio of two magnetic dipoles as
\(\frac{M_{2}}{M_{1}}=\frac{B_{2}}{B_{1}} \times \frac{T_{1}}{T_{2}}\)
∴ M₂ = \(\frac{B_{2} T_{1} M_{1}}{B_{1} T_{2}}\) = \(\frac{0.98 \times 4.2 \times 4.5}{2.8 \times 0.64}\) = 10 336 JT^-1
Therefore, 10.336 J T^-1 is the total dipole moment of the sample for a magnetic field of 0.98 T and a temperature of 2.8 K.

Question 24.
A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetising current of 1.2 A?
Answer:
Mean radius of the Rowland ring, r = 15 cm = 0.15 m
Number of turns on the ferromagnetic core, N = 3500
Relative permeability of the core material, μ_r = 800
Magnetising current, I = 1.2 A
The magnetic field is given by the relation
B = \(\frac{\mu_{r} \mu_{0} I N}{2 \pi r}\)
B = \(\frac{800 \times 4 \pi \times 10^{-7} \times 1.2 \times 3500}{2 \pi \times 0.15}\) = 4.48T
Therefore, the magnetic field in the core is 4.48 T.

Question 25.
The magnetic moment vectors μ_s and μ_l associated with the intrinsic spin angular momentum S and orbital angular momentum l, respectively, of an electron are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by
μ_g = -(e/m)S, μ_l = -(e/2m)l
Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.
Answer:
Out of these two relations, \(\overrightarrow{\mu_{l}}=-\frac{e}{2 m} \vec{l}\) is in accordance with classical physics and can be derived as follows

We know that electrons revolving around the nucleus of an atom in circular orbits behave as tiny current loops having angular momentum \(\vec{\imath}\) given in magnitude as
\(\vec{\imath}\) = mvr …………. (1)
where m = mass of an electron
v = its orbital velocity
r = radius of the circular orbit.
or v_r = \(\frac{l}{m}\) …………… (2)

\(\vec{\imath}\) acts along the normal to the plane of the orbit in upward direction. The orbital motion of electron is taken as equivalent to the flow of conventional current I given by

where – ve sign shows that the electron is negatively
charged. The eqn. (3) shows that μ_e and \(\vec{\imath}\) are opposite to each other i. e., antiparallel and both being normal to the plane of the orbit as shown in the figure

∴ \(\overrightarrow{\mu_{l}}=-\frac{e}{2 m} \cdot \vec{l}\)
\(\frac{\mu_{s}}{S}\) in contrast to \(\frac{\mu_{l}}{\vec{l}}\) is \(\frac{e}{m}\) i.e., twice the classically
expected value. This latter result is an outstanding consequence of modern quantum theory and cannot be obtained classically.

Punjab State Board PSEB 12th Class Physics Book Solutions Chapter 11 Dual Nature of Radiation and Matter Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

PSEB 12th Class Physics Guide Dual Nature of Radiation and Matter Textbook Questions and Answers

Question 1.
Find the
(a) maximum frequency, and
(b) minimum wavelength of X-rays produced by 30 kV electrons.
Answer:
Potentialoftheelectrons, V=30 kV= 3O x 10³ V=3 x 10⁴ V
Hence, energy of the electrons, E = 3 x 10⁴ eV
where, e = Charge on an electron = 1.6 x 10^-19C
(a) Maximum frequency produced by the X-rays = v
The energy of the electrons is given by the relation
E=eV=hv
where, h = Planck’s constant = 6.63 x 10^-34 Js
∴ v = \(\frac{e V}{h} \) (∵ E = eV)
= \(\frac{1.6 \times 10^{-19} \times 3 \times 10^{4}}{6.63 \times 10^{-34}}\) = 7.24 x 10¹⁸ Hz
Hence, the maximum frequency of X-rays produced is 7.24 x 10¹⁸ Hz

(b) The minimum wavelength produced by the X-rays is given as
λ = \(\frac{c}{v}\)
= \(\frac{3 \times 10^{8}}{7.24 \times 10^{18}}\)
= 0.414 x 10^-10
= 0.0414 x 10^-9 m
= 0.0414 nm
Hence, the minimum wavelength of X-rays produced is 0.0414 nm.

Question 2.
The work function of caesium metal is 2.14 eV. When light of frequency 6 x 10¹⁴ Hz is incident on the metal surface, photoemission of electrons occurs.
What is the
(a) maximum kinetic energy of the emitted electrons,
(b) stopping potential, and
(c) maximum speed of the emitted photoelectrons?
Answer:
Work function of caesium metal, Φ₀ = 2.14 eV
Frequency of light, v = 6.0 x 10¹⁴ Hz
The maximum kinetic energy is given by the photoelectric effect as
K = hv- Φ₀
where, h = Planck’s constant = 6.63 x 10^-34 Js .
∴ k = \(\frac{6.63 \times 10^{-34} \times 6 \times 10^{14}}{1.6 \times 10^{-19}}-2.14 \)
( ∵ e=1.6 x 10^-19)
= 2.485-2.140 =0.345eV
Hence, the maximum kinetic energy of the emitted electrons is 0.345 eV.

(b) For stopping potential V₀, we can write the equation for kinetic energy
as K=eV₀
∴ V₀ = \(\frac{K}{e}\) (∵ e=1.6×10¹⁹)
= \(\frac{0.345 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-19}}\) =0.345V
Hence, the stopping potential of the material is 0.345 V.

(c) Maximum speed of the emitted photoelectrons = y
Hence, the relation for kinetic energy can be written as
K = \(\frac{1}{2}\) mv²
where, m = mass of an electron = 9.1 x 10^-31 kg
(∴ e=1.6 x 10^-19)
= \(\frac{2 \times 0.345 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}\) = 0.1104 x 10¹²
∴ v = 3.323 x 10⁵ m/s = 332.3 km/s
Hence, the maximum speed of the emitted photoelectrons is 332.3 km/s.

Question 3.
The photoelectric cut off voltage n a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
Answer:
Cut-off voltage, V₀ = 1.5 V
Maximum kinetic energy of photoelectrons
E_K =eV₀ =1.5eV=1.5 x 1.6 x 10^-19J
=2.4 x 10^-19J.

Question 4.
Monochromatic light of wavelength 632.8 mn is produced by a helium-neon laser. The power emitted is 9.42 mW.
(a) Find the energy and momentum of each photon in the light beam,
(b) How many photons per second, on the average, arrive at a target irradiated by this beam? (Assume the beam to have uniform cross-section which is less than the target area), and
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
Answer:
Wavelength of the monochromatic light, 632.8 nm = 632.8 x 10^-9 m
Power emitted by the laser, P = 9.42 mW = 9.42 x 10^-3 W
Planck’s constant, h = 6.63 x 10^-34Js
Speed of light, c=3 x 10⁸ m/s
Mass of a hydrogen atom, m =1.66 x 10^-27 kg
(a) The energy of each photon is given as
E = \(\frac{h c}{\lambda}\)
= \(\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{632.8 \times 10^{-9}}\)
= 3.141 x 10^-19

The momentum of each photon is given as
p = \(\frac{h}{\lambda}\)
= \(\frac{6.63 \times 10^{-34}}{632.8 \times 10^{-9}}=1.047 \times 10^{-27} \mathrm{~kg} \mathrm{~ms}^{-1} \)

(b) Number of photons arriving per second, at a target irradiated by the beam = n.
Assume that the beam has a uniform cross-section that is less than the
target area.
Hence, the equation for power can be written as
P=nE
∴ n= \(\frac{P}{E}\)
= \(\frac{9.42 \times 10^{-3}}{3.141 \times 10^{-19}}\) = 3 x 10¹⁶

(c) Momentum of the hydrogen atom is the same as the momentum of the photon, .
p=1.047 x 10²⁷ kgms^-1
Momentum is given as
p = mv
where, v = speed of the hydrogen atom
v = \(\frac{p}{m}\)
= \(=\frac{1.047 \times 10^{-27}}{1.66 \times 10^{-27}}\) = 0.630m/s

Question 5.
The enery flux of sunlight reaching the surface of the earth is 1.388 x 10³ W/m².
How many photons (nearly) per square metre are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of 550 rims.
Answer:
Energy flux of sunlight reaching the surface of earth,
Φ = 1.388 x 10³ W/m²
Hence, power of sunlight per square metre, P = 1.388 x 10³W
Speed of light, c = 3 x 10⁸ m/s
Planck’s constant, h = 6.63 x 10^-34 Js
Average wavelength of photons present in sunlight, λ = 550 nm.
=550 x 10^-9m
Number of photons per square metre incident on earth per second = n

Hence, the equation for power can be written as
P = nE
∴ n = \(\frac{P}{E}=\frac{P \lambda}{h c}=\frac{1.388 \times 10^{3} \times 550 \times 10^{-9}}{6.63 \times 10^{-34} \times 3 \times 10^{8}}\)
= 3.84 x 10²¹ photons/m²/s
Therefore, every second, 3.84×1021 photons are incident per square metre on earth.

Question 6.
In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 x 10^-15 Vs. Calculate the value of Planck’s constant.
Answer:
The slope of the cut-off voltage (V) versus frequency (v) of an incident light is given as
\(\frac{V}{v}\) = 4.12 x 10^-15 Vs
V is related to frequency by the equation
hv = eV

where, e = charge on an electron = 1.6 x 10^-19
h = Planck’s constant
∴ h = e x \(\frac{V}{v}\)
= 1.6 x 10^-19 x 4.12 x 10^-15
= 6.592 x 10^-34 Js
Therefore, the value of Planck’s constant is 6.592 x 10^-34 Js.

Question 7.
A 100 W sodium lamp radiates energy uniformly in all directions. The lamp is located at the centre of a large sphere that absorbs all the sodium light which is incident on it. The wavelength of the sodium light is 589 nm.
(a) What is the energy per photon associated with the sodium light?
(b) At what rate are the photons delivered to the sphere?
Answer:
Power of the sodium lamp,. P = 100 W
Wavelength of the emitted sodium light, λ = 589 nm = 589 x 10⁹ m
Planck’s constant, h = 6.63 x 10^-34 Js
Speed of light, c = 3 x 10⁸ m/s
The energy per photon associated with the sodium light is given as
E= \(\frac{h c}{\lambda}\)

(b) Number of photons delivered to the sphere = n
The equation for power can be written as
P=nE
∴ n = \(\frac{P}{E}=\frac{100}{3.37 \times 10^{-19}}\) = 2.96 x 10²⁰ photons/s
Therefore, every second, 2.96 x 10²⁰ photons are delivered to the sphere.

Question 8.
The threshold frequency for a certain metal is 3.3 x 10¹⁴ Hz. If light of frequency 8.2 x 10¹⁴ Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.
Answer:
Threshold frequency of the metal, v₀ = 3.3 x 10¹⁴ Hz
Frequency of light incident on the metal, v = 82 x 10¹⁴ Hz
Charge on’an electron, e = 1.6 x 10^-19 C .
Planck’s constant, h = 6.63 x 10^-34 Js
Cut-off voltage for the photoelectric emission from the metal = V₀
The equation for the cut-off energy is given as
eV₀ = h(v-v₀)
V_o = \(\frac{h\left(v-v_{0}\right)}{e}\)
= \(\frac{6.63 \times 10^{-14} \times\left(8.2 \times 10^{14}-3.3 \times 10^{14}\right)}{1.6 \times 10^{-19}}\)
= 2.0292 V
Therefore, the cut-off voltage for the photoelectric emission is 2.0292 V.

Question 9.
The work function for a certain metal Is 4.2 eV. Will this metal give photoelectric emission for incident radiation of
wavelength 330 nm?
Answer:
The energy of incident radiations
E = \(\frac{h c}{\lambda}\) = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{330 \times 10^{-9}}\) J
= 6.03 x 10^-19 J
= \(\frac{6.03 \times 10^{-19}}{1.6 \times 10^{-19}}\)eV = 3.77 eV
The work function of photometal, Φ₀ = 4.2 eV
As energy of incident photon is less than work function, photoemission is not possible.

Question 10.
Light of frequency 7.21 x 10¹⁴ Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 x10s m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
Answer:
Frequency of light incident on the metal surface, v = 7.21 x 10¹⁴ Hz
Maximum speed of the electrons, v = 6.0 x 10⁵ m/s
Planck’s constant, h = 6.63 x 10^-34 Js
Mass of an electron, m = 9.1 x 10^-31 kg
For threshold frequency v₀, the relation for kinetic energy is written asFor threshold frequency y0, the relation for kinetic energy is written as
\(\frac{1}{2} m v^{2}\) = h(v-v₀)
v₀ = v – \(\frac{m v^{2}}{2 h}\)

Therefore, the threshold frequency for the photoemission of electrons is 4.738 x 10¹⁴ Hz.

