Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter: NCERT Questions and Answers with PDF

Q.11.1 Find the

(a) maximum frequency, and

(b) minimum wavelength of X-rays produced by 30 kV electrons.

Ans.11.1 Potential of the electrons, V = 30 kV = 3 $\times {10}^4$  V

Energy of the electron, E = e $\times V$  where e = charge of an electron = 1.6 $\times {10}^{-19}C$

• Maximum frequency produced by the X-ray = $\nu$

The energy of the electron is given by the relation, E = h $\nu$ ,

where h = Planck’s constant = 6.626 $\times {10}^{-34}$  Js

$\nu =\frac{E}{h}$ = $\frac{3\mathrm{}\times {10}^4\times 1.6\times {10}^{-19}}{6.626\times {10}^{-34}}$  = 7.244 $\times {10}^{18}$  Hz

$\mathrm{H}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{x}\mathrm{i}\mathrm{u}\mathrm{m}\mathrm{}\mathrm{f}\mathrm{r}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{y}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{X}-\mathrm{r}\mathrm{a}\mathrm{y}\mathrm{}\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{e}\mathrm{d}\mathrm{}\mathrm{i}\mathrm{s}\mathrm{}$ 7.244 $\times {10}^{18}$  Hz

• The minimum wavelength produced is given as

  $\lambda =\frac{c}{\nu }$ where c = Speed of light in air, c = 3 $\times {10}^{18}$  m/s

  $\lambda =\frac{3\mathrm{}\times {10}^8}{7.244\times {10}^{18}}$ = 4.14 $\times {10}^{-11}$  m = 0.0414 nm

Hence, the minimum wavelength produced is 0.414 nm.

Q.11.2 The work function of cesium metal is 2.14 eV. When light of frequency 6  $\times {10}^{14}$ 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.11.2 Work function of cesium metal,  ${\varnothing }_0$ = 2.14 eV

Frequency of light,  $\nu$ = 6  $\times {10}^{14}$ Hz

The maximum kinetic energy is given by the photoelectric effect,  $K$ =  $h\nu -{\varnothing }_0$ ,

Where  $h$ = Planck’s constant = 6.626  $\times {10}^{-34}$ Js, 1 eV = 1.602  $\times {10}^{-19}$ J

$K$ =  $\frac{6.626\times {10}^{-34}\times 6\times {10}^{14}}{1.602\times {10}^{-19}}-2.14$ = 0.345 eV

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For stopping potential  $V_0$ , we can write the equation for kinetic energy as:

$K$ =  $V_0$ , where e = charge of an electron =  $1.6\times {10}^{-19}$

or  $V_0=\frac{K}{e}$ =  $\frac{0.345\times 1.602\times {10}^{-19}}{1.6\times {10}^{-19}}$ = 0.345 V

Hence, the stopping potential is 0.345 V

Maximum speed of the emitted photoelectrons = v

Kinetic energy K =  $\frac{1}{2}$ m  $v^2$ , where m = mass of the electron = 9.1  $\times {10}^{-31}$ kg

Hence, v =  $\sqrt{\frac{2K}{m}}$ =  $\sqrt{\frac{2\times 0.345\times 1.602\times {10}^{-19}}{9.1\mathrm{}\times {10}^{-31}}}$ = 348.5  $\times {10}^3$ m/s = 346.8 km/s

Therefore, the maximum speed of the emitted photoelectrons is 346.8 km/s

Ans.11.3 Photoelectric cut-off voltage,  $V_0$ = 1.5 V

Maximum kinetic energy of the photoelectrons emitted is given as

$K=e{V}_0$ , where e = charge of an electron = 1.6  $\times {10}^{-19}$ C

K = 1.6  $\times {10}^{-19}\times 1.5$ = 2.4  $\times {10}^{-19}$ J

Hence, maximum kinetic energy of the photoelectrons emitted is 2.4  $\times {10}^{-19}$ J

Q.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.11.4 Wavelength of the monochromatic light,  $\lambda =632.8nm$ = 632.8  $\times {10}^{-9}$ m

Power emitted by laser, P = 9.42 mW = 9.42  $\times {10}^{-3}$ W

Planck’s constant, h = 6.626  $\times {10}^{-34}$ Js

Speed of light, c = 3  $\times {10}^8$ m/s

Mass of hydrogen atom, m = 1.66  $\times {10}^{-27}$ kg

The energy of each photon is given as,  $E$ =  $\frac{hc}{\lambda }$ =  $\frac{6.626\mathrm{}\times {10}^{-34}\times 3\mathrm{}\times {10}^8}{632.8\times {10}^{-9}}$ J = 3.141  $\times {10}^{-19}$ J

The momentum of each photon is given by  $p$ =  $\frac{h}{\lambda }$ =  $\frac{6.626\mathrm{}\times {10}^{-34}}{632.8\times {10}^{-9}}$ = 1.047  $\times {10}^{-27}$ kg-m/s

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,  $P=nE$

n =  $\frac{P}{E}$ =  $\frac{9.42\times {10}^{-3}}{3.141\times {10}^{-19}}$ = 3.0  $\times {10}^{16}$

Momentum of the hydrogen atom = momentum of the proton = 1.047  $\times {10}^{-27}$ kg-m/s

The momentum of the hydrogen atom ,  $p=mv$ , where m = mass of hydrogen atom and v = velocity

Hence, v =  $\frac{p}{m}$ =  $\frac{1.047\times {10}^{-27}}{1.66\mathrm{}\times {10}^{-27}}$ = 0.631 m/s