# NCERT Solution : Dual nature of matter & radiation

Q:1. Find the (a) maximum frequency and (b) minimum wavelength of X – rays produced by 30 kV electrons.

Sol. (a)Energy = eV = hv

or , $\large \nu = \frac{e V}{h}$

$\large \nu = \frac{1.6 \times 10^{-19} \times 3 \times 10^{4}}{6.63 \times 10^{-34}}$

= 7.24×1018 Hz

(b) $\large e V = \frac{h c}{\lambda_{min}}$

$\large \lambda_{min} = \frac{h c}{e V}$

$\large \lambda_{min} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8 }{1.6 \times 10^{-19} \times 3 \times 10^4}$

= 0.0414 nm;

Q: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 maximum kinetic energy of the emitted electrons, stopping potential and maximum speed of the emitted photoelectrons ?

Sol.
Given, work function of caesium metal,

φ0 = 2.14 eV

Frequency of light, v = 6×1014 Hz

K.Emax = h ν – φ0

KEmax = 0.35 eV;

Let stopping potential be V0

We know that,

KEmax = e V0

0.35 eV = e V0

V0 = 0.35 V

maximum kinetic energy, $\large K.E_{max} = \frac{1}{2}m v_{max}^2$

$\large 0.35 \times 1.6 \times 10^{-19} = \frac{1}{2}\times 9.1 \times 10^{-31} \times v_{max}^2$

(where, vmax is the maximum speed and m is the mass of electron)

or, vmax = 349 km/s

Q:3. The photoelectric cut – off voltage in certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted ?

Sol. Given, cut – off voltage, V0 = 1.5 V Maximum kinetic energy is given by

KEmax = eV0

= 1.5 eV = 1.5 × 1.6 × 10-19

= 2.4 × 10-19 J

Q: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 ) (c) How fast does a hydrogen atom have to travel in order to have the same momentum as that of the photon .

Sol: (a) $\large E = \frac{h c}{\lambda}$

$\large E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{632.8 \times 10^{-9}}$

= 3.14 × 10-19 J

$\large p = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34}}{632.8 \times 10^{-9}}$

= 1.05 × 10-27 kg m/s

(b) $\large n = \frac{Power}{E} = \frac{9.42 \times 10^{-3}}{3.14 \times 10^{-19}}$

= 3 × 1016 photon/sec

(c) $\large v = \frac{p}{m} = \frac{1.05 \times 10^{-27}}{1.66 \times 10^{-27}}$

= 0.63 m/s

Q:5. The energy flux of sunlight reaching the surface of the earth is 1.388 × 103 W/m2. How many photons (nearly) per square meter are incident on the earth per second? Assume that the photons in the sunlight have an average wavelength of 550 nm.

Sol. Energy of a photon , E = hc/λ

Number of photons incident per square meter per second,

$\large n = \frac{P}{E} = \frac{P}{hc/\lambda}$

$\large n = \frac{P \lambda}{h c}$

$\large = \frac{1.388 \times 10^3 \times 550 \times 10^{-9}}{6.63 \times 10^{-34} \times 3 \times 10^8}$

= 3.84 × 1021 photons/m2-s

Q:6. In a experiment on photoelectric effect, the slope of the cur – off voltage versus frequency of incident light is found to be 4.12 × 10-15 V-s .

Sol: Slope of graph = 4.12 × 10-15 V-s

Slope of graph = h/e

h = e × Slope of graph

= 1.6 × 10-19 × 4.12 × 10-15

= 6.592 × 10-34 J-s

Q: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.
(i) What is the energy per photon associated with the sodium light?
(ii) At what rate are the photons delivered to the sphere?

Sol: $\large E = \frac{h c}{\lambda}$

$\large E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{589 \times 10^{-9}}$

= 3.38 × 10-19 J

$\large n = \frac{P}{E} = \frac{100}{3.38 \times 10^{-19}}$

= 3.0 × 1020 photons/s

Q: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 cut – off voltage for the photoelectric emission.

Sol. Using the formula for kinetic energy,

$\large KE = eV_0 = h\nu – h\nu_0$

$\large V_0 = \frac{h(\nu – \nu_0)}{e}$

$\large V_0 = \frac{6.63 \times 10^{-34}(8.2 \times 10^{14} – 3.3 \times 10^{14} )}{1.6 \times 10^{-19}}$

= 2.03 V

Q:9. The work function for a certain metal is 4.2 eV. Will this metal given photoelectric emission of incident radiation of wavelength 330 nm ?

Sol. Given, φ0 = 4.2 eV = 4.2 × 1.6 × 10-19 J

= 6.72 × 10-19 J

and , λ = 330 nm = 330 × 10-9 m

Energy of incident photon, E = hc/λ

$\large = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{330 \times 10^{-9}}$

= 6.027 × 101-9 J

As energy of incident photon E < φ0, hence no photoelectric emission will take place.

Q:10. Light of frequency 7.21 × 1014 Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 105 ms-1 are ejected from the surface. What is the threshold frequency for photoemission of electrons? ]

h = 6.63 × 10-34 Js, me = 9·1 × 10-31 kg.

Sol. $\large \frac{1}{2} m v_{max}^2 = h \nu – h \nu_0$

$\large \nu_0 = \nu – \frac{m v_{max}^2}{2 h}$

$\large \nu_0 = 7.21 \times 10^{14} – \frac{9.1 \times 10^{-31} \times (6 \times 10^5)^2}{2 \times 6.63 \times 10^{-34}}$

= 4.74 × 1014 Hz

Q: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 cathode, the stopping potential of photoelectrons is 0.38 eV. Find the work function of the material from which the cathode is made.
Given : h = 6.63 × 10-34 Js, 1 eV = 1·6 × 10-19 J.

Sol: $\large e V_0 = \frac{h c}{\lambda} – \phi_0$

$\large \phi_0 = \frac{h c}{\lambda} – e V_0$

$\large \phi_0 = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{488 \times 10^{-9} \times 1.6 \times 10^{-19}} – 0.38$

= 2.17 e V

Q:12. Calculate the (a) momentum and (b) de–Broglie wavelength of the electron acceleration through a potential difference of 56 V. Given ; h = 6.63 × 10-34 Js, me = 9 × 10-31 kg; e = 1·6 × 10-19 C.

Sol: $\large e V = \frac{1}{2} m v^2 = \frac{(m v)^2}{2 m}$

$\large m v = \sqrt{2 m e V} = \sqrt{2 \times 9 \times 10^{-31} \times 1.6 \times 10^{-19}\times 56}$

Momentum = 4.02 x 10-24 kg m/s

$\large \lambda = \frac{12.27}{\sqrt{V}} A^o$

$\large \lambda = \frac{12.27}{\sqrt{56}}$

= 1.64 A°

Q:13. What is the (a) momentum (b) speed and (c) de – Broglie wavelength of an electron with kinetic energy of 120 ev.

Given h = 6.63 × 10-34 Js, me = 9 × 10-31 kg; 1 eV = 1·6 × 10-19 J

Sol:(a) $\large p = \sqrt{2 m K }$

$\large p = \sqrt{2 \times 9 \times 10^{-31} \times 120 \times 1.6 \times 10^{-19} }$

= 5.88 x 10-24 kg m/s

(b) $\large v = \sqrt{\frac{2 K}{m}}$

$\large v = \sqrt{\frac{2 \times 120 \times 1.6 \times 10^{-19}}{9 \times 10^{-31}}}$

= 6.53 x 106 m/s

(c) $\large \lambda = \frac{h}{m v}$

$\large \lambda = \frac{6.63 \times 10^{-34}}{5.88 \times 10^{-24}}$

= 1.13 x 10-10 m = 1.13 A°