Question 11.
Light of wavelength 488 mn is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 V. Find the work function of the material from which the emitter is made.
Answer:
Wavelength of light produced by the argon laser,
λ = 488 nm = 488 x 10^-9 m
Stopping potential of the photoelectrons, V₀ = 0.38 V
1 eV=l.6 x 10^-19 J
∴ V₀= \(\frac{0.38}{1.6 \times 10^{-19}}\) eV
Planck’s constant, h = 6.63 x 10^-34 Js
Charge on an electron, e = 1.6 x 10^-19 C
Speed of light, c =3 x 10 m/s
From Einstein’s photoelectric effect, we have the relation involving the work function Φ₀ of the material of the emitter as

Therefore, the material with which the emitter is made has the work function of 2.16 eV.

Question 12.
Calculate the
(a) momentum, and
(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
Answer:
Potential difference, V = 56 V
Planck’s constant, h = 6.63 x 10^-34 Js
Mass of an electron, m = 9.1 x 10^-31 kg
Charge on an electron, e = 1.6 x 10^-19 C
(a) At equilibrium, the kinetic energy of each electron is equal to the accelerating potential, i.e., we can write the relation for velocity (v) of each electron as

The momentum of each accelerated electron is given as
P = mv
= 9.1 x 10^-31 x 4.44 x 10⁶
= 4.04 x 10^-24 kg m s^-1
Therefore, the momentum of each electron is 4.04 x 10^-24 kg m s^-1.

(b) de Broglie wavelength of an electron accelerating through a potential V is given by the relation

Therefore, the de Broglie wavelength of each electron is 0.1639 nm.

Question 13.
What is the
(a) momentum,
(b) speed, and
(c) de Broglie wavelength of an electron with kinetic energy of 120 eV.
Answer:
Kinetic energy of the electron, E_k = 120 eV
Planck’s constant, h = 6.63 x 10^-34 Js
Mass of an electron, m = 9.1 x 10^-31
Charge on an electron, e = 1.6 x 10^-19 C

(a) For the electron, we can write the relation for kinetic energy as
Ek = \(\frac{1}{2}\) mv²
where, v = speed of the electron
∴ v² = \(\sqrt{\frac{2 e E_{k}}{m}}=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 120}{9.1 \times 10^{-31}}}\)
= \(\sqrt{42.198 \times 10^{12}}\) = 6.496 x 10⁶ m/s
Momentum of the electron, p = mv = 9.1 x 10^-31 x 6.496 x 10⁶
=5.91 x 10²⁴ kg ms^-1
Therefore, the momentum of the electron is 5.91 x 10²⁴ kg ms^-1.

Question 14.
The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which an electron, and a neutron, would have the same de Broglie wavelength.
Answer:
Wavelength of light of sodium line, λ = 589 nm = 589 x 10^-9 m
Mass of an electron, m_e = 9.1 x 10^-31 kg
Mass of a neutron, m_n = 1.66 x 10^-27 kg
Planck’s constant, h = 6.63 x 10^-34 Js

(a) For the kinetic energy K, of an electron accelerating with a velocity v, we have the relation
K = \(\frac{1}{2}\) m_ev² ………………………… (1)
We have the relation for de Broglie wavelength as
λ = \(\frac{h}{m_{e} v}\)
∴ v² = \(\frac{h^{2}}{\lambda^{2} m_{e}^{2}}\) ………………………… (2)
Substituting equation (2) in equation (1), we get the relation
K = \(\frac{1}{2} \frac{m_{e} h^{2}}{\lambda^{2} m_{e}^{2}}=\frac{h^{2}}{2 \lambda^{2} m_{e}}\) ………….. (3)

Hence, the kinetic energy of the electron is 6.9 x 10^-25 J or 4.31 µeV.

(b) Using equation (3), we can write the relation for the kinetic energy of the neutron as = \(\frac{h^{2}}{2 \lambda^{2} m_{n}}\)
= \(\frac{\left(6.63 \times 10^{-34}\right)^{2}}{2 \times\left(589 \times 10^{-9}\right)^{2} \times 1.66 \times 10^{-27}}\)
= 3.78 x 10^-28
= \(\frac{3.78 \times 10^{-28}}{1.6 \times 10^{-19}} \) = 2.36 x 10^-9 eV
= 2.36 neV
Hence, the kinetic energy of the neutron is 3.78 x 10^-28 J or 2.36 neV.

Question 15.
What is the de Broglie wavelength of
(a) a bullet of mass 0.040 kg travelling at the speed of 1.0 km/s,
(b) a ball of mass 0.060 kg moving at a speed of 1.0 m/s, and
(c) a dust particle of mass 1.0 x 10^-9 kg drifting with a speed of 2.2 m/s?
Answer:
(a) Mass of the bullet, m = 0.040 kg
Speed of the bullet, v = 1.0 km/s = 1000 m/s
Planck’s constant, h = 6.63 x 10^-34 Js
de Broglie wavelength of the bullet is given by the relation
λ = \(\frac{h}{m v}\)
= \(\frac{6.63 \times 10^{-34}}{0.040 \times 1000} \) = 1.65 x 10^-35 m

(b) Mass of the ball, m = 0.060 kg
Speed of the ball, v =1.0 m/s
de Brogue wavelength of the ball is given by the relation
λ = \(\frac{h}{m v}\)
= \(\frac{6.63 \times 10^{-34}}{0.060 \times 1}\) = 1.1 x 10^-32 m

(c) Mass of the dust particle, m = 1 x 10^-9 kg
Speed of the dust particle, v = 2.2 m/s
de Brogue wavelength of the dust particle is given by the relation
λ = \(\frac{h}{m v}\)
= \(\frac{6.63 \times 10^{-34}}{2.2 \times 1 \times 10^{-9}}\) = 3.0 x 10^-25 m.

Question 16.
An electron and a photon each have a wavelength of 1.00 run. Find
(a) their momenta,
(b) the energy of the photon, and
(c) the kinetic energy of electrons.
Answer:
Wavelength of an electron (λe) and a photon (λp),λe = λp = λ = 1 nm
= 1 x 10^-9 m
Planck’s constant, h = 6.63 x 10^-34 Js

(a) The momentum of an elementary particle is given by de Broglie relation
λ = \(\frac{h}{p}\)
p = \(\frac{h}{\lambda}\)
It is clear that momentum depends only on the wavelength of the particle. Since the wavelengths of an electron and a photon are equal, both have an equal momentum.
∴ p= \(\frac{6.63 \times 10^{-34}}{1 \times 10^{-9}} \) =6.63 x 10^-25 kgms^-1

(b) The energy of a photon is given by the relation
E= \(\frac{h c}{\lambda}\)
where, speed of light, c =3 x 10⁸ m/s
∴ E = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{1 \times 10^{-9} \times 1.6 \times 10^{-19}}\)
= 1243.1 eV = 1.243 keV
Therefore, the energy of the photon is 1.243 keV.

