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

NCERT
SOLUTIONS
CLASS - 12th

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Class : 12th
Subject : Physics
Chapter : 11
Chapter Name : Dual Nature Of Radiation And Matter

Q11.1 Find the (a) maximum frequency, and (b) minimum wavelength of X-rays produced by 30 kV
electrons

Answer. Potential of the electrons, v = 30kV = 3 × 10 4v
Hence, energy of the electrons, E = 3 × 10 4eV
Where, e = Charge on an electron = 1.6 × 10 − 19C
(a) Maximum frequency produced by the X-rays = v The energy of the electrons is given by the
relation:
E = hv
where,
h = Planck's constant = 6.626 × 10 − 34Js
E
∴v=
h
1.6 × 10 − 19 × 3 × 10 4
= = 7.24 × 10 18Hz
6.626 × 10 − 34
Hence, the maximum frequency of X-rays produced is 7.24 × 10 18Hz.
(b) The minimum wavelength produced by the X-rays is given as:
c
λ= v
3 × 10 8
= = 4.14 × 10 − 11m = 0.0414nm
7.24 × 10 18
Hence, the minimum wavelength of X-rays produced is 0.0414 nm.

Page : 407 , Block Name : Exercise

Q11.2 The work function of caesium metal is 2.14 eV. When light of frequency 6 × 10 14Hz is
incident on the metal surface, photo emission of electrons occurs. What is the (a) maximum
kinetic energy of the emitted electrons,(b)Stopping potential, and (c) maximum speed of the
emitted photo electrons?

Answer. Work function of caesium metal, ϕ 0 = 2.14eV
Frequency of light. v = 6.0 × 10 14Hz

(a) The maximum kinetic energy is given by the photoelectric effect as:

K = hv − ϕ 0
Where,
h = Planck's constant = 6.626 × 10 − 34 Js
6.626 × 10 34 × 6 × 10 14
∴K= − 2.14
1.6 × 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 V 0, we can write the equation for kinetic energy as:
K = eV 0
K
∴ V0 = e
0.345 × 1.6 × 10 − 19
= = 0.345V
1.6 × 10 − 19
Hence, the stopping potential of the material is 0.345 V.

(c) Maximum speed of the emitted photoelectrons = v
Hence, the relation for kinetic energy can be written as:
1
K = 2 mv 2
Where,
m = Mass of an electron = 9.1 × 10 − 31kg
2K
v2 = m
2 × 0.345 × 1.6 × 10 − 19
= = 0.1104 × 10 12
9.1 × 10 − 31

∴ v = 3.323 × 10 5m / s = 332.3km / s
Hence, the maximum speed of the emitted photoelectrons is 332.3 km/s.

Page : 407 , Block Name : Exercise

Q11.3 The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum
kinetic energy of photoelectrons emitted?

Answer. Photoelectric cut-off voltage, V 0 = 1.5V
The maximum kinetic energy of the emitted photoelectrons is given as:
K e = eV 0
Where,
e = Charge on an electron = 1.6 × 10 − 19C
∴ K e = 1.6 × 10 − 19 × 1.5

= 2.4 × 10 − 19J
Therefore, the maximum kinetic energy of the photoelectrons emitted in the given experiment is

2.4 × 10 − 19J .

Page : 407 , Block Name : Exercise

Q11.4 Monochromatic light of wavelength 632.8 nm 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 monochromatic light, λ = 632.8nm = 632.8 × 10 − 9m
Power emitted by the laser, P = 9.42mW = 9.42 × 10 − 3W
Planck's constant, h = 6.626 × 10 − 34Js
Speed of light, c = 3 × 10 8m / s
Mass of a hydrogen atom, m = 1.66 × 10 − 27kg

(a) The energy of each photon is given as:
hc
E= λ
6.626 × 10 − 34 × 3 × 10 8
= = 3.141 × 10 − 19J
632.8 × 10 − 9
The momentum of each photon is given as:
h
P= λ
6.626 × 10 − 34
= 632.8
= 1.047 × 10 − 27kgms − 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
P
∴n= E
9.42 × 10 − 3
= ≈ 3 × 10 16 photon/s
3.141 × 10 − 19

(c) Momentum of the hydrogen atom is the same as the momentum of the photon,

p = 1.047 × 10 − 27kgms − 1
Momentum is given as:
p = mv
Where,
v = speed of the hydrogen atom
p
∴v= m
p
∴v= m
1.047 × 10 − 27
= = 0.621m / s
1.66 × 10 − 27

Page : 407 , Block Name : Exercise

Q11.5 The energy ux of sunlight reaching the surface of the earth is1.388 × 10 3W / m 2. 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 nm.

Answer. Energy ux of sunlight reaching the surface of earth, Φ = 1.388 × 10 3W / m 2
Hence, power of sunlight per square metre, P = 1.388 × 10 3W
Speed of light, c = 3 × 10 8m / s
Planck's constant, h = 6.626 × 10 − 34Js
Average wavelength of photons present in sunlight, λ = 550nm
= = 550 × 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
P Pλ
∴n= =
E hc
1.388 × 10 3 × 550 × 10 − 9
= − 34 8 = 3.84 × 10 21 photons/m 2 / s
6.626 × 10 × 3 × 10
Therefore, every second, 3.84 × 10 21.photons are incident per square metre on earth.

Page : 407 , Block Name : Exercise

Q11.6 In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency
of incident light is found to be 4.12 × 10 − 15Vs. 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:
V
v
= 4.12 × 10 − 15Vs
V is related to frequency by the equation:
hv = eV
Where,

e = Charge on an electron = 1.6 × 10 − 19C
h = Planck's constant
V
∴h=e× v

= 1.6 × 10 − 19 × 4.12 × 10 − 15 = 6.592 × 10 − 34J S
Therefore, the value of Planck's constant is 6.592 × 10 − 34J S.

