Page No 408: - Chapter 11 Dual Nature Of Radiation & Matter Exercise Solutions class 12 ncert solutions Physics - SaraNextGen [2024-2025]
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On April 24, 2024, 11:35 AM
Question 11.8:
The threshold frequency for a certain metal is 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hz is incident on the metal, predict the cutoff voltage for the photoelectric emission.
Answer:
Threshold frequency of the metal, 
Frequency of light incident on the metal, 
Charge on an electron, e = 1.6 × 10−19 C
Planck’s constant, h = 6.626 × 10−34 Js
Cut-off voltage for the photoelectric emission from the metal = 
The equation for the cut-off energy is given as:
Therefore, the cut-off voltage for the photoelectric emission is 
Question 11.9:
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, 
Charge on an electron, e = 1.6 × 10−19 C
Planck’s constant, h = 6.626 × 10−34 Js
Wavelength of the incident radiation, λ = 330 nm = 330 × 10−9 m
Speed of light, c = 3 × 108 m/s
The energy of the incident photon is given as:
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.
Question 11.10:
Light of frequency 7.21 × 1014 Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 105 m/s are ejected from the surface. What is the threshold frequency for photoemission of electrons?
Answer:
Frequency of the incident photon, 
Maximum speed of the electrons, v = 6.0 × 105 m/s
Planck’s constant, h = 6.626 × 10−34 Js
Mass of an electron, m = 9.1 × 10−31 kg
For threshold frequency ν0, the relation for kinetic energy is written as:
Therefore, the threshold frequency for the photoemission of electrons is 
Question 11.11:
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, λ = 488 nm
= 488 × 10−9 m
Stopping potential of the photoelectrons, V0 = 0.38 V
1eV = 1.6 × 10−19 J
∴ V0 = 
Planck’s constant, h = 6.6 × 10−34 Js
Charge on an electron, e = 1.6 × 10−19 C
Speed of light, c = 3 × 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 11.12:
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 = 6.6 × 10−34 Js
Mass of an electron, m = 9.1 × 10−31 kg
Charge on an electron, e = 1.6 × 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 × 10−31 × 4.44 × 106
= 4.04 × 10−24 kg m s−1
Therefore, the momentum of each electron is 4.04 × 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 11.13:
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, Ek = 120 eV
Planck’s constant, h = 6.6 × 10−34 Js
Mass of an electron, m = 9.1 × 10−31 kg
Charge on an electron, e = 1.6 × 10−19 C
(a) For the electron, we can write the relation for kinetic energy as:
Where,
v = Speed of the electron
Momentum of the electron, p = mv
= 9.1 × 10−31 × 6.496 × 106
= 5.91 × 10−24 kg m s−1
Therefore, the momentum of the electron is 5.91 × 10−24 kg m s−1.
(b) Speed of the electron, v = 6.496 × 106 m/s
(c) De Broglie wavelength of an electron having a momentum p, is given as:
Therefore, the de Broglie wavelength of the electron is 0.112 nm.
Question 11.14:
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 = 589 nm = 589 × 10−9 m
Mass of an electron, me= 9.1 × 10−31 kg
Mass of a neutron, mn= 1.66 × 10−27 kg
Planck’s constant, h = 6.6 × 10−34 Js
(a) For the kinetic energy K, of an electron accelerating with a velocity v, we have the relation:
We have the relation for de Broglie wavelength as:
Substituting equation (2) in equation (1), we get the relation:
Hence, the kinetic energy of the electron is 6.9 × 10−25 J or 4.31 μeV.
(b) Using equation (3), we can write the relation for the kinetic energy of the neutron as:
Hence, the kinetic energy of the neutron is 3.78 × 10−28 J or 2.36 neV.
Question 11.15:
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 × 10−9 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 = 6.6 × 10−34 Js
De Broglie wavelength of the bullet is given by the relation:
(b) Mass of the ball, m = 0.060 kg
Speed of the ball, v = 1.0 m/s
De Broglie wavelength of the ball is given by the relation:
(c)Mass of the dust particle, m = 1 × 10−9 kg
Speed of the dust particle, v = 2.2 m/s
De Broglie wavelength of the dust particle is given by the relation:
Question 11.16:
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 
= 1 × 10−9 m
Planck’s constant, h = 6.63 × 10−34 Js
(a) The momentum of an elementary particle is given by de Broglie relation:
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.
(b) The energy of a photon is given by the relation:
Where,
Speed of light, c = 3 × 108 m/s
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:
Where,
m = Mass of the electron = 9.1 × 10−31 kg
p = 6.63 × 10−25 kg m s−1
Hence, the kinetic energy of the electron is 1.51 eV.
Question 11.17:
(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 × 10−10 m?
(b) Also find the de Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of (3/2) kT at 300 K.
Answer:
(a) De Broglie wavelength of the neutron, λ = 1.40 × 10−10 m
Mass of a neutron, mn = 1.66 × 10−27 kg
Planck’s constant, h = 6.6 × 10−34 Js
Kinetic energy (K) and velocity (v) are related as:
… (1)
De Broglie wavelength (λ) and velocity (v) are related as:
Using equation (2) in equation (1), we get:
Hence, the kinetic energy of the neutron is 6.75 × 10−21 J or 4.219 × 10−2 eV.
(b) Temperature of the neutron, T = 300 K
Boltzmann constant, k = 1.38 × 10−23 kg m2 s−2 K−1
Average kinetic energy of the neutron:
The relation for the de Broglie wavelength is given as:
Therefore, the de Broglie wavelength of the neutron is 0.146 nm.
Question 11.18:
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 (hν) is given as:
Where,
λ = Wavelength of the electromagnetic radiation
c = Speed of light
h = Planck’s constant
De Broglie wavelength of the photon is given as:
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.
Question 11.19:
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 × 14.0076 = 28.0152 u
But 1 u = 1.66 × 10−27 kg
∴m = 28.0152 ×1.66 × 10−27 kg
Planck’s constant, h = 6.63 × 10−34 Js
Boltzmann constant, k = 1.38 × 10−23 J K−1
We have the expression that relates mean kinetic energy 
of the nitrogen molecule with the root mean square speed 
as:
Hence, the de Broglie wavelength of the nitrogen molecule is given as:
Therefore, the de Broglie wavelength of the nitrogen molecule is 0.028 nm.
Also Read : Page-No-409:-Chapter-11-Dual-Nature-Of-Radiation-&-Matter-Additional-Exercise-Solutions-class-12-ncert-solutions-Physics