(c) The kinetic energy (K) of an electron having momentum p, is given by the relation
K = \(\frac{1}{2} \frac{p^{2}}{m}\)
where, m = mass of the electron = 9.1 x 10^-31 kg;
p = 6.63 x 10^-25 kgm s^-1

∴ K = \(\frac{1}{2} \times \frac{\left(6.63 \times 10^{-25}\right)^{2}}{9.1 \times 10^{-31}}\) = 2.415 x 10^-19 J
= \(\frac{2.415 \times 10^{-19}}{1.6 \times 10^{-19}}\) = 1.51 eV
Hence, the kinetic energy of the electron is 1.51 eV.

Question 17.
(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40x 10^-10 m?
(b) Also, find the de Brogue wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) kT at 300 K.
Answer:
(a) de Brogue wavelength of the neutron, λ =1.40 x 10^-10 m
Mass of a neutron,m_n =1. 66 x 10^-27 kg
Planck’s constant, h = 6.63 x 10^-34 Js
Kinetic energy (K) and velocity ( v) are related as
K = \( \frac{1}{2} m_{n} v^{2}\) ……………………………… (1)
de Brogue wavelength (λ) and velocity (v) are related as
λ= \(\frac{h}{m_{n} v}\) ……………………………….. (2)
Using equation (2) in equation (1), we get


Hence, the kinetic energy of the neutron is 6.75 x 10^-21 J or 4.219 x 10^-2 eV.

(b) Temperature of the neutron, T = 300 K
Boltzmann’s constant, k =1.38 x 10^-23 kg m² s^-2 K^-1
Average kinetic energy of the neutron,
K’ = \(\frac{3}{2} \) kT.
= \(\frac{3}{2} \) x 1.38 x 10^-23 x 300 = 6.21 x 10^-21 J
The relation for the de Broglie wavelength is given as
λ ‘ = \(\frac{h}{\sqrt{2 K^{\prime} m_{n}}}\)

where, m_n = 1.66 x 10^-27 kg
h = 6.63 x 10^-34 Js
K’ = 6.21 x 10^-21 J
∴ λ’ = \(\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 6.21 \times 10^{-21} \times 1.66 \times 10^{-27}}}\)
=1.46 x 10^-10
m = 0.146 nm
Therefore, the de Broglie wavelength of the neutron is 0.146 nm.

Question 18.
Show that the wavelength cf electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
Answer:
The momentum of a photon having energy (hv) is given as ’
p = \(\frac{h v}{c}=\frac{h}{\lambda}\)
λ = \(\frac{h}{p}\) ………………………….. (1)
where, λ = wavelength of the electromagnetic radiation
c = speed of light
h = Planck’s constant
de Broglie wavelength of the photon is given as
λ = \(\frac{h}{m v}\)
But p = mv
∴ λ =\(\frac{h}{p}\) ………………………………….. (2)
where, m = mass of the photon
v = velocity of the photon
Hence, it can be inferred from equations (1) and (2) that the wavelength of the electromagnetic radiation is equal to the de Broglie wavelength of the photon.

Question 19.
What is the de Broglie wavelength of a nitrogen molecule in air at 300 K?
Assume that the molecule is moving with the root mean square speed of molecules at this temperature. (Atomic mass of nitrogen = 140076 u)
Answer:
Temperature of the nitrogen molecule, T = 300 K
Atomic mass of nitrogen = 14.0076 u
Hence, mass of the nitrogen molecule, m = 2 x 14.0076 = 28.0152 u
But 1 u = 1.66 x 10^-27 kg
∴ m=28.0152 x 1.66 x 10^-27 kg
Planck’s constant, h = 6.63 x 10^-34 Js
Boltzmann’s constant, k = 1.38 x 10^-23 K^-1
We have the expression that relates mean kinetic energy \(\left(\frac{3}{2} k T\right)\) of the nitrogen molecuLe with the root mean square speed (v_rms) as

Hence, the de Broglie wavelength of the nitrogen molecule is given as
λ = \(\frac{h}{m v_{\text {rms }}}=\frac{h}{\sqrt{3 m k T}}\)
= \(\frac{6.63 \times 10^{-34}}{\sqrt{3 \times 28.0152 \times 1.66 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300}}\)
= 0.028 x 10^-9 m
= 0.028 nm
Therefore, the de Broglie wavelength of the nitrogen molecule is 0.028 nm.

[Additional Exercises]

Question 20.
(a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter.
Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its elm, is given to be 1.76 x 10¹¹ Ckg^-1.
(b) Use the same formula you employ in (a) to obtain electron speed for an collector potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?
Answer:
(a) Potential difference across the evacuated tube, V = 500 V
Specific charge of the electron, e/m = 1.76 x 10¹¹ C kg^-1
The speed of each emitted electron is given by the relation for kinetic energy as

KE = \(\frac{1}{2}\) mv²
Therefore, the speed of each emitted electron is
KE =\(\frac{1}{2}\) mv² = eV
∴ v = \(\left(\frac{2 e V}{m}\right)^{1 / 2}=\left(2 V \times \frac{e}{m}\right)^{1 / 2} \)
= (2x 500 xl.76 x 10¹¹)^1/2
= 1.327 x 10⁷ m/s

(b) Potential of the anode, V = 10 MV = 10 x 10⁶ = 10⁷ V
The speed of each electron is given as
v = \(\left(2 V \frac{e}{m}\right)^{1 / 2}\)
= (2 x 10⁷x 1.76 x 10¹¹)^1/2
= 1.88 x 10⁹ m/s .
This result is wrong because nothing can move faster than light. In the above formula, the expression (mv² / 2) for energy can only be used in the non-relativistic limit, i. e., for v < < c.
For very high speed problems, relativistic equations must be considered for solving them. In the relativistic limit, the total energy is given as E = mc²
where, m = relativistic mass
m₀ = \(\left(1-\frac{v^{2}}{c^{2}}\right)^{1 / 2}\) = mass of the particle at rest
Kinetic energy is given as
K = mc² – m₀c²

