NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

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Class 12 : NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

The document NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev is a part of the Class 12 Course NCERT Textbooks (Class 6 to Class 12).
All you need of Class 12 at this link: Class 12

NCERT QUESTION

(Dual Nature of Radiation and Matter)

Ques 11.1:
Find the
(a) maximum frequency, and
(b) minimum wavelength of X-rays produced by 30 kV electrons.
Ans:  Potential of the electrons, V = 30 kV = 3 × 104 V
Hence, energy of the electrons, E = 3 × 104 eV
Where,
e = Charge on an electron = 1.6 × 10−19 C
(a)Maximum frequency produced by the X-rays = ν
The energy of the electrons is given by the relation:
E =
Where,
h = Planck’s constant = 6.626 × 10−34 Js

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Hence, the maximum frequency of X-rays produced is NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

(b)The minimum wavelength produced by the X-rays is given as:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Hence, the minimum wavelength of X-rays produced is 0.0414 nm.

Ques 11.2:
The work function of caesium metal is 2.14 eV. When light of frequency 6 ×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?
Ans:  Work function of caesium metal, NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Frequency of light, NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

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

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Where,
h = Planck’s constant = 6.626 × 10−34 Js

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Hence, the maximum kinetic energy of the emitted electrons is
0.345 eV.

(b)For stopping potential Vo , we can write the equation for kinetic energy as:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

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:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Where,
m = Mass of an electron = 9.1 × 10−31 kg

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Hence, the maximum speed of the emitted photoelectrons is 332.3 km/s.

Ques 11.3:
The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?
Ans:  Photoelectric cut-off voltage, V0 = 1.5 V

The maximum kinetic energy of the emitted photoelectrons is given as:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev
Where,
e = Charge on an electron = 1.6 × 10−19 C

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

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

Ques 11.4:
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?
Ans:  Wavelength of the monochromatic light, λ = 632.8 nm = 632.8 × 10−9 m
Power emitted by the laser, P = 9.42 mW = 9.42 × 10−3 W
Planck’s constant, h = 6.626 × 10−34 Js
Speed of light, c = 3 × 108 m/s
Mass of a hydrogen atom, m = 1.66 × 10−27 kg

(a)The energy of each photon is given as:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

The momentum of each photon is given as:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

(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:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

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

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Momentum is given as:

p = mv 

Where,

v = Speed of the hydrogen atom

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Ques 11.5:
The energy flux of sunlight reaching the surface of the earth is 1.388 × 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 nm.
Ans:  Energy flux of sunlight reaching the surface of earth, Φ = 1.388 × 103 W/m2
Hence, power of sunlight per square metre, P = 1.388 × 103 W
Speed of light, c = 3 × 108 m/s
Planck’s constant, h = 6.626 × 10−34 Js
Average wavelength of photons present in sunlight, NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Number of photons per square metre incident on earth per second = n
Hence, the equation for power can be written as:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Therefore, every second, NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRevphotons are incident per square metre on earth.

Ques 11.6:
In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 × 10−15 V s. Calculate the value of Planck’s constant.
Ans:  The slope of the cut-off voltage (V) versus frequency (ν) of an incident light is given as:
NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

v is related to frequency by the equation:
hv = ev
Where,
e = Charge on an electron = 1.6 × 10−19 C
h = Planck’s constant
e = Charge on an electron = 1.6 x 10-19 C
h = Planck`s constant

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev
NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Therefore, the value of Planck’s constant is NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Ques 11.7:
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?
Ans:  Power of the sodium lamp, P = 100 W
Wavelength of the emitted sodium light, λ = 589 nm = 589 × 10−9 m
Planck’s constant, h = 6.626 × 10−34 Js
Speed of light, c = 3 × 108 m/s

(a)The energy per photon associated with the sodium light is given as:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

(b) Number of photons delivered to the sphere = n
The equation for power can be written as:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Therefore, every second, NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRevphotons are delivered to the sphere.

Ques 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.
Ans:  Threshold frequency of the metal, NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev
Frequency of light incident on the metal, NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev
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 = Vo
The equation for the cut-off energy is given as:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Therefore, the cut-off voltage for the photoelectric emission is NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Ques 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?
Ans:  No
Work function of the metal,  φ0 = 4.2 eV
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:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

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.


Ques 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?
Ans:  Frequency of the incident photon, NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev
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:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Therefore, the threshold frequency for the photoemission of electrons is NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Ques 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.
Ans:  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 = NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev
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:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

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

Ques 11.12:
Calculate the
(a) momentum, and
(b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
Ans:  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:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

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:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

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

Ques 11.13:
What is the
(a) momentum,
(b) speed, and
(c) de Broglie wavelength of an electron with kinetic energy of 120 eV.
Ans:  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:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Where,
v = Speed of the electron

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

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:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Therefore, the de Broglie wavelength of the electron is 0.112 nm.

Ques 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.
Ans:  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:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev
We have the relation for de Broglie wavelength as:
NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Substituting equation (2) in equation (1), we get the relation:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

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:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Hence, the kinetic energy of the neutron is 3.78 × 10−28 J or 2.36 neV.

Ques 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?
Ans:  (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:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

(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:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

(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:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Ques 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.
Ans:  Wavelength of an electron 

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

= 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:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

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.

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

(b) The energy of a photon is given by the relation:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Where,
Speed of light, c = 3 × 108 m/s

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev
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:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Where,
m = Mass of the electron = 9.1 × 10−31 kg
p = 6.63 × 10−25 kg m s−1
NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev
Hence, the kinetic energy of the electron is 1.51 eV.

Ques 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.
Ans:  (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:
NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

De Broglie wavelength (λ) and velocity (v) are related as:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev ..... (2)

Using equation (2) in equation (1), we get:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

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:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

The relation for the de Broglie wavelength is given as:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

where,
mn = 1.66 x 10-27 kg
h = 6.6 x 10-34 Js
K` = 6.75 x 10-21 J

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

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

Ques 11.18:
Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
Ans:  The momentum of a photon having energy () is given as:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

Where,
λ = Wavelength of the electromagnetic radiation
c = Speed of light
h = Planck’s constant
De Broglie wavelength of the photon is given as:

NCERT Solutions (Part - 1)- Dual Nature of Radiation and Matter Class 12 Notes | EduRev

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.

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