Page : 407 , Block Name : Exercise

Q11.7 A 100W 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, l = 589 nm = 589 × 10 − 9m
Planck's constant, h = 6.626 × 10 − 34js
Speed of light, c = 3 × 10 8m / s

(a) The energy per photon associated with the sodium light is given as:
hc
E= λ
6.626 × 10 − 34 × 3 × 10 8
= = 3.37 × 10 − 19J
589 × 10 − 9
3.37 × 10 − 19
= = 2.11eV
1.6 × 10 − 19

(b) Number of photons delivered to the sphere = n
The equation for power can be written as: P = nE
P
∴n=
E
100
= − 19
= 2.96 × 10 20 photons/s
3.37 × 10
Therefore, every second, 2.96 × 10 20photons are delivered to the sphere.

Page : 407 , Block Name : Exercise

Q11.8 The threshold frequency for a certain metal is 3.3 × 10 14Hz. If light of frequency 8.2 × 10 14 is
incident on the metal, predict the cutoff voltage for the photoelectric emission.

Answer. Threshold frequency of the metal, v 0 = 3.3 × 10 14Hz
Frequency of light incident on the metal. v = 8.2 × 10 14Hz
Charge on an electron, e = 1.6 × 10 − 19C
Planck's constant, h = 6.626 × 10 − 34Js

Cut-off voltage for the photoelectric emission from the metal = v 0
The equation for the cut-off energy is given as:

( )
eV 0 = h v − v 0

h (v − v 0 )
V0 =
e

(
6.626 × 10 − 34 × 8.2 × 10 14 − 3.3 × 10 14 )
= = 2.0292V
1.6 × 10 − 19
Therefore, the cut-off voltage for the photoelectric emission is 2.0292 V.

Page : 408 , Block Name : Exercise

Q11.9 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. No , Work function of the metal, ϕ 0 = 4.2eV
Charge on an electron, e = 1.6 × 10 − 19C
Planck's constant, h = 6.626 × 10 − 34Js
Wavelength of the incident radiation, λ = 330nm = 330 × 10 − 9m
Speed of light, c = 3 × 10 8m / s
The energy of the incident photon is given as:
hc
E=
λ
6.626 × 10 − 34 × 3 × 10 8
= = 6.0 × 10 − 19J
330 × 10 − 9
6.0 × 10 − 19
= 3.76eV
1.6 × 10 − 19
It can be observed that the energy of the incident radiation is less than the work function of the
metal. Hence, no photoelectric emission will take place.

Page : 408 , Block Name : Exercise

Q11.10 Light of frequency 7.21 × 10 14 is incident on a metal surface. Electrons with a maximum
speed of 6.0 × 10 5m / s are ejected from the surface. What is the threshold frequency for
photoemission of electrons?

Answer. Frequency of the incident photon, v = 488nm = 488 × 10 − 9m
Maximum speed of the electrons, v = 6.0 × 10 5m / s
Planck's constant, h = 6.626 × 10 − 34Js
Mass of an electron, m = 9.1 × 10 − 31kg
For threshold frequency v 0 the relation for kinetic energy is written as:

1
2 (
mv 2 = h v − v 0 )
mv 2
v 0 = v − 2h

14
( 9.1 × 10 ) × ( 6 × 10 )− 31 5 2

= 7.21 × 10 −
2 × ( 6.626 × 10 ) − 34

= 7.21 × 10 14 − 2.472 × 10 14
= 4.738 × 10 14Hz
Therefore, the threshold frequency for the photoemission of electrons is 4.738 × 10 14Hz.

Page : 408 , Block Name : Exercise

Q11.11 Light of wavelength 488 nm 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, 488 nm = 488 × 10 − 9m
Stopping potential of the photoelectrons, V 0 = 0.38 V
1eV = 1.6 × 10 − 19J
0.38
∴ V0 = eV
1.6 × 10 − 19
Planck's constant, h = 6.6 × 10 − 34Js
Charge on an electron, e = 1.6 × 10 − 19C
Speed of light, c = 3 × 10m / s
From Einstein's photoelectric effect, we have the relation involving the work function Φ 0 of the
material of the emitter as:
hc
eV 0 = λ − ϕ 0
hc
ϕ 0 = λ − eV 0
6.6 × 10 − 34 × 3 × 10 8 1.6 × 10 − 19 × 0.38
= −
1.6 × 10 − 34 × 488 × 10 − 9 1.6 × 10 − 19
= 2.54 − 0.38 = 2.16eV
Therefore, the material with which the emitter is made has the work function of 2.16 eV.

Page : 408 , Block Name : Exercise

Q11.12 Calculate the
(a) momentum, and
(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.

Answer. Potential difference, V = 56V

Planck’s constant, h = 6.6 × 10 − 34Js
Mass of an electron, m = 9.1 × 10 − 31kg
Charge on an electron, e = 1.6 × 10 − 19C

(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:
1
2
mv 2 = eV
2eV
v2 = m

2 × 1.6 × 10 − 19 × 56
∴v=
√ 9.1 × 10 − 31

= √19.69 × 10 12 = 4.44 × 10 6m / s
The momentum of each accelerated electron is given as:
p = mv
= 9.1 × 10 − 31 × 4.44 × 10 6
= 4.04 × 10 − 24kgms − 1
Therefore, the momentum of each electron is 4.04 × 10 − 24kgms − 1

(b) De Broglie wavelength Of an electron accelerating through a potential V, is given by the
relation:
12.27
λ= A
√V
12.27
= × 10 − 10m
√56
= 0.1639nm
Therefore, the de Broglie wavelength of each electron is 0.1639 nm.