Question 21.
(a) A monoenergetic electron beam with electron speed of 5.20x 10⁶ ms^-1 is subject to a magnetic field of 1.30 x 10 ⁴T normal to the beam velocity. What is the radius of the circle traced by the beam, given elm for electron equals 1.76x 10¹¹ C kg^-1
(b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified? [Note: Exercises 11.20 (b) and 11.21 (b) take you to relativistic mechanics which is beyond the scope of this book. They have been inserted here simply to emphasise the point that the formulas you use in part (a) of the exercises are not valid at very high speeds or energies. See answers at the end to know what ‘very high speed or energy’ means.]
Answer:
(a) Speed of the electron, v = 5.20 x 10⁶ m/s
Magnetic field experienced by the electron, B = 1.30 x 10^-4 T
Specific charge of the electron, e/m = 1.76 x 10¹¹ C kg’
where, e = charge on the electron = 1.6 x 10^-19 C
m = mass of the electron = 9.1 x 10^-31 kg
The force exerted on the electron is given as
F = e\(|\vec{v} \times \vec{B}|\)
= evBsinθ
θ = angle between the magnetic field and the beam velocity.

The magnetic field is normal to the direction of beam.
∴ θ = 90°
F = evB ……………………………. (1)
The beam traces a circular path of radius, r. It is the magnetic field, due to its bending nature, that provides the centripetal force \(\left(F=\frac{m v^{2}}{r}\right)\) for the
beam.
Hence, equation (1) reduces to
evB = \(\frac{m v^{2}}{r}\)
∴ r = \(\frac{m v}{e B}=\frac{v}{\left(\frac{e}{m}\right) B}=\frac{5.20 \times 10^{6}}{\left(1.76 \times 10^{11}\right) \times 1.30 \times 10^{-4}}\)
= 0.227 m
= 0.227 x 100 cm = 22.7 cm
Therefore, the radius of the circular path is 22.7 cm.

(b) Energy of the electron beam, E = 20 MeV = 20 x 10⁶ x 1.6 x 10^-19 J
The energy of the electron is given as
E = \(\frac{1}{2} \) mv²
∴ v = \(\left(\frac{2 E}{m}\right)^{1 / 2}\)
= \(\sqrt{\frac{2 \times 20 \times 10^{6} \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}}\) = 2.652 x 10⁹ m/s

This result is incorrect because nothing can move faster than light. In the above formula, the expression (mv² / 2) for energy can only be used in the non-relativistic limit, i. e., for v« c.

When very high speeds are concerned, the relativistic domain comes into consideration. In the relativistic domain, mass is given as
m = m₀ \(\left[1-\frac{v^{2}}{c^{2}}\right]^{1 / 2}\)
where, m₀ = mass of the particle at rest

Hence, the radius of the circular path is given as
r = mv/eB = \(\frac{m_{0} v}{e B \sqrt{\frac{c^{2}-v^{2}}{c^{2}}}}\).

Question 22.
An electron gun with its collector at a potential of 100 V fires out electrons in a spherical hulh containing hydrogen gas at low pressure (~ 10^-2 nun of Hg). A magnetic field of 2.83 x 10^-4 T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method). Determine elm from the data.
Answer:
Potential of the anode, V = 100 V
Magnetic field experienced by the electrons, B = 2.83 x 10,sup>-4 T
Radius of the circular orbit r = 12.0 cm = 12.0 x 10^-2 m
Mass of each electron = m
Charge on each electron = e ‘
Velocity of each electron = v

The energy of each electron is equal to its kinetic energy, i. e.,
\(\frac{1}{2}\) mv² = eV
v² = \(\frac{2 e V}{m}\)
It is the magnetic field, due to its bending nature, that provides the \(\left(F=\frac{m v^{2}}{r}\right) \) for the beam.
Hence, we can write Centripetal force = Magnetic force mv²
\(\frac{m v^{2}}{r}\) = evB
eB = \(\frac{m v}{r}\)
v = \(\frac{e B r}{m}\) ………………………………. (2)
Putting the value of y in equation (1), we get

Therefore, the specific charge ratio (e/ m) is 1.73 x 10¹¹ C kg^-1.

Question 23.
(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 A. What is the maximum energy of a photon in the radiation?
(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?
Answer:
(a) Wavelength produced by the X-ray tube, λ= 0.45Å= 0.45 x 10^-10 m
Planck’s constant, h = 6.63 x 10^-34 Js
Speed of light, c = 3 x 10⁸ m/s
The maximum energy of a photon is given as
E = \(=\frac{h c}{\lambda}\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{0.45 \times 10^{-10} \times 1.6 \times 10^{19}} \) = 27.6 x 10³ eV = 27.6 keV
Therefore, the maximum energy of the X-ray photon is 27.6 keV.

(b) Accelerating voltage provides energy to the electrons for producing X-rays.
To get an X-ray of 27.6 keV, the incident electrons must possess at least 27.6 keV of kinetic electric energy. Hence, an accelerating voltage of the order of 30 keV is required for producing X-rays.

Question 24.
In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as an annihilation of an electron-positron pair of total energy 10.2 BeV into two y-rays of equal energy. What is the wavelength associated with each γ-ray? (1 BeV = 10⁹ eV)
Answer:
Total energy of two γ-rays
E = 10.2 BeV
= 10.2 x 10⁹ eV
= 10.2 x 10⁹ x 1.6 x 10^-19 J
= 10.2 x 1.6 x 10^-10

Hence, the energy of each γ-ray
E’ = \(\frac{E}{2}\)
= \(\frac{10.2 \times 1.6 \times 10^{-10}}{2}\)
= 8.16 x 10^-10 J
Planck’s constant, h = 6.63 x 10^-34 Js
Speed of light, c = 3 x 10⁸ m/s
Energy is related to wavelength as
E ‘ = \(\frac{h c}{\lambda}\)
∴ λ = \(\frac{h c}{E^{\prime}}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{8.16 \times 10^{-10}}\)
= 2.436 x 10^-16 m
Therefore, the wavelength associated with each γ-ray is 2.436 x 10^-16 m.

