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

1. Find the:

(a) Maximum frequency and

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

Sol. Given, voltage V = 30 kV = 30 × 103 V and e = 1.6 × 10–19C

(a) E = eV = hv

$$\text{or \space v =}\frac{\text{eV}}{\text{h}}=\frac{1.6×10^{-19}×30×10^{3}}{6.63×10^{\normalsize-34}}$$

= 7.24 × 1018 Hz

Maximum frequency v = 7.24 × 1018 Hz

$$\text{(b)}\space\lambda_{\text{min}}=\frac{c}{v}=\frac{3×10^{8}}{7.24×10^{18}}$$

∴ where c = 3 × 108 m/s (speed of light)

λ = 0.414 × 10–10

= 0.0414 × 10–9 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 and

(c) maximum speed of the emitted photoelectrons?

Sol. (a) Maximum kinetic energy of emitted electrons.

KEmax = hv – Φ0

$$=\frac{6.63×10^{\normalsize-34}×6×10^{14}}{1.6×10^{\normalsize-19}}=2.14$$

= 0.35 eV

(b) Let stopping potential be V0.

We have

KEmax = eV0

0.35 eV = eV0

V0 = 0.35V

(c) Maximum kinetic energy,

$$\text{KE}_{\text{max}}=\frac{1}{2}\text{mv}^{2}_{\text{max}}\\0.35\text{eV}=\frac{1}{2}\text{mv}^{2}_{\text{max}}$$

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

$$\text{or}\space\frac{0.35×2×1.6×10^{\normalsize-19}}{9.1×10^{\normalsize-31}}=v^{2}_{max}\\(\because e=1.6×10^{\normalsize-19})$$

or v2max = 0.123 × 1012

or vmax = 350713.55 m/s

vmax = 350.7 km/s

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

Sol. Given, cut-off voltage V0 = 1.5V

Maximum kinetic energy

KEmax = eV0 = 1.5eV

= 1.5 × 1.6 × 10–19 = 2.4 × 10–19J

4. Monochromatic light of wavelength 632.8 nm is produced by a helium neon laser. The power emitted is 9.42mW.

(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. Given, wavelength of monochromatic light, λ = 632.8 nm = 632.8 × 10–9 m

Power = 9.42 mW = 9.42 × 10–3W

$$\text{(a) Energy E =}\frac{hc}{\lambda}\\=\frac{6.63×10^{\normalsize-34}×10^{8}}{632.8×10^{-9}}$$

= 3.14 × 10–19J

The momentum of each photon,

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

= 1.05 × 10–27 kg-mg/s

(b) Let n be the number of photons per second, So,

$$n=\frac{\text{Power}}{\text{Energy of each photon}}\\=\frac{9.42×10^{\normalsize-3}}{3.14×10^{\normalsize-19}}$$

= 3 × 1016 photon/s

(c) Momentum p = mv

Velocity of hydrogen atom,

$$v=\frac{p}{m}=\frac{1.05×10^{-27}}{1.66×10^{-27}}=0.63\space\text{m/s}$$

[∵ m = 1.66 × 10–27 kg (mass of electron)]

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.

Sol. Given, energy per unit area per second, P = 1.388 × 103 W/m2
Let n be the number of photons incident on the earth per metre2.

Wavelength = 350 nm = 550 × 10–8 m

$$\text{Energy of each photon, E =}\frac{hc}{\lambda}\\=\frac{6.63×10^{\normalsize-34}×3×10^{8}}{550×10^{\normalsize-8}}$$

= 3.616 × 10–19J

Number of photons incident on the earth’s surface

$$n=\frac{\text{P}}{\text{E}}=\frac{\text{1.388×10}^{3}}{3.616×10^{\normalsize-19}}$$

= 3.838 × 1021

= 3.838 × 1021 photon/m2-s

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. Sol. Given, slope of graph tan θ = 4.12 × 10–15 V-s and charge on electron e = 1.6 × 10–19C.