Page : 408 , Block Name : Exercise

Q11.13 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 = 120eV
Planck's constant, h = 6.6 × 10 − 34Js
Mass of an electron, m = 9.1 × 10 − 31kg
Charge on an electron, e = 1.6 × 10 − 19C
(a) For the electron, we can write the relation for kinetic energy as:

1
E k = 2 mv 2

Where,
v = speed of the electron
2eE k
∴ v2 =
√ m

2 × 1.6 × 10 − 19 × 120
=
√ 9.1 × 10 − 31

= √42.198 × 10 12 = 6.496 × 10 6m / s
Momentum of the electron, p = mv
= 9.1 × 10 − 31 × 6.496 × 10 6
= 5.91 × 10 − 24kgms − 1
Therefore, the momentum of the electron is 5.91 × 10 − 24kgms − 1.

(b) Speed of the electron, v = 6.496 × 10 6m / s

(c) De Broglie wavelength of an electron having a momentum p, is given as:
h
λ=
p
6.6 × 10 − 34
= = 1.116 × 10 − 10m
5.91 × 10 − 34
= 0.112nm
Therefore, the de Broglie wavelength of the electron is 0.112 nm.

Page : 408 , Block Name : Exercise

Q11.14 The wavelength of light from the spectral emission line of sodium is 589 nm. Find the
kinetic energy at which
(a) an electron, and
(b) a neutron, would have the same de Broglie wavelength.

Answer. Wavelength of light of a sodium line, λ = 589nm = 589 × 10 − 9m
Mass of an electron, m e = 9.1 × 10 − 31kg
Mass of a neutron, m n = 1.66 × 10 − 27kg
Planck's constant, h = 6.6 × 10 − 34Js

(a) For the kinetic energy K, of an electron accelerating with a velocity v, we have the
relation :
1
K = 2 m ev 2 ….(1)
We have the relation for de Broglie wavelength as:

h
λ= mv
e
h2
2
∴v = 2 ….(2)
λ 2m e
Substituting equation (2) in equation (1), we get the relation:
2
1 m eh h2
K= 2 2 2 = …(3)
λ me 2λ 2m e

( 6.6 × 10 ) − 34 2

=
2 × ( 589 × 10 ) × 9.1 × 10
−9 2 − 31

≈ 6.9 × 10 − 25J
6.9 × 10 − 25
= = 4.31 × 10 − 6eV = 4.31μeV
1.6 × 10 − 19
Hence, the kinetic energy of the electron is 6.9 × 10 − 25J or 4.31 μ eV.

(b) Using equation (3), we can write the relation for the kinetic energy of the neutron as:
h2
2λ 2m n

( 6.6 × 10 ) −3 2

=
2 × ( 589 × 10 ) × 1.66 × 10
−9 2 − 27

= 3.78 × 10 − 28J
3.78 × 10 − 28
= = 2.36 × 10 − 9eV = 2.36neV
1.6 × 10 − 19
Hence, the kinetic energy of the neutron is 3.78 × 10 − 28J or 2.36nev.

Page : 408 , Block Name : Exercise

Q11.15 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 × 10 − 3 kg drifting with a speed of 2.2 m/s?

Answer. (a)
Mass of the bullet, m = 0.040kg
speed of the bullet, v = 1.0km / s = 1000m / s
Planck's constant, h = 6.6 × 10 − 34Js
De Broglie wavelength of the bullet is given by the relation:
h
λ = mv
6.6 × 10 − 34
= 0.040 × 1000 = 1.65 × 10 − 35m

(b)
Mass of the ball, m = 0.060kg
speed of the ball, v = 1.0m / s
De Broglie wavelength of the ball is given by the relation:
h
λ = mv
6.6 × 10 − 34
= 0.060 × 1 = 1.1 × 10 − 32m

(c)
Mass of the dust particle, m = 1 × 10 − 9kg
speed of the dust particle, v = 2.2m / s
De Broglie wavelength of the dust particle is given by the relation:
h
λ = mv
6.6 × 10 − 34
= = 3.0 × 10 − 25m
2.2 × 1 × 10 − 9

Page : 408 , Block Name : Exercise

Q11.16 An electron and a photon each have a wavelength of 1.00 nm. Find
(a) their momenta,
(b) the energy of the photon, and
(c) the kinetic energy of electron.

( ) ( )
Answer. Wavelength of an electron λ e and a photon λ p , λ v = λ p = λ = 1nm
= 1 × 10 − 9m
Planck's constant, h = 6.63 × 10 − 34Js

(a) The momentum of an elementary particle is given by de Broglie relation:
h
λ= p
h
p= λ
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.
6.63 × 10 − 34
∴p= = 6.63 × 10 − 25kgms − 1
1 × 10 − 9

(b) The energy of a photon is given by the relation:
hc
E= λ
where,
Speed of light, c = 3 × 10 8m / s

6.63 × 10 − 34 × 3 × 10 8
∴E=
1 × 10 − 9 × 1.6 × 10 − 19
= 1243.1eV = 1.243keV
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:
1 p2
K= 2m
Where,
m = Mass of the electron = 9.1 × 10 − 31kg
p = 6.63 × 10 − 25kgms − 1

1 ( ) − 25 2
6.63 × 10
∴K= × = 2.415 × 10 − 19J
2 9.1 × 10 − 31
2.415 × 10 − 19
= = 1.51eV
1.6 × 10 − 19
Hence, the kinetic energy of the electron is 1.51 eV.

Page : 408 , Block Name : Exercise

Q11.17 (a) For what kinetic energy of a neutron will the associated de Broglie wavelength be
1.40 × 10 − 10m?
(b) Also nd the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having
an average kinetic energy of (3/2) k T at 300 K.

Answer. (a) De Broglie wavelength of the neutron, λ = 1.40 × 10 − 10m
Mass of a neutron, m n = 1.66 × 10 − 27kg
Planck's constant,h = 6.6 × 10 − 34Js
Kinetic energy (K) and velocity (v) are related as:
1
K = 2 m nv 2 ….(1)
De Broglie wavelength () and velocity (v) are related as:
h
λ = m v ….(2)
n
Using equation (2) in equation (1), we get:
2
1 m nh h2
K= =
2 λ 2m 2 2λ 2m n
n

(6.63 × 10 ) − 34 2

= = 6.75 × 10 − 21J
2 × (1.40 × 10 ) × 1.66 × 10
− 10 2 − 27

6.75 × 10 − 21
= = 4.219 × 10 − 2eV
1.6 × 10 − 19

Hence, the kinetic energy of the neutron is 6.75 × 10 − 21J or 4.219 × 10 − 2eV.

(b) Temperature of the neutron, T = 300 K
Boltzmann constant, k = 1.38 × 10 − 23kgm 2s − 2K − 1
Average kinetic energy of the neutron:
h
λ′ =
√2K m
′
n

Where,
m n = 1.66 × 10 − 27kg

h = 6.6 × 10 − 34Js
K = 6.75 × 10 − 21J
6.63 × 10 − 34
∴ λ′ = = 1.46 × 10 − 10m = 0.146nm
√ 2 × 6.21 × 10 − 21 × 1.66 × 10 − 27

Therefore, the de Broglie wavelength of the neutron is 0.146 nm.

Page : 408 , Block Name : Exercise

Q11.18 Show that the wavelength of 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:
hv h
p= c = λ
h
λ = p …(i)
h
λ = mv

But p = mv
h
∴ λ = p ..(ii)
Where,
m = Mass of the photon
v = Velocity of the photon
Hence, it can be inferred from equations (i) and (ii) that the wavelength of the
electromagnetic radiation is equal to the de Broglie wavelength of the photon.

Page : 408 , Block Name : Exercise

Q11.19 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 = 14.0076 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 × 10 − 27kg
∴ m = 28.0152 × 1.66 × 10 − 27kg
Planck's constant, h = 6.63 × 10 − 34Js
Boltzmann constant, k = 1.38 × 10 − 23JK − 1

( )
3
We have the expression that relates mean kinetic energy 2 kT of the nitrogen molecule with the

root mean square speed v rms as : ( )
1 3
2
mv 2rms = 2 kT

3kT
v rms =
√ m

Hence, the de Broglie wavelength Of the nitrogen molecule is given as:
h h
λ= =
mv rms √3mkT
6.63 × 10 − 34
=
√3 × 28.0152 × 1.66 × 10 − 27 × 1.38 × 10 − 23 × 300
= 0.028 × 10 − 9m
= 0.028nm
Therefore, the de Broglie wavelength of the nitrogen molecule is 0.028 nm.

Page : 408 , Block Name : Exercise

(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 modi ed?
Q11.20 (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 speci c charge of the electron, i.e., its
e/m is given to be 1.76 × 10 11Ckg − 1.

Answer. (a) Potential difference across the evacuated tube, V = 500 V
Speci c charge of an electron, e/m = 1.76 × 10 11Ckg − 1
The speed of each emitted electron is given by the relation for kinetic energy as:
1
KE = mv 2 = eV
2
1 1

∴v=
( ) (
2eV
m
2
= 2V ×
e
m ) 2

1

(
= 2 × 500 × 1.76 × 10 11
) = 1.327 × 10 m / s
2 7

Therefore, the speed of each emitted electron is 1.327 × 10 7m / s

(b) Potential of the anode, V = V = 10MV = 10 × 10 6V
The speed of each electron is given as:
1

( )
v = 2V
e
m
2

1

( 7
= 2 × 10 × 1.76 × 10 11
) 2

= 1.88 × 10 9m / s

This result is wrong because nothing can move faster than light. In the above formula,

( )
the expression mv 2 / 2 for energy can only be used in the non-relativistic limit, i.e., for v < 2
For very high speed problems, relativistic equations must be considered for solving them. In the
relativistic limit, the total energy is given as:
E = mc 2
Where,
m = Relativistic mass

( )
1
v2 2
= m0 1 −
c2

m 0 = Mass of the particle at rest
Kinetic energy is given as:
K = mc 2 − m 0c 2

Page : 409 , Block Name : Additional Exercise

Q11.21 (a) A monoenergetic electron beam with electron speed of 5.20 × 10 6ms − 1 is subject to a
magnetic eld of 1.30 × 10 − 4T normal to the beam velocity. What is the radius of the circle traced
by the beam, given e/m for electron equals 1.76 × 10 11Ckg − 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 modi ed?

Answer. Speed of an electron, v = 5.20 × 10 6m / s
Magnetic eld experienced by the electron , B = 1.30 × 10 − 4T
Speci c charge of an electron, e / m = 1.76 × 10 11Ckg − 1
Where,
e = Charge on the electron = 1.6 × 10 − 19C
m = Mass of the electron = 9.1 × 10 − 31kg − 1
The force exerted on the electron is given as:
→
F = e|→
v × 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 ..(i)
The beam traces a circular path of radius, r. It is the magnetic eld, due to its bending

nature, that provides the centripetal force F = r
( )
mv 2
for the beam.