Question 25.
Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons. The second number tells you why our eye can never ‘count photons’, even in barely detectable light.
(a) The number of photons emitted per second by a Mediumwave transmitter of 10 kW power, emitting radio waves of wavelength 500 m.
(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (~ 10^-10 W m^-2). Take the area of the pupil to be about 0.4 cm², and the average frequency of white light to be about 6 x 10¹⁴ Hz.
Answer:
(a) Power of the medium wave transmitter,
P = 10kW = 10⁴ W = 10⁴ J/s
Hence, the energy emitted by the transmitter per second, E = 10⁴
The wavelength of the radio wave, λ = 500 m
The energy of the wave is given as E₁ = \(\frac{h c}{\lambda}\)
where, h = Planck’s constant = 6.63 x 10^-34 Js
c = speed of light = 3 x 10⁸ m/s
∴ E₁ = \(\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{500}\) = 3.96 x 10^-28

Let n be the number of photons emitted by the transmitter.
∴ nE₁ = E
n = \(\frac{E}{E_{1}}\) =\(\frac{10^{4}}{3.96 \times 10^{-28}}\) = 2.525 x 10³¹
≈ 3 x 10³¹
The energy (E₁) of a radio photon is very less, but the number of photons (n) emitted per second in a radio wave is very large. The existence of a minimum quantum of energy can be ignored and the total energy of a radio wave can be treated as being continuous.

(b) Intensity of light perceived by the human eye, I = 10^-10 W m^-2
Area of the pupil, A = 0.4 cm² = 0.4 x 10^-4 m²
Frequency of white light, v = 6 x 10¹⁴ Hz
The energy emitted by a photon is given as
E = hv
where, h = Planck’s constant = 6.63 x 10^-34 Js
∴ E = 6.63 x 10^-34 x 6 x 10¹⁴
= 3.96 x 10^-19 J
Let n be the total number of photons falling per second, per unit area of the pupil. The total energy per unit for n falling photons is given as
E = n x 3.96 x 10^-19 Js^-1 m^-2
The energy per unit area per second is the intensity of light.
∴ E = I
n x 3.96 x 10^-19 = 10^-10
n = \(\frac{10^{-10}}{3.96 \times 10^{-19}}\)
= 2.52 x 10⁸ m² s^-1

The total number of photons entering the pupil per second is given as
n_A =n x A
= 2.52 x 10⁸ x 0.4 x 10^-4
= 1.008 x 10⁴ s^-1
This number is not as large as the one found in problem (a), but it is large enough for the human eye to never see the individual photons.

Question 26.
Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is -1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity (~10⁵ Wm^-2) red light of wavelength 6328 Å produced by a He-Ne laser?
Answer:
Wavelength of ultraviolet light, λ = 2271 Å = 2271 x 10^-10 m
Stopping potential of the metal, V₀ = 1.3 V
Planck’s constant, h = 6.63 x 10^-34 Js
Charge on an electron, e = 1.6 x 10^-19 C
Work function of the metal = Φ₀
Frequency of light = v
We have the photo-energy relation from the photoelectric effect as
Φ₀ =hv₀
v₀ = \(\frac{\phi_{0}}{h}\)
= \(\frac{6.64 \times 10^{-19}}{6.63 \times 10^{-34}} \)
= 1.006 x 10¹⁵ Hz = 4.15 eV
Let v₀ be the threshold frequency of the metal.
∴ Φ₀ = hv₀
v₀ = \(\frac{\phi_{0}}{h}\)
= \(\frac{6.64 \times 10^{-19}}{6.63 \times 10^{-34}}\)
= 1.006 x 10¹⁵ Hz

Wavelength of red light, λ_r = 6328 Å = 6328 x 10^-10
∴ Frequency of red light, v_r = \(\frac{c}{\lambda_{r}}=\frac{3 \times 10^{8}}{6328 \times 10^{-10}} \)
= 4.74 x 10¹⁴ Hz
Since v₀ > v_r, the photocell will not respond to the red light produced by the laser.

Question 27.
Monochromatic radiation of wavelength 640.2 nm (1 nm = 10^-9 m) from a neon lamp irradiates photosensitive material made of caesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.
Answer:
Wavelength of the monochromatic radiation, λ = 640.2 nm = 640.2 x 10^-9 m
Stopping potential of the neon lamp, V₀ = 0.54 V
Charge on an electron, e = 1.6 x 10^-19 C
Planck’s constant, h = 6.63 x 10^-34 Js
Let Φ₀ the work function and v be the frequency of emitted light. We have the photo-energy relation from the photoelectric effect as

Wavelength of the radiation emitted from an iron source, λ’ = 427.2 nm = 427.2 x 10^-9 m
Let V’₀ be the new stopping potential. Hence, photo-energy is given as
eV’₀ = \(\frac{h c}{\lambda^{\prime}}-\phi_{0}\)

Hence, the new stopping potential is 1.50 eV.

Question 28.
A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used :
λ₁ =3650 Å,
λ₂ = 4047 Å,
λ₃ =4358 Å,
λ₄ =5461 Å,
λ₅ =6907 Å,
The stopping voltages, respectively, were measured to be :
V₀₁ = 1.28 V,
V₀₂ = 0.95 V,
V₀₃ = 0.74 V,
V₀₄ = 0.16 V,
V₀₅ = 0 V.
Determine the value of Planck’s constant h, the threshold frequency and work function for the material.
[Note: You will notice that to get h from the data, you will need to know e (which you can take to be 1.6 x 10^-19 C). Experiments of this kind on Na, Li, K, etc. were performed by Millikan, who, using his own value of e (from the oil-drop experiment) confirmed Einstein’s photoelectric equation and at the same time gave an independent estimate of the value of h.]
Answer:
Given, the following wavelength from a mercury source were used
λ₁ =3650 Å, = 3650 x 10^-10 m
λ₂ = 4047 Å, = 4047 x 10^-10m
λ₃ =4358 Å, = 4358 x 10^-10 m
λ₄ =5461 Å, = 5461 x 10^-10 m
λ₅ =6907 Å, = 6907 x 10^-10m
The stopping voltages are as follows
V01 =1.28 V,
V02 = 0.95V,
V03 =0.74V,
V04 =0.16 V,
V5 =0

Frequencies corresponding to wavelengths

As we know that
eV₀ = hv-Φ₀
v₀ = \(\frac{h v}{e}-\frac{\phi_{0}}{e}\)

As the graph between V₀ and frequency v is a straight line.
The slope of this graph gives the values of \(\frac{h}{e} \)

(b) Work function, Φ₀ = hv₀
= 6.574 x 10^-34 x 5 x 10¹⁴
= 32.870 x 10^-20J
= 2.05eV.

Question 29.
The work function for the following metals is given: Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV.
Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 Å from a He-Cd laser placed 1 m away from the photo-cell? What happens if the laser is brought nearer and placed 50 cm away?
Answer:
Mo and Ni will not show photoelectric emission in both cases.
Wavelength for the radiation, λ – 3300 Å = 3300 x 10^-10 m
Speed of light, c = 3 x 10⁸ m/s
Planck’s constant, h = 6.63 x 10^-34 Js
The energy of incident radiation is given as
E = \(\frac{h c}{\lambda}\)

= 3.758eV
It can be observed that the energy of the incident radiation is greater than the work function of Na and K only. It is less for Mo and Ni. Hence, Mo and Ni will not show photoelectric emission. If the source of light is brought near the photocells and placed 50 cm away from them, then the intensity of radiation will increase. This does not affect the energy of the radiation. Hence, the result will be the same as before. However, the photoelectrons emitted from Na and K will increase in proportion to intensity.

Question 30.
Light of intensity 10^-5 W m^-2 falls on a sodium photo-cell of surface area 2 cm². Assuming that the top 5 layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about 2 eV. What is the implication of your answer?
Answer:
Given, intensity of light = 10^-5 W/m²
Area = 2 cm² = 2 x 10^-4 m²
Work function for the metal Φ₀ = 2 eV
Let t be the time.
The effective atomic area of Na = 10^-20 m² and it contains one conduction electron per atom.
Number of conduction electrons in five layers
= \(\frac{5 \times \text { Area of one layer }}{\text { Effective atomic area }}=\frac{5 \times 2 \times 10^{-4}}{10^{-20}}\)
= 10⁷
We know that sodium has one free electron (or conduction electron) per atom.
Incident power on the surface area of photocell
= Incident intensity x Area on the surface area of photocell
= 10^-5 x 2 x 10^-4
= 2 x 10^-9W
The electron present in all the 5 layers of sodium will share the incident energy equally.
Energy absorbed per second per electron,
E = \(\frac{\text { Incident power }}{\text { Number of electrons in five layers }} \)
= \(\frac{2 \times 10^{-9}}{10^{17}}\) = 2 x 10^-26W
Time required for emission by each electron, t = \(\frac{\text { Energy required per electron }}{\text { Energy absorbed per second }} \) = \(\frac{2 \times 1.6 \times 10^{-19}}{2 \times 10^{-26}} \) = 1.6 x 10⁷s
which is about 0.5 yr.
The answer obtained implies that the time of emission of electrons is very large and is not agreement with the observed time of emission. There is no time lag between the incidence of light and the emission of photoelectrons. Thus, it is implied that the wave theory cannot be applied in this experiment.

Question 31.
Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Å, which is of the order of interatomic spacing in the lattice) (m_e = 9.11 x 10^-31 kg).
Answer:
An X-ray probe has a greater energy than an electron probe for the same wavelength.
Wavelength of light emitted from the probe, λ = 1 Å = 10^-10 m
Mass of an electron, me = 9.11 x 10^-31 kg
Planck’s constant, h = 6.63 x 10^-34 Js
Charge on an electron, e = 1.6 x 10^-19 C
The kinetic energy of the electron is given as
K = \(\frac{1}{2} m_{n} v^{2}\)
m_nv = \(\sqrt{2 K m_{n}}\)
where, v = velocity of the electron
m_nv = momentum (p) of the electron

According to the de Brogue principle, the de Brogue wavelength is given as

The energy of a photon,

Hence, a photon has a greater energy than an electron for the same wavelength.

Question 32.
(a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have seen in Exercise 11.31, an electron beam of this energy is suitable for crystal diffraction experiments. Would a neutron beam of the same energy be equally suitable? Explain. (m = 1.675x 10^-27 kg)
(b) Obtain the de Brogue wavelength associated with thermal neutrons at room temperature (27C). Hence, explain why a fast neutron beam needs to be thermalised with the environment before it can be used for neutron diffraction experiments.
Answer:
(a) de Brogue wavelength= 2.327 x 10^-12 m; neutron is not suitable for
the diffraction experiment Kinetic energy of the neutron, K = 150 eV
= 150 x 1.6 X 10^-19
=2.4 x 10^-17 J
Mass of a neutron, m_n = l.675 x 10^-27 kg
The kinetic energy of the neutron is given by the relation
K = \(\frac{1}{2} m_{n} v^{2}\)
m_nv = \(\sqrt{2 K m_{n}}\)
where, y = velocity of the neutron
m_nv = momentum of the neutron

deBroglie wavelength of the neutron is given as
λ = \(\frac{h}{m_{n} v}=\frac{h}{\sqrt{2 K m_{n}}}\)
It is clear that wavelength is inversely proportional to the square root of mass.
Hence, wavelength decreases with increase in mass and vice versa.
∴ λ = \(\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 2.4 \times 10^{-17} \times 1.675 \times 10^{-27}}}\)
= 2.327 x 10^-12 m
It is given in the previous problem that the interatomic spacing of a crystal is about 1 Å, i.e., 10^-10 m.
Hence, the interatomic spacing is about a hundred times greater. Hence, a neutron beam of energy 150 eV is not
suitable for diffraction experiments.

(b) de Brogue wavelength =1.447 x 10^-10 m
Room temperature, T = 27°C = 27+ 273 = 300 K
The average kinetic energy of the neutron is given as
E=\(\frac{3}{2}\) kT
where, k = Boltzmann’s constant = 1.38 x10^-23 J mol^-1K^-1
The wavelength of the neutron is given as
λ = \(\frac{h}{\sqrt{2 m_{n} E}}=\frac{h}{\sqrt{3 m_{n} k T}}\)
= \(\frac{6.63 \times 10^{-34}}{\sqrt{3 \times 1.675 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300}}\)
=1.447 x 10^-10 m
This wavelength is comparable to the interatomic spacing of a crystal. Hence, the high energy neutron beam should first be thermalised, before using it for diffraction.

Question 33.
An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Brogue wavelength associated with the electrons, if other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?
Answer:
Electrons are accelerated by a voltage, V = 50 kV = 50 x 10³ V
Charge on an electron, e 1.6 x 10^-19 C
Mass of an electron, me = 9.11 x 10^-31 kg
Wavelength of yellow light = 5.9 x 10^-7 m

The kinetic energy of the electron is given as
E=eV
=l.6 x 10^-19x 50x 10³
= 8 x 10^-15 J
de Brogue wavelength is given by the relation
λ = \(\frac{h}{\sqrt{2 m_{e} E}}\)
= \(\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 8 \times 10^{-15}}}\)
= 5.467 x 10^-12 m
This wavelength is nearly 10⁵ times less than the wavelength of yellow light. The resolving power of a microscope is inversely proportional to the wavelength of light used. Thus, the resolving power of an electron microscope is nearly 10⁵ times that of an optical microscope.