$$\text{tan}\space\theta=\frac{V}{v}$$

We know that

hv = eV

$$\frac{\text{V}}{v}=\frac{\text{h}}{e}\\\therefore\space\frac{h}{e}=4.12×10^{\normalsize-15}$$

h = 1.6 × 10–19 × 4.12 × 10–15

= 6.592 × 10–34 J-s

7. A 100W 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?

Sol. Given, power of sodium lamp, P = 100W

Wavelength of the sodium lamp, λ = 589 nm = 589 × 10–9 m

(a) Energy of each photon

$$\text{E}=\frac{hc}{v}=\frac{6.63×10^{\normalsize-34}×3×10^{8}}{589×10^{\normalsize-9}}\\(\because c=3×10^{8}\space m/s)\\=3.38×10^{\normalsize-19}\text{J}\\=\frac{3.38×10^{\normalsize-19}}{1.9×10^{\normalsize-19}}\text{eV}$$

= 2.11 eV

(b) Let n photons are delivered per second.

$$\therefore\space n=\frac{\text{Power}}{\text{Energy of each photon}}\\(\text{From P = E-n})\\=\frac{100}{3.38×10^{-19}}$$

= 3 × 1020 photon/s

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. Given, threshold frequency for a metal, v0 = 3.3 × 1014 Hz

Frequency of light, v = 8.2 × 1014 Hz

The kinetic energy,

KE = eV0 = hv – hv0

$$\text{V}_{0}=\frac{h(v-v_{0})}{e}\\=\frac{6.63×10^{\normalsize-34}(8.2×10^{14}-3.3×10^{14})}{1.6×10^{\normalsize-19}}\\=\frac{6.63×10^{\normalsize-34}×10^{14}×4.9}{1.6×10^{\normalsize-19}}=2.03\text{V}$$

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

Sol. Given, work function Φ0 = 4.2 eV

= 4.2 × 1.6 × 10–19 J = 6.72 × 10–19J

Wavelength of given radiation, λ = 330 nm = 330 × 10–9 m

For the energy of each photon is more than the work function, the photoelectric emission takes place.

Energy of each photon,

$$\text{E}=\frac{hc}{\lambda}=\frac{6.63×10^{\normalsize-34}×3×10^{8}}{330×10^{\normalsize-9}}$$

= 6.027 × 10–19J

Since value of energy vE = 6.027 × 10–19 J is less than the work function, Φ0 = 6.72 × 10–19 J no photoelectric emission takes place.

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?

Sol. Given, frequency of light, v = 7.21 × 1014 Hz

Mass of electron, m = 9.1 × 10–31 kg

Maximum speed of electrons, vmax = 6 × 105 m/s

Let v0 be the threshold frequency.

The kinetic energy

$$\text{K.E}=\frac{1}{2}mv^{2}_{\text{max}}=hv-hv_{0}\\\text{i.e.\space}\frac{1}{2}×9.1×10^{\normalsize-31}×6×10^{5}×6×10^{5}\\=6.63 × 10^{\normalsize–34} (v – v_0)\\\text{or}\space v-v_{0}=\frac{36×9.1×10^{\normalsize-21}}{2×6.63×10^{\normalsize-34}}=2.47×10^{14}$$

or  v0 = 7.21 × 1014 – 2.47 × 1014

= 4.74 × 1014 Hz

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 0.38V. Find the work function of the material from which the emitter is made.
Sol. incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38V. Find the work function of the material from which the emitter is made.