Hence, equation (1) reduces to:
mv 2
evB = r
mv v
∴ r = eB =

( ) e
m
B

5.20 × 10 6
= = 0.227m = 22.7cm
( )
1.76 × 10 11 × 1.30 × 10 − 4

Therefore the radius of the circular path is 22.7 cm.

(b) Energy of the electron beam, E = 20MeV = 20 × 10 6 × 1.6 × 10 − 19J
The energy of the electron is given as:
1
E = 2 mv 2
1

∴v=
( )
2E
m
2

2 × 20 × 10 6 × 1.6 × 10 − 19
=
√ 9.1 × 10 − 31
= 2.652 × 10 9m / s

This result is incorrect because nothing can move faster than light. In the above formula,the

( )
expression mv 2 / 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:

[ ]
1
v2 2
m = m0 1 −
c2

Where,
m 0 = Mass of the particle at rest
Hence, the radius of the circular path is given as:
r = mv / eB
m 0v
=
c2 − v2

√
eB
c2

Page : 409 , Block Name : Additional Exercise

Q11.22 An electron gun with its collector at a potential of 100 V res out electrons in a spherical
bulb containing hydrogen gas at low pressure (∼10 − 2 of Hg). A magnetic eld of 2.83 × 10 − 4T
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 ‘ ne beam tube’ method.) Determine e/m from the
data.

Answer. Potential of an anode, V 100 V
Magnetic eld experienced by the electrons, B = 2.83 × 10 − 4T
Radius of the circular orbit r = 12.0 cm = 12.0 × 10 − 2m
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.,
1
2
mv 2 = eV
2eV
v 2 = m … (1)

( ) mv 2
It is the magnetic eld, due to its bending nature, that provides the centripetal force F = r for

the beam. Hence, we can write:
Centripetal force = Magnetic force
mv 2
r
= evB
mv
eB = r
eBr
v = m …(2)

Putting the value of v in equation (1), we get:
2eV e 2B 2r 2
m
=
m2
e 2V
m
=
B 2r 2
2 × 100
= = 1.73 × 10 11Ckg − 1
( 2.83 × 10 ) × ( 12 × 10 )
−4 2 −2 2

11 −1
Therefore, the specific charge ratio (e / m) is 1.73 × 10 Ckg

Page : 409 , Block Name : Additional Exercise

Q11.23 (a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength
end at 0.45 Å. 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. Wavelength produced by an X-ray tube, λ = 0.45A = 0.45 × 10 − 10m
Planck’s constant, h = 6.626 × 10 − 34Js
Speed of light, c = 3 × 10 8m / s
The maximum energy of a photon is given as:
hc
E=
λ
6.626 × 10 − 34 × 3 × 10 8
=
0.45 × 10 − 10 × 1.6 × 10 19
= 27.6 × 10 3cV = 27.6keV
Therefore, the maximum energy of an 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.

Page : 409 , Block Name : Additional Exercise

Q11.24 In an accelerator experiment on high-energy collisions of electrons with positrons, a
certain event is interpreted as annihilation of an electron-positron pair of total energy 10.2 BeV
into two γ-rays of equal energy. What is the wavelength associated with each γ-ray? (1BeV = 10 9
eV)

Answer.

Total energy of two y-rays:
E = 10.2BeV
= 10.2 × 10 9eV
= 10.2 × 10 9 × 1.6 × 10 − 10J
Hence, the energy of each y-ray:
E
E′ = 2
10.2 × 1.6 × 10 − 10
= 2
= 8.16 × 10 − 10J

Planck's constant, h = 6.626 × 10 − 34Js
speed of light, c = 3 × 10 sm / s
Energy is related to wavelength as:
hc
E′ =
λ
hc
∴λ= ′
E
6.626 × 10 − 34 × 3 × 10 8
= = 2.436 × 10 − 16m
8.16 × 10 − 10
Therefore, the wavelength associated with each γ- ray is 2.436 × 10 − 16m

Page : 409 , Block Name : Additional Exercise

Q11.25 Estimating the following two numbers should be interesting. The rst 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 Medium wave 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 − 10Wm − 2 ). Take the area of
the pupil to be about 0.4cm 2 , and the average frequency of white light to be about 6 × 10 14Hz.

Answer. (a) power of the medium wave transmitter, P = 10kW = 10 4w = 10 4J / s
Hence, energy emitted by the transmitter per second, E = 10 4
Wavelength of the radio wave, λ = 500m
The energy of the wave is given as:
hc
E1 = λ
Where,
h = Planck's constant = 6.6 × 10 − 34Js
c = speed of light = 3 × 10 sm / s
6.6 × 10 − 34 × 3 × 10 8
∴ E1 = 500
= 3.96 × 10 − 28J
Let n be the number of photons emitted by the transmitter.

∴ nE 1 = E
E
n= E
1
10 4
= = 2.525 × 10 31
3.96 × 10 − 28

≈ 3 × 10 31

( )
The energy E 1 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 − 10Wm − 2
Area of a pupil, A = 0.4cm 2 = 0.4 × 10 − 4m 2
Frequency of white light, V = 6 × 10 14Hz
The energy emitted by a photon is given as:
E = hv
Where,
h = Planck's constant = 6.6 × 10 − 34 js
∴ E = 6.6 × 10 − 34 × 6 × 10 14
= 3.96 × 10 − 19J
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 × 3.96 × 10 − 19Js − 1m − 2
The energy per unit area per second is the intensity of light.
∴E=I
n × 3.96 × 10 − 19 = 10 − 10
10 − 10
n=
3.96 × 10 − 19

= 2.52 × 10 8m 2s − 1
The total number of photons entering the pupil per second is given as:
nA = n × A

= 2.52 × 10 8 × 0.4 × 10 − 4
= 1.008 × 10 4s − 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.

Page : 409 , Block Name : Additional Exercise

Q11.26 Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photocell
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 (∼10 5Wm − 2) red light of wavelength
6328 Å produced by a He-Ne laser ?

Answer. Wavelength of ultraviolet light, A — 2271 Å
Stopping potential of the metal, Vo = 1.3 V
Planck's constant, h = 6.6 × 10 − 343
Charge on an electron, e = 1.6 × 10 − 19C
Work function of the metal = ϕ 0
Frequency of light = v
We have the photo-energy relation from the photoelectric effect as:
ϕ 0 = hv − eV 0
hc
= λ − eV 0
6.6 × 10 − 34 × 10 8
= − 1.6 × 10 − 19 × 1.3
2271 × 10 − 10

= 8.72 × 10 − 19 − 2.08 × 10 − 19
= 6.64 × 10 − 19J
6.64 × 10 − 19
= = 4.15eV
1.6 × 10 − 19
Let ϕ 0 be the threshold of the metal.
∴ ϕ 0 = hv 0
ϕ0
v0 =
h
6.64 × 10 − 19
= = 1.006 × 10 15Hz
6.6 × 10 − 34
Wavelength of red light, λ r = 6328A = 6328 × 10 − 10m

c
∴ Frequency of red light, v r = λ
r

∴ Frequency of red light,
3 × 10 8
= = 4.74 × 10 14Hz
6328 × 10 − 10
Since v 0 > v r, the photocell will not respond to the red light produced by the laser.

Page : 410 , Block Name : Additional Exercise

Q11.27 Monochromatic radiation of wavelength 640.2 nm (1nm = 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
photocell. Predict the new stopping voltage.

Answer. Wavelength of the monochromatic radiation, λ = 640.2nm = 640.2 × 10 − 9m
Stopping potential of the neon lamp, V 0 = 0.54V
Charge on an electron, e = 1.6 × 10 − 19C

Planck's constant, h = 6.6 × 10 − 34Js
Let ϕ 0 be the work function and v be the frequency of emitted light.
We have the photo-energy relation from the photoelectric effect as:
eV 0 = hv − ϕ 0
hc
ϕ0 = − eV 0
λ
6.6 × 10 − 34 × 3 × 10 8
= − 1.6 × 10 − 19 × 0.54
640.2 × 10 − 9
= 3.093 × 10 − 19 − 0.864 × 10 − 19
= 2.229 × 10 − 19J
2.229 × 10 − 19
= = 1.39eV
1.6 × 10 − 19
Wavelength of the radiation emitted from an iron source, λ ′ = 427.2nm = 427.2 × 10 − 9m
Let V ′0 be the new stopping potential. Hence, photo-energy is given as:
hc
eV ′0 = − ϕ0
λ′
6.6 × 10 − 34 × 3 × 10 8
= − 2.229 × 10 − 19
427.2 × 10 − 9
= 4.63 × 10 − 19 − 2.229 × 10 − 19
= 2.401 × 10 − 19J
2.401 × 10 − 19
= = 1.5eV
1.6 × 10 − 19
Hence, the new stopping potential is 1.50 eV.

Page : 410 , Block Name : Additional Exercise

Q11.28 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 photocell, the following lines from a mercury source
were used:
λ 1 = 3650 Å
λ 2 = 4047 Å
λ 3 = 4358 Å
λ 4 = 5461 Å
λ 5 = 6907 Å
The stopping voltages respectively were measured to be:
V 01 = 1.28V, V 02 = 0.95V, V 03 = 0.74V, V 04 = 0.16V, V 05 = 0V-
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 × 10 − 19C ). 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) con rmed Einstein’s photoelectric

equation and at the same time gave an independent estimate of the value of h.]

Answer. Einstein's photoelectric equation is given as:
eV 0 = hv − ϕ 0
h ϕ0
V 0 = e v − e …(1)
Where,
V 0 = Stopping potential
h = Planck's constant
e = Charge on an electron
v = Frequency of radiation
ϕ 0 = Work function of a material
It can be concluded from equation (1) that potential Vo is directly proportional to frequency v.
Frequency is also given by the relation:
Speed of light ( c )
v= Wavelength ( λ )
This relation can be used to obtain the frequencies of the various lines of the given wavelengths.
c 3 × 10 8
v1 = λ = = 8.219 × 10 14Hz
1 3650 × 10 − 10
c 3 × 10 8
v2 = λ = = 7.412 × 10 14Hz
2 4047 × 10 − 10
c 3 × 10 s
v3 = λ = = 6.884 × 10 14Hz
3 4358 × 10 − 10
c 3 × 10 8
v4 = λ = = 5.493 × 10 14Hz
4 5461 × 10 − 10
c 3 × 10 8
v5 = λ = = 4.343 × 10 14Hz
5 6907 × 10 − 10
The given quantities can be listed in tabular form as:

The following gure shows a graph between v and V 0.

It can be observed that the obtained curve is a straight line. It intersects the v-axis at 5 × 10 14Hz,
which is the threshold frequency V 0 of the material. Point D corresponds to a frequency less than

the threshold frequency. Hence, there is no photoelectric emission for the line λ 5 and therefore, no
stopping voltage [s required to stop the current.
AB 1.28 − 0.16
Slope of the current line = CB =
( 8.214 − 5.493 ) × 10 14
From equation (1) the slope h/e can be written as :
h 1.28 − 0.16
=
e (8.214 − 5.493) × 10 14
1.12 × 1.6 × 10 − 19
∴h=
2.726 × 10 14
= 6.573 × 10 − 34Js
The work function of the metal is given as:
ϕ 0 = hv 0
= 6.573 × 10 − 34 × 5 × 10 14
= 3.286 × 10 − 19J
3.286 × 10 − 19
= = 2.054eV
1.6 × 10 − 18

Page : 410 , Block Name : Additional Exercise

Q11.29 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 photocell? 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 a radiation, λ = 3300 Å = 3300 × 10 − 10m
Speed of light, c = 3 × 10 8m / s
Planck's constant, h = 6.6 × 10 − 34js
The energy of incident radiation is given as:
hc
E=
λ
6.6 × 10 − 34 × 3 × 10 8
= = 6 × 10 − 19J
3300 × 10 − 10
6 × 10 − 19
= = 3.158eV
1.6 × 10 − 19
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.

Page : 410 , Block Name : Additional Exercise

Q11.30 Light of intensity 10 − 5Wm − 2 falls on a sodium photocell of surface area 2cm 2 . 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.
Intensity of incident light, I = 10 − 5Wm − 2
surface area of a sodium photocell, A = 2cm 2 = 2 × 10 − 4m 2
Incident power of the light, P = I × A
= 10 − 5 × 2 × 10 − 4
= 2 × 10 − 9W
Work function of the metal, ϕ 0 = 2eV

= 2 × 1.6 × 10 − 19
= 3.2 × 10 − 19J
Number of layers of sodium that absorbs the incident energy, n = 5
We know that the effective atomic area of a sodium atom, A e is 10 − 20m 2
Hence, the number of conduction electrons in n layers is given as:
A
n′ = n ×
Ae
2 × 10 − 4
=5× = 10 17
10 − 20
The incident power is uniformly absorbed by all the electrons continuously. Hence, the
amount of energy absorbed per second per electron is:
P
E= ′
n
2 × 10 − 9
= = 2 × 10 − 26J / s
10 17
Time required for photoelectric emission:
ϕ0
t=
E
3.2 × 10 − 19
= = 1.6 × 10 7s ≈ 0.507 years
2 × 10 − 19
The time required for the photoelectric emission is nearly half a year, which is not practical.
Hence, the wave picture is in disagreement with the given experiment.

Page : 410 , Block Name : Additional Exercise

Q11.31 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 × 10 − 31kg . )
Answer. An X-ray probe has a greater energy than an electron probe for the same wavelength.
Wavelength of light emitted from the probe, A = 1 Å = 10 − 10m
Mass of an electron, m e = 9.11 × 10 − 31kg
Planck's constant, h = 6.6 × 10 − 34Js
Charge on an electron, e = 1.6 × 10 − 19C
The kinetic energy Of the electron is given as:
1
E = 2 m ev 2

m ev =
√2Eme
Where,
v = Velocity of the electron
m eV = Momentum (p) of the electron
According to the de Broglie principle, the de Broglie wavelength is given as:
h h h
λ= = =
p m ev 2Em
√ e

h2
∴E=
2λ 2m e

(6.6 × 10 ) − 34 2

= = 2.39 × 10 − 17J
2 × (10 ) × 9.11 × 10
− 10 2 − 31

2.39 × 10 − 17
= = 149.375eV
1.6 × 10 − 19
hc
Energy of a photon, E ′ = λe eV
6.6 × 10 − 34 × 3 × 10 8
=
10 − 10 × 1.6 × 10 − 19

= 12.375 × 10 3eV = 12.375keV
Hence, a photon has a greater energy than an electron for the same wavelength.

Page : 410 , Block Name : Additional Exercise

Q11.32 (a) Obtain the de Broglie wavelength of a neutron of kinetic energy 150 eV. As you have
seen in Q31, 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 n = 1.675 × 10 − 27kg .
(b) Obtain the de Broglie wavelength associated with thermal neutrons at room temperature (27
°C). 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 Broglie wavelength = 2.327 × 10 − 12m; neutron is not suitable for the diffraction
experiment
Kinetic energy of the neutron, K = 150 eV
= 150 × 1.6 × 10 − 19
= 2.4 × 10 − 17J
Mass of a neutron, m n = 1.675 × 10 − 27kg
The kinetic energy of the neutron is given by the relation:
1
K = 2 m nv 2

m nv =
√2Kmn
Where,
v = Velocity of the neutron
m nv = Momentum of the neutron
De-Broglie wavelength of the neutron is given as:
h h
λ= mv =
e √2Kmn
It is clear that wavelength is inversely proportional to the square root of mass, Hence, wavelength
decreases with increase in mass and vice versa.
6.6 × 10 − 44
∴λ=
2 × 2.4 × 10 − 17 × 1.675 × 10 − 27
√
= 2.327 × 10 − 12m
It is given in the previous problem that the inter-atomic spacing of a crystal is about 1 Å, i.e.,
10 − 10m. Hence, the inter-atomic spacing is about a hundred times greater.
Hence,a neutron beam of energy 150 eV is not suitable for diffraction experiments.

(b) De Broglie wavelength = 1.447 × 10 − 10m
Room temperature, T = 27 ∘ C = 27 + 273 = 300K
The average kinetic energy of the neutron is given as:
3
E = 2 kT
Where,
k = Boltzmann constant = 1.38 × 10 − 23Jmol − 1K − 1
The wavelength of the neutron is given as:
h h
λ= =
2m nE 3m nkT
√ √
6.6 × 10 − 34
=
√3 × 1.675 × 10 − 27 × 1.38 × 10 − 23 × 300
= 1.447 × 10 − 10m
This wavelength is comparable to the inter-atomic spacing of a crystal. Hence, the high-

energy neutron beam should rst be thermalised, before using it for diffraction.