Question 34.
The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length scale of 10^-15 m or less. This structure was first, probed in early 1970s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV.)
Answer:
Wavelength of a proton or a neutron, λ ≈ 10^-15 m
Rest mass energy of an electron,
m₀c² =0.511 MeV
= 0.511 x 10⁶ x 1.6 x 10^-19
= 0.8176 x 10^-13 J
Planck’s constant, h = 6.63 x 10^-34 Js
Speed of light, c = 3 x 10⁸ m/s ,

The momentum of a proton or a neutron is given as
p = \(\frac{h}{\lambda}\)
= \(\frac{6.63 \times 10^{-34}}{10^{-15}}\) = 6.6 x 10^-19 kg m/s
The relativistic relation for energy (E) is given as
E² = p²c² +m₀²c⁴
= (6.6x 10^-19 x 3 x 10⁸)² + (0.8176 x 10^-13)²
= 392.04 x 10^-22 +0.6685 x 10^-26
≈ 392.04 x 10^-22
∴ E = 1.98x 10^-10 J
Thus, the electron energy emitted from the accelerator at Stanford, USA might be of the order of 1.24 BeV.

Question 35.
Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27°C) and 1 atm pressure, and compare it with the mean separation between two atoms under these conditions.
Answer:
de Broglie wavelength associated with He atom = 0.7268 x 10^-10 m .
Room temperature, T = 27°C =27 + 273 = 300 K
Atmospheric pressure, P = 1 atm = 1.01 x 10⁵ Pa
Atomic weight of a He atom = 4
Avogadro’s number, N_A = 6.023 x 10²³
Boltzmann’s constant, k = 1.38 x 10^-23 J mol^-1 K^-1

Average energy of a gas at temperature T, is given as
E = \(\frac{3}{2}\) kT
de Broglie wavelength is given by the relation
λ = \(\frac{h}{\sqrt{2 m E}}\)
where, m = mass of a He atom

We have the ideal gas formula
PV = RT
PV = kNT
∵ \(\frac{V}{N}=\frac{k T}{P}\)
where V = volume of the gas
N = number of moles of the gas
Mean separation between two atoms of the gas is given by the relation
r = \(\left(\frac{V}{N}\right)^{1 / 3}=\left(\frac{k T}{P}\right)^{1 / 3}=\left[\frac{1.38 \times 10^{-23} \times 300}{1.01 \times 10^{5}}\right]^{1 / 3}\)
= 3.35 x 10^-9 m
Hence, the mean separation between the atoms is much greater than the de Broglie wavelength.

Question 36.
Compute the typical de Broglie wavelength of an electron in a metal at 2 7° C and compare it with the mean separation between two electrons in a metal which is given to be about 2 x 10¹⁰ m. [Note: Exercises 11.35 and 11.36 reveal that while the wave-packets associated with gaseous molecules under ordinary conditions are non-overlapping, the electron wave- packets in a metal strongly overlap with one another. This suggests that whereas molecules in an ordinary gas can be distinguished apart, electrons in a metal cannot be distinguished apart from one another. This indistinguishability has many fundamental implications which you will explore in more advanced Physics courses.]
Answer:
Temperature, T = 27°C = 27 +273 = 300 K
Mean separation between two electrons, r = 2 x 10^-10 m
de Broglie wavelength of an electron is given as
λ = \(\frac{h}{\sqrt{3 m k T}}\)
where,
h = Planck’s constant = 6.63 x 10^-34 Js
m = Mass of an electron = 9.11 x 10 ^-31 kg
k = Boltzmann’s constant = 1.38 x 10^-23 J mol^-1 K^-1
∴ λ = \(\frac{6.63 \times 10^{-34}}{\sqrt{3 \times 9.11 \times 10^{-31} \times 1.38 \times 10^{-23} \times 300}}\)
= 6.2 x 10⁹ m
Hence, the de Broglie wavelength is much greater than the given inter-electron separation.

Question 37.
Answer the following questions :
(a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3)e ; (-1/3)e]. Why do they not show up in Millikan’s oil-drop experiment?
(b) What is so special about the combination e/m? Why do we not simply talk of e and m separately?
(c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures?
(d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons?
(e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations :
E = hv, p = \(\frac{\boldsymbol{h}}{\lambda}\)
But while the value of λ is physically significant, the value of v (and therefore, the value of the phase speed vλ) has no physical significance. Why?
Answer:
(a) Quarks inside protons and neutrons carry fractional charges. This is because nuclear force increases extremely if they are pulled apart. Therefore, fractional charges may exist in nature; observable charges are siti the integral multiple of an electrical charge.

(b) Thè basic relations for electric field and magnetic field are \(\left(e V=\frac{1}{2} m v^{2}\right)\) and \(\left(e B v=\frac{m v^{2}}{r}\right) \) respectively.

These relations include e (electric charge), y (velocity), m (mass), V (potential), r (radius), and B (magnetic field)._These relations give the value of velocity of an electron as \(\left(v=\sqrt{2 v\left(\frac{e}{m}\right)}\right) \text { and }\left(v=B r\left(\frac{e}{m}\right)\right)\) respectively. It can be observed from these relations that the dynamics of an electron is determined not by e and m separately, but by the ratio e / m.

(c) At atmospheric pressure, the ions of gases have no chance of reaching their respective electrons because of collision and recombination with other gas molecules. Hence, gases are insulators at atmospheric pressure. At low pressure, ions have a chance of reaching their respective electrodes and constitute a current. Hence, they conduct electricity at these pressures.

(d) The work function of a metal is the minimum energy required for a conduction electron to get out of the metal surface. All the electrons in an atom do not have the same energy level. When a ray having some photon energy is incident on a metal surface, the electrons come out from different levels with different energies. Hence, these emitted electrons show different energy distributions.

(e) The absolute value of energy of a particle is arbitrary within the additive constant. Hence, wavelength (λ) is significant, but the frequency (v) associated with an electron has no direct physical significance. Therefore, the product vλ (phase speed) has no physical significance. Group speed is given as

This Quantity has a physical meaning.