NCERT Solutions Class 12 Physics Chapter 11 Dual Nature Of Radiation And Matter
1. Find the:
(a) Maximum frequency, and
(b) The minimum wavelength of X-rays produced by 30 kV electrons.
Answer:
Electron potential, V = 30 kV = 3 × 104 V
Hence, electron energy, E = 3 × 104 eV
Where, e = Charge on one electron = 1.6 × 10-19 C
(a) Maximum frequency by the X-rays = ν
The energy of the electrons:
E = hν
Where,
h = Planck’s constant = 6.626 × 10-34 Js
Therefore, \(v = \frac{E}{h}\)
\(= \frac{1.6\times 10^{-19}\times 3 \times 10^{4}}{6.626\times 10^{-34}}\) = 7.24 x 1018 Hz
Hence, 7.24 x 1018 Hz is the maximum frequency of the X-rays.
(b) The minimum wavelength produced:
\(\lambda =\frac{c}{v}\)
\(= \frac{3\times 10^{8}}{7.24\times 10^{18}}\) = 4.14 x 10-11 m = 0.0414 nm
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
(c) maximum speed of the emitted photoelectrons?
Answer:
Work function of caesium, Φo = 2.14eV
Frequency of light, v = 6.0 x 1014 Hz
(a) The maximum energy (kinetic) by the photoelectric effect:
K = hν – Φo
Where,
h = Planck’s constant = 6.626 x 10-34 Js
Therefore,
K = \(\frac{6.626\times 10^{-34}\times 6\times 10^{14}}{1.6\times 10^{-19}}\) – 2.14
= 2.485 – 2.140 = 0.345 eV
Hence, 0.345 eV is the maximum kinetic energy of the emitted electrons.
(b) For stopping potential Vo, we can write the equation for kinetic energy as:
K = eVo
Therefore, Vo = \(\frac{K}{e}\)
= \(\frac{0.345\times 1.6\times 10^{-19}}{1.6\times 10^{-19}}\)
= 0.345 V
Hence, 0.345 V is the stopping potential of the material.
(c) Maximum speed of photoelectrons emitted = ν
Following is the kinetic energy relation:
K = \(\frac{1}{2}mv^{2}\)
Where,
m = mass of electron = 9.1 x 10-31 Kg
\(v^{2}=\frac{2K}{m}\)
= \(\frac{2\times 0.345\times 1.6\times 10^{-19}}{9.1\times 10^{-31}}\)
=0.1104 x 1012
Therefore, ν = 3.323 x 105 m/s = 332.3 km/s
Hence, 332.3 km/s is the maximum speed of the emitted photoelectrons.
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, Vo = 1.5 V
For emitted photoelectrons, the maximum kinetic energy is:
Ke = eVo
Where,
e = charge on an electron = 1.6 x 10-19 C
Therefore, Ke = 1.6 x 10-19 x 1.5 = 2.4 x 10-19 J
Therefore, 2.4 x 10-19 J is the maximum kinetic energy emitted by the photoelectrons.
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 a uniform cross-section which is less than the target area)
(c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon?
Answer:
Monochromatic light having a wavelength, λ = 632.8 nm = 632.8 × 10-9 m
Given that the laser emits the power of, 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 photons having the energy as:
E = \(\frac{hc}{\lambda }\)
= \(\frac{6.626\times 10^{-34}\times 3\times 10^{8}}{632.8\times 10^{-9}}\)
= 3.141 x 10-19 J
Therefore, each photon has a momentum of :
P = \(\frac{h}{\lambda}\)
= \(\frac{6.626\times 10^{-34}}{632.8\times 10^{-9}}\)
= 1.047 x 10-27 kg m/s
(b) Number of photons/second arriving at the target illuminated by the beam = n
Assuming the uniform cross-sectional area of the beam is less than the target area.
Hence, equation for power is written as:
P = nE
Therefore, n = \(\frac{P}{E}\)
= \(\frac{9.42\times 10^{-3}}{3.141\times 10^{-19}}\)
= 3 x 1016 photons/s
(c) Given that, momentum of the hydrogen atom is equal to the momentum of the photon,
P = 1.047 x 10-27 kg m/s
Momentum is given as:
P = mv
Where,
ν = speed of hydrogen atom
Therefore, ν = \(\frac{p}{m}\)
= \(\frac{1.047\times 10^{-27}}{1.66\times 10^{-27}}\) = 0.621 m/s
5. The energy flux of sunlight reaching the surface of the earth is 1.388 × 103 W/m2. How many photons are incident on the Earth per second/square meter? Assume an average wavelength of 550 nm.
Answer:
Sunlight reaching the surface of the earth has an energy flux of
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
λ = 550nm = 550 x 10-9m is the average wavelength of the photons from the sunlight
Number of photons per square metre incident on earth per second = n
Hence, the equation for power be written as:
P = nE
Therefore, n = \(\frac{P}{E}\)
= \(\frac{P\lambda }{hc}\)
= \(\frac{1.388\times 10^{3}\times 550\times 10^{-9}}{6.626\times 10^{-34}\times 3\times 10^{8}}\) = 3.84 x 1021 photons/m2/s
Therefore, 3.84 x 1021 photons are incident on the earth per square meters.