Page : 410 , Block Name : Additional Exercise

Q11.33 An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de
Broglie 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 × 10 3V
Charge on an electron, e = 1.6 × 10 − 19C
Mass of an electron, m e = 9.11 × 10 − 31kg
Wavelength of yellow light = 5.9 × 10 − 7m
The kinetic energy of the electron is given as:
E = eV
= 1.6 × 10 − 19 × 50 × 10 3
= 8 × 10 − 15J
De Broglie is given by the relation:
h
λ=
2m eE
√
6.6 × 10 − 34
=
√2 × 9.11 × 10 − 31 × 8 × 10 − 15
= 5.467 × 10 − 12m
This wavelength is nearly 10 5 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.

Page : 411 , Block Name : Additional Exercise

Q11.34 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 or less. This structure was rst probed in early 1970’s 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 − 15m Rest mass energy of an electron:
moc 2 = 0.511MeV
= 0.511 × 10 6 × 1.6 × 10 − 19
= 0.8176 × 10 − 13J
Planck's constant, h = 6.6 × 10 − 34Js
Speed of light, c = 3 × 10 8m / s
The momentum of a proton and neutron is given as:
h
p=
λ
6.6 × 10 − 34
= = 6.6 × 10 − 19kgm / s
10 − 19
The relativistic relation for energy (E) is given as:
2
E 2 = p 1c 2 + m 0 c 4

(
= 6.6 × 10 − 19 × 3 × 10 6 ) + (0.8176 × 10 )
2 − 13 2

= 392.04 × 10 − 21 + 0.6685 × 10 − 38
= 392.04 × 10 − 22
∴ E = 1.98 × 10 − 10J
1.98 × 10 − 10
=
1.6 × 10 − 19
= 1.24 × 10 9cV = 1.24BcV
Thus, the electron energy emitted from the accelerator at Stanford, USA might be of the order of
1.24 BeV.

Page : 411 , Block Name : Additional Exercise

Q11.35 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.
Wavelength of a proton or a neutron, λ ≈ 10 − 15m
Rest mass energy of an electron:
m 0c 2 = 0.511MeV

= 0.511 × 10 6 × 1.6 × 10 − 19
= 0.8176 × 10 − 13J
Planck's constant, h = 6.6 × 10 − 34Js
speed of light, c = 3 × 10 8m / s
The wavelength or proton or a neutron is given as :

h
p=
λ
6.6 × 10 − 34
= = 6.6 × 10 − 19kgm / s
10 − 15
The relativistic relation for energy ( E ) is given as :
2
E 2 = p 2c 2 + m 0 c 4

(
= 6.6 × 10 − 19 × 3 × 10 8 ) + (0.8176 × 10 )
2 − 13 2

= 392.04 × 10 − 22 + 0.6685 × 10 − 26
≈ 392.04 × 10 − 22
∴ E = 1.98 × 10 − 10J
1.98 × 10 − 10
=
1.6 × 10 − 19
= 1.24 × 10 9eV = 1.24BeV
Thus, the electron energy emitted from the accelerator at Stanford, tJSA might be of the order of
1.24 BeV.

Page : 411 , Block Name : Additional Exercise

Q11.36 Compute the typical de Broglie wavelength of an electron in a metal at 27 °C and compare
it with the mean separation between two electrons in a metal which is given to be about
2 × 10 − 10m.
[Note: Exercises Q35 and Q36 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. De Broglie wavelength associated with He atom = 0.7268 × 10 − 10m
Room temperature, T = 27 ∘ C = 27 + 273 = 300k
Atmospheric pressure, P = 1 atm = 1.01 × 10 5Pa
Atomic weight of a He atom = 4
Avogadro's number, N A = 6.023 × 10 23
Boltzmann constant, k = 1.38 × 10 − 23Jmol − 1K − 1
Average energy of a gas at temperature T, is given as:
3
E = 2 kT
De Broglie wavelength is given by the relation:

h
λ=
√2mE
Where,
m = Mass of a He atom
Atomic weight
= NA
4
=
6.023 × 10 23

= 6.64 × 10 − 24g = 6.64 × 10 − 27kg
h
∴λ=
√3mkT
6.6 × 10 − 34
=
√3 × 6.64 × 10 − 27 × 1.38 × 10 − 23 × 300
= 0.7268 × 10 − 10m
We have the ideal gas formula:
PV = RT
PV = kNT
V kT
N
= P
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:
1 1

r= () ( )
V
N
3
=
kT
P
3

[ ]
1
1.38 × 10 − 23 × 300 3
=
1.01 × 10 5

= 3.35 × 10 − 9m
Hence, the mean separation between the atoms is much greater than the de Broglie wavelength.

Page : 411 , Block Name : Additional Exercise

Q11.37 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 de nite 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
h
associated matter wave by the relations: E = hv, p = λ
But while the value of λ is physically signi cant, the value of ν (and therefore, the value of the
phase speed ν λ) has no physical signi cance. 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 still the integral multiple of an electrical charge.

(b) The basic relations for electric eld and magnetic eld are

( 1
) (
eV = 2 mv 2 and eBv = r
mv 2
)
These relations include e (electric charge), v (velocity), m (mass), V (potential), r (radius), and B

( √ ( ))
(magnetic eld). These relations give the value of velocity of an electron as v = 2V m
e
and

( ( ))
v = Br m
e
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 pressures, 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 signi cant, but the frequency (v) associated with an electron has no
direct physical signi cance.
Therefore, the product vA(phase speed)has no physical signi cance.
Group speed is given as:

dv
vG =
dk

=
dv
=
dE
=
()
p2
d 2m

=
p
dp dp m
()
d λ
1

This quantity has a physical meaning.

Page : 411 , Block Name : Additional Exercise