NCERT · Class 12 · Physics

Chapter 11: Dual Nature of Radiation and Matter

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32 questions

Q11.16Written answer

An electron and a photon each have a wavelength of 1.00 nm. Find the following:

(a) Momenta

(b) The energy of the photon

(c) The kinetic energy of the electron

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Worked solution

Wavelength of an electron

\(\begin{array}{l}\lambda _{e}\end{array} \)

\(\begin{array}{l}\lambda _{p}\end{array} \)

\(\begin{array}{l}\lambda _{e}\end{array} \)

\(\begin{array}{l}\lambda _{p}\end{array} \)

\(\begin{array}{l}\lambda\end{array} \)

= 1 x 10^-9 m

Planck’s constant, h = 6.63 x 10^-34 Js

(a) The momentum of an elementary particle is given by de Broglie relation:

\(\begin{array}{l}\lambda = \frac{h}{p}\end{array} \)

\(\begin{array}{l}p=\frac{h}{\lambda }\end{array} \)

It is clear that momentum depends only on the wavelength of the particle. Since the wavelengths of an electron and a photon are equal, both have an equal momentum.

Therefore, p =

\(\begin{array}{l}\frac{6.63\times 10^{-34}}{1\times 10^{-9}}\end{array} \)

= 6.63 x 10^-25 Kg m/s

Therefore, both the electron and photon have an equal momentum of 6.63 x 10^-25 Kg m/s.

(b) The energy of a photon is given by the relation:

E =

\(\begin{array}{l}\frac{hc}{\lambda }\end{array} \)

Where,

Speed of light, c = 3 x 10⁸ m/s

Therefore, E =

\(\begin{array}{l}\frac{6.63\times 10^{-34}\times 3\times 10^{8}}{1\times 10^{-9}\times 1.6\times 10^{-19}}\end{array} \)

= 1243.1 eV = 1.243 keV

Therefore, the energy of the photon is 1.243 keV.

(c) The kinetic energy (K) of an electron having momentum p, is given by the relation:

K =

\(\begin{array}{l}\frac{1}{2}\frac{p^{2}}{m}\end{array} \)

Where, m = Mass of the electron = 9.1 x 10^-31Kg

p = 6.63 x 10^-25 Kg m/s

Therefore, K =

\(\begin{array}{l}\frac{1}{2}\times \frac{(6.63\times 10^{-25})^{2}}{9.1\times 10^{-31}}\end{array} \)

= 2.415 x 10^-19 J

\(\begin{array}{l}\frac{2.415\times 10^{-19}}{1.6\times 10^{-19}}\end{array} \)

= 1.51 eV

Therefore, 1.51eV is the kinetic energy of the electron.

Q11.17Written answer

(a) For what kinetic energy of a neutron will the associated de Broglie wavelength be 1.40 x 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 300 K.

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Worked solution

(a) de Broglie wavelength of the neutron,

\(\begin{array}{l}\lambda\end{array} \)

^-10

Mass of a neutron, m_n = 1.66 x 10^-27 Kg

Planck’s constant, h = 6.63 x 10^-34 Js

Kinetic energy (K) and velocity (v) are related as:

K =

\(\begin{array}{l}\frac{1}{2}m_{n}v^{2}\end{array} \)

……… (1)

de Broglie wavelength (

\(\begin{array}{l}\lambda\end{array} \)

\(\begin{array}{l}\lambda = \frac{h}{m_{n}v}\end{array} \)

………. (2)

Using equation (2) and equation (1), we get

K =

\(\begin{array}{l}\frac{1}{2}\frac{m_{n}h^{2}}{\lambda ^{2}m_{n}^{2}}\end{array} \)

\(\begin{array}{l}\frac{h^{2}}{2\lambda ^{2}m_{n}}\end{array} \)

\(\begin{array}{l}\frac{(6.63 \times 10^{-34})^{2}}{2\times (1.40\times 10^{-10})^{2} \times 1.66\times 10^{-27} }\end{array} \)

= 6.75 x 10^-21 J

\(\begin{array}{l}\frac{6.75\times 10^{-21}}{1.6\times 10^{-19}}\end{array} \)

= 4.219 x10^-2 eV

Hence, the kinetic energy of the neutron is 6.75 x 10^-21 J or 4.219 x 10^-2 eV.

(b) Temperature of the neutron, T = 300K

Boltzmann constant, k = 1.38 x 10^-23 Kg m² s^-2 K^-1

The average kinetic energy of the neutron:

K’ =

\(\begin{array}{l}\frac{3}{2}\end{array} \)

^-23

= 6.21 x 10^-21 J

The relation for the de Broglie wavelength is given as:

\(\begin{array}{l}\lambda ‘=\frac{h}{\sqrt{2K’m_{n}}}\end{array} \)

Where,

m_n = 1.66 x 10^-27 Kg

h = 6.63 x 10^-34 Js

K’ = 6.21 x 10^-21 J

Therefore,

\(\begin{array}{l}\lambda ‘=\frac{6.63\times 10^{-34}}{\sqrt{2\times 6.21\times 10^{-21}\times 1.66\times 10^{-27}}}\end{array} \)

= 1.46 × 10^-10 m = 0.146 nm

Therefore, 0.146nm is the de Broglie wavelength of the neutron.

Q11.18Written answer

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

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Worked solution

The momentum of a photon having energy (hv) is given as:

p =

\(\begin{array}{l}\frac{hv}{c}\end{array} \)

\(\begin{array}{l}\frac{h}{\lambda }\end{array} \)

\(\begin{array}{l}\lambda = \frac{h}{p}\end{array} \)

………. (i)

Where,

\(\begin{array}{l}\lambda\end{array} \)

c = speed of light

h = Planck’s constant

De Broglie wavelength of the photon is given as:

\(\begin{array}{l}\lambda = \frac{h}{mv}\end{array} \)

But, p = mv

Therefore,

\(\begin{array}{l}\lambda = \frac{h}{p}\end{array} \)

……………(ii)

Where, m = mass of the photon

v = velocity of the photon

From equations (i) and (ii), it can be concluded that the wavelength of the electromagnetic radiation and the de Broglie wavelength of the photon is equal.

Q11.19Written answer

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)

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Worked solution

The temperature of the nitrogen molecule, T = 300 K

The atomic mass of nitrogen = 14.0076 u

Hence, the mass of the nitrogen molecule, m = 2 × 14.0076 = 28.0152 u

But, 1 u = 1.66 × 10^-27 kg

Therefore, m = 28.0152 ×1.66 × 10^-27 kg

Planck’s constant, h = 6.63 × 10^-34 Js

Boltzmann constant, k = 1.38 × 10^-23 J/K

We have the expression that relates mean kinetic energy

\(\begin{array}{l}(\frac{3}{2}kT)\end{array} \)

_rms

\(\begin{array}{l}\frac{1}{2}mv_{rms}^{2}\end{array} \)

\(\begin{array}{l}(\frac{3}{2}kT)\end{array} \)

V_rms =

\(\begin{array}{l}\sqrt{\frac{3kT}{m}}\end{array} \)

For nitrogen molecule, the de Broglie wavelength is given as:

\(\begin{array}{l}\lambda =\frac{h}{mv_{rms}}\end{array} \)

\(\begin{array}{l}\frac{h}{\sqrt{3mkT}}\end{array} \)

\(\begin{array}{l}\frac{6.63\times 10^{-34}}{\sqrt{3\times 28.0152\times 1.66\times 10^{-27}\times 1.38\times 10^{-23}\times 300}}\end{array} \)

= 0.028 x 10^-9 m = 0.028 nm

Therefore, the de Broglie wavelength of the nitrogen molecule is 0.028 nm.

Q11.20Written answer

(a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the low initial speeds of the electrons. The specific charge of the electron, i.e. its e/m, is given to be 1.76 × 10¹¹ C kg^–1.
(b) Use the same formula you employ in (a) to obtain electron speed for a collector potential of 10 MV. Do you see what is wrong? In what way is the formula to be modified?

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Worked solution

(a) Potential difference of the evacuated tube = 500 V

Specific charge of the electron, e/m = 1.76 × 10¹¹ C kg^–1

Kinetic energy = (1/2) mv² = eV

Speed of the emitted electron, v = (2Ve/m)^1/2

= (2 x 500 x 1.76 x 10¹¹)^1/2

= 1.32 x 10⁷ m/s.

(b) Collector potential, V = 10 MV = 10 x 10⁶ V.

Speed of electron = v = (2Ve/m)^1/2

= (2 x 10⁷ x 1.76 x 10¹¹)^1/2

= 1.88 x 10⁹ m/s.

This answer is not correct. Since the value is greater than the speed of light (c), the expression (1/2) mv² for energy should be used in the non -relativistic limit, i.e., v << c.

In the relativistic limits, the total energy is given as

E = mc²

Here,

m is the relativistic mass

m = m₀ (1- v²/c²)^1/2

m₀ = mass of the particle at rest

Kinetic energy is given as

K = mc² – m₀c².

Q11.21Written answer

(a) A monoenergetic electron beam with an electron speed of 5.20 × 10⁶ m s^–1 is subject to a magnetic field of 1.30 × 10^–4 T normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals 1.76 × 10¹¹C kg^–1?
(b) Is the formula you employ in (a) valid for calculating the radius of the path of a 20 MeV electron beam? If not, in what way is it modified?

[Note: Exercises 11.20 (b) and 11.21 (b) take you to relativistic mechanics, which is beyond the scope of this book. They have been inserted here simply to emphasise the point that the formulas you use in part (a) of the exercises are not valid at very high speeds or energies. See answers at the end to know what ‘very high speed or energy’ means.]

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Worked solution

Speed of the electron, v = 5.20 × 10⁶ m s^–1

Magnetic field experienced by the electron, B = 1.30 × 10^–4 T

Specific charge, e/m = 1.76 x 10¹¹ Ckg^-1

Here,

e = charge on the electron = 1.6 x 10^-19 C

m = mass of the electron = 9.1 x 10^-31 kg^-1

Force exerted on the electron is given as

\(\begin{array}{l}F = e\left | \vec{v}\times \vec{B} \right |\end{array} \)

= evBsinθ

θ is the angle between the magnetic field and the velocity of the beam. The magnetic field is normal in the direction of the beam.

θ = 90⁰

F = evB

The normal magnetic field provides the centripetal force.

Therefore, evB = