Chapter 11 Dual Nature of Radiation and Matter NCERT Solutions Class 12 Physics

Question 11.1

Find the (a) maximum frequency, and (b) minimum wavelength of X-rays produced by 30 kV electrons.

Given Data

Accelerating voltage, $V = 30 \text{ kV} = 30 \times 10^3 \text{ V}$.
Electron energy $E = eV$.

(a) Maximum Frequency

The maximum energy of the photon corresponds to the kinetic energy of the electron.

$$h \nu_{max} = eV \implies \nu_{max} = \frac{eV}{h}$$ $$\nu_{max} = \frac{1.6 \times 10^{-19} \times 30 \times 10^3}{6.63 \times 10^{-34}} = \frac{48 \times 10^{-16}}{6.63 \times 10^{-34}}$$ $$\nu_{max} \approx 7.24 \times 10^{18} \text{ Hz}$$

(b) Minimum Wavelength

$$\lambda_{min} = \frac{c}{\nu_{max}} = \frac{3 \times 10^8}{7.24 \times 10^{18}}$$ $$\lambda_{min} \approx 0.0414 \times 10^{-10} \text{ m} = 0.0414 \text{ nm}$$

(a) $7.24 \times 10^{18} \text{ Hz}$
(b) $0.041 \text{ nm}$

Question 11.2

The work function of caesium metal is 2.14 eV. When light of frequency $6 \times 10^{14} \text{ Hz}$ is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy, (b) Stopping potential, and (c) maximum speed of the emitted photoelectrons?

[Image of Photoelectric Effect Process]

Given Data

$\phi_0 = 2.14 \text{ eV}$. Frequency $\nu = 6 \times 10^{14} \text{ Hz}$.

(a) Maximum Kinetic Energy

Energy of incident photon $E = h\nu$.

$$E = 6.63 \times 10^{-34} \times 6 \times 10^{14} = 39.78 \times 10^{-20} \text{ J}$$ $$\text{In eV: } E = \frac{39.78 \times 10^{-20}}{1.6 \times 10^{-19}} \approx 2.486 \text{ eV}$$ $$K_{max} = E – \phi_0 = 2.486 – 2.14 = 0.346 \text{ eV}$$

(b) Stopping Potential

$$V_0 = \frac{K_{max}}{e} = 0.346 \text{ V}$$

(c) Maximum Speed

$$K_{max} = \frac{1}{2} m v^2 \implies v = \sqrt{\frac{2 K_{max}}{m}}$$ $$K_{max} = 0.346 \times 1.6 \times 10^{-19} = 0.554 \times 10^{-19} \text{ J}$$ $$v = \sqrt{\frac{2 \times 0.554 \times 10^{-19}}{9.11 \times 10^{-31}}} = \sqrt{0.1216 \times 10^{12}} \approx 3.48 \times 10^5 \text{ m/s}$$

(a) 0.346 eV, (b) 0.346 V, (c) $3.48 \times 10^5 \text{ m/s}$

Question 11.3

The photoelectric cut-off voltage in a certain experiment is 1.5 V. What is the maximum kinetic energy of photoelectrons emitted?

Formula

Maximum Kinetic Energy is equal to the work done by stopping potential.

$$K_{max} = e V_0$$ $$K_{max} = 1.6 \times 10^{-19} \text{ C} \times 1.5 \text{ V} = 2.4 \times 10^{-19} \text{ J}$$

$K_{max} = 1.5 \text{ eV}$ or $2.4 \times 10^{-19} \text{ J}$.

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

(a) Energy and Momentum

$\lambda = 632.8 \text{ nm} = 6.328 \times 10^{-7} \text{ m}$. Power $P = 9.42 \text{ mW}$.

$$E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{6.328 \times 10^{-7}} \approx 3.14 \times 10^{-19} \text{ J}$$ $$p = \frac{h}{\lambda} = \frac{6.63 \times 10^{-34}}{6.328 \times 10^{-7}} \approx 1.05 \times 10^{-27} \text{ kg m/s}$$

(b) Photons per Second

$$N = \frac{P}{E} = \frac{9.42 \times 10^{-3}}{3.14 \times 10^{-19}} = 3.0 \times 10^{16} \text{ photons/s}$$

(c) Speed of Hydrogen Atom

Mass of H atom $m_H \approx 1.67 \times 10^{-27} \text{ kg}$.

$$v = \frac{p}{m_H} = \frac{1.05 \times 10^{-27}}{1.67 \times 10^{-27}} \approx 0.63 \text{ m/s}$$

(a) $3.14 \times 10^{-19} \text{ J}$, $1.05 \times 10^{-27} \text{ kg m/s}$
(b) $3 \times 10^{16} / \text{s}$
(c) $0.63 \text{ m/s}$

Question 11.5

In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be $4.12 \times 10^{-15} \text{ V s}$. Calculate the value of Planck’s constant.

Formula

Einstein’s photoelectric equation: $K_{max} = h\nu – \phi_0 \implies eV_0 = h\nu – \phi_0$.
$V_0 = \left(\frac{h}{e}\right)\nu – \frac{\phi_0}{e}$.
The graph of $V_0$ vs $\nu$ is a straight line with slope $m = \frac{h}{e}$.

Calculation

$$h = e \times \text{slope} = 1.6 \times 10^{-19} \times 4.12 \times 10^{-15}$$ $$h = 6.592 \times 10^{-34} \text{ J s}$$

Planck’s Constant $h \approx 6.59 \times 10^{-34} \text{ J s}$.

Question 11.6

The threshold frequency for a certain metal is $3.3 \times 10^{14} \text{ Hz}$. If light of frequency $8.2 \times 10^{14} \text{ Hz}$ is incident on the metal, predict the cutoff voltage for the photoelectric emission.

Formula

$V_0 = \frac{h}{e} (\nu – \nu_0)$.

Calculation

$$V_0 = \frac{6.63 \times 10^{-34}}{1.6 \times 10^{-19}} (8.2 \times 10^{14} – 3.3 \times 10^{14})$$ $$V_0 = 4.14 \times 10^{-15} \times (4.9 \times 10^{14})$$ $$V_0 \approx 2.03 \text{ V}$$

Cutoff voltage = 2.03 V

Question 11.7

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

Calculation

Energy of incident photon $E = \frac{hc}{\lambda}$.

$$E = \frac{1240 \text{ eV nm}}{330 \text{ nm}} \approx 3.76 \text{ eV}$$

Conclusion

The incident energy ($3.76 \text{ eV}$) is less than the work function ($4.2 \text{ eV}$).

No, photoelectric emission will not occur.

Question 11.8

Light of frequency $7.21 \times 10^{14} \text{ Hz}$ is incident on a metal surface. Electrons with a maximum speed of $6.0 \times 10^5 \text{ m/s}$ are ejected. What is the threshold frequency?

1. Kinetic Energy

$$K_{max} = \frac{1}{2} m v^2 = \frac{1}{2} \times 9.11 \times 10^{-31} \times (6.0 \times 10^5)^2$$ $$K_{max} = 1.64 \times 10^{-19} \text{ J}$$

2. Threshold Frequency

$h\nu_0 = h\nu – K_{max} \implies \nu_0 = \nu – \frac{K_{max}}{h}$.

$$\nu_0 = 7.21 \times 10^{14} – \frac{1.64 \times 10^{-19}}{6.63 \times 10^{-34}}$$ $$\nu_0 = 7.21 \times 10^{14} – 2.47 \times 10^{14} = 4.74 \times 10^{14} \text{ Hz}$$

Threshold Frequency = $4.74 \times 10^{14} \text{ Hz}$.

Question 11.9

Light of wavelength 488 nm is produced by an argon laser. When light from this line is incident on the emitter, the stopping potential is 0.38 V. Find the work function.

Formula

$\phi_0 = E – eV_0 = \frac{hc}{\lambda} – eV_0$.

Calculation

$$E = \frac{1240 \text{ eV nm}}{488 \text{ nm}} \approx 2.54 \text{ eV}$$ $$\phi_0 = 2.54 \text{ eV} – 0.38 \text{ eV} = 2.16 \text{ eV}$$

Work Function = 2.16 eV.

Question 11.10

What is the de Broglie wavelength of:
(a) a bullet of mass 0.040 kg travelling at 1.0 km/s,
(b) a ball of mass 0.060 kg moving at 1.0 m/s,
(c) a dust particle of mass $1.0 \times 10^{-9} \text{ kg}$ drifting with 2.2 m/s?

Formula

$\lambda = \frac{h}{mv}$.

Calculations

$$\text{(a) } \lambda = \frac{6.63 \times 10^{-34}}{0.040 \times 1000} \approx 1.66 \times 10^{-35} \text{ m}$$ $$\text{(b) } \lambda = \frac{6.63 \times 10^{-34}}{0.060 \times 1.0} \approx 1.1 \times 10^{-32} \text{ m}$$ $$\text{(c) } \lambda = \frac{6.63 \times 10^{-34}}{1.0 \times 10^{-9} \times 2.2} \approx 3.0 \times 10^{-25} \text{ m}$$

(a) $1.7 \times 10^{-35} \text{ m}$
(b) $1.1 \times 10^{-32} \text{ m}$
(c) $3.0 \times 10^{-25} \text{ m}$

Question 11.11

Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

Proof

For a photon, energy $E = h\nu = \frac{hc}{\lambda_{em}}$.

Momentum of a photon is given by $p = \frac{E}{c} = \frac{h\nu}{c} = \frac{h}{\lambda_{em}}$.

According to de Broglie hypothesis, the wavelength associated with a particle of momentum $p$ is:

$$\lambda_{dB} = \frac{h}{p}$$

Substituting $p = h/\lambda_{em}$:

$$\lambda_{dB} = \frac{h}{(h/\lambda_{em})} = \lambda_{em}$$

Thus, de Broglie wavelength equals EM wavelength.