Sol. Given, wavelength, λ = 488 nm = 488 × 10–9 m

cut-off potential V0 = 0.38V, e = 1.6 × 10–19C

kinetic energy

$$\text{KE}=\text{eV}_{0}=\frac{\text{hc}}{\lambda}-\phi_{0}\\1.6×10^{\normalsize-19}×0.38=\frac{6.63×10^{\normalsize-34}×3×10^{8}}{488×10^{\normalsize-9}}-\phi_{0}$$

or  6.08 × 10–20 = 40.75 × 10–20– Φ0

or Φ0 = (40.75 – 6.08) × 10–20

= 34.67 × 10–20J

$$=\frac{34.67×10^{\normalsize-20}}{1.6×10^{\normalsize-19}}\text{eV}$$

= 2.17eV

12. Calculate the

(a) momentum and

(b) de-Broglie wavelength of the electrons accelerated through a potential difference of 56V.

Sol. (a) The kinetic energy

$$\text{eV}=\frac{1}{2}mv^{2}\\\frac{2\text{eV}}{m}=v^{2}\\v=\sqrt{\frac{2\text{eV}}{m}}\\\text{Momentum accelerated electron,}\\\text{p = mv=m}\sqrt{\frac{2\text{eV}}{m}}=\sqrt{2\text{eVm}}\\=\sqrt{2×1.6×10^{\normalsize-9}×56×9×10^{\normalsize-31}}\\\text{= 4.02 × 10}^{\normalsize–24} \text{kg-m/s}\\\text{(b) de-Broglie wavelength,}\\\lambda=\frac{12.27}{\sqrt{\text{V}}}=6.164×10^{\normalsize-9}m$$

= 0.164 nm

13. What is the

(a) momentum,

(b) speed and

(c) de-Broglie wavelength of an electron with kinetic energy of 120eV?

$$\textbf{Sol.}\text{(a) Momentum, p =}\sqrt{2eVm}\\=\sqrt{2\text{K.E.m}}\space \text{and}\\\text{e = 1.6 × 10}^{\normalsize–19}\space (\because \text{KE}=e\text{V})\\\sqrt{2×120×1.6×10^{\normalsize-19}×9.1×10^{\normalsize-31}}$$

= 5.91 × 10–24 kg-m/s

(b) Momentum p = mv

$$\therefore\space v=\frac{p}{m}=\frac{5.91×10^{\normalsize-24}}{9.1×10^{\normalsize-31}}=6.5×10^{6}\space\text{m/s}$$

(c) de-Broglie wavelength associated with electrons,

$$\lambda=\frac{12.27}{\sqrt{v}}\text{Å}=\frac{12.27}{\sqrt{120}}\text{Å}$$

= 0.112 × 10–9m

= 0.112 nm

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.

Sol. Given, wavelength of light = 589 nm = 589 × 10–9 m

Mass of electron me = 9.1 × 10–31kg

Mass of neutron mn = 1.67 × 10–34kg

Planck’s constant h = 6.62 × 10–34 J-s

$$\text{(a) Wavelength,}\space\lambda =\frac{h}{\sqrt{2\text{KE}m_e}}\\\text{Kinetic energy of electron,}\\\text{KE}_{e}=\frac{h^{2}}{2\lambda^{2}m_{e}}\\=\frac{(6.63×10^{\normalsize-34})^{2}}{2×(589×10^{\normalsize-9})^{2}×9.1×10^{-31}}$$

= 6.96 × 10–25J

(b) Kinetic energy of neutron

$$\text{K.E}_{n}=\frac{h^{2}}{2\lambda^{2}m_{n}}\\=\frac{(6.63×10^{\normalsize-34})^{2}}{2×(589×10^{\normalsize-9})^{2}×1.66×10^{\normalsize-27}}$$

= 3.81 × 10–28J

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?

Sol. (a) de-Broglie wavelength

$$\lambda=\frac{h}{mv}=\frac{6.63×10^{\normalsize-34}}{0.040×1×10^{3}}$$

(b) Mass of the ball, m = 0.060 kg and speed by which the ball is moving, v = 1 m/s

$$\lambda=\frac{h}{mv}=\frac{6.63×10^{\normalsize-34}}{0.060×1}$$

= 1.1 × 10–32m

(c) Mass of a dust particle, m = 1 × 10–9 kg and speed of the dust particle, v = 2.2 m/s

$$\lambda=\frac{h}{mv}=\frac{6.63×10^{\normalsize-34}}{1×10^{\normalsize-9}×2.2}$$

= 3.0 × 10–25 m

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.

Sol. (a) Momentum of electron,

$$p_{e}=\frac{h}{\lambda}=\frac{6.63×10^{\normalsize-34}}{10^{\normalsize-9}}\\\text{Momentum of photon}\\\text{p}_{ph}=\frac{h}{\lambda}=\frac{6.63×10^{\normalsize-34}}{10^{\normalsize-9}}$$

= 6.63 × 10–25 m

(b) Energy of photon,

$$\text{E}=\frac{hc}{\lambda}=\frac{6.63×10^{\normalsize-34}×3×10^{8}}{10^{\normalsize-9}×1.6×10^{\normalsize-19}}\\=\frac{19.86×10^{\normalsize-17}}{1.6×10^{\normalsize-19}}\text{eV}=1243\space\text{eV}\\\text{(c) Energy of electron}\\\text{E}=\frac{p^{2}}{2me}\\=\frac{(6.63×10^{\normalsize-25})^{2}}{2×9.1×10^{\normalsize-31}×1.6×10^{\normalsize-19}}\text{eV}$$

= 1.51 eV

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 300K.

Sol. (a) de-Broglie wavelength l = 1.40 × 10–10 m

Mass of neutron, mn = 1.675 × 10–27 kg

The wavelength associated with kinetic energy

$$\lambda=\frac{h}{\sqrt{2\text{mKE}}}\\\text{or}\space\text{KE}=\frac{h^{2}}{2\lambda^{2}m_n}\\=\frac{(6.63×10^{\normalsize-34})^{2}}{2×(1.40×10^{\normalsize-10})^{2}×1.675×10^{\normalsize-27}}$$

= 6.686 × 10–21J

(b) Kinetic energy associated with temperature

$$\text{KE}=\frac{3}{2}\text{kT}=\frac{3}{2}(1.38×10^{\normalsize-23})×300$$

= 6.21 × 10–21J

KE = 6.21 × 10–21J

de-Broglie wavelength associated with kinetic energy

$$\lambda=\frac{h}{\sqrt{2\text{mKE}}}\\=\frac{6.63×10^{\normalsize-34}}{\sqrt{2×1.675× 10^{\normalsize-27}×6.21×10^{\normalsize-21}}}$$

= 1.45 × 10–10m = 1.45Å

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

Sol. For the momentum of a photon of frequency n, wavelength λ is given by

$$p=\frac{hv}{c}=\frac{h}{\lambda}\\\lambda=\frac{h}{p}\\\text{de-Broglie wavelength of photon}\\\lambda=\frac{h}{mv}\\\Rarr\space\frac{h}{p}=\frac{h}{hv/c}=\frac{c}{v}$$

Thus, the wavelength of electromagnetic radiation is equal to the de-Broglie wavelength.

19. What is the de-Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)

Sol. Molecular weight of nitrogen molecule  m = 28.0152u

= 28.0152 × 1.67 × 10–27 kg

Mean kinetic energy of molecules

$$\frac{1}{2}mv^{2}=\frac{3}{2}\text{kT}\\\text{or}\space v=\sqrt{\frac{3KT}{m}}\\=\sqrt{\frac{3×1.38×10^{-23}×300}{28.0152×1.66×10^{-27}}}$$

= 516.78 m/s

de-Broglie wavelength

$$\lambda=\frac{h}{mv}\\=\lambda=\frac{6.63×10^{\normalsize-34}}{28.0152×1.66×10^{\normalsize-27}×516.78}$$

= 2.75 × 10–11 m

= 0.0275 × 10–9 m = 0.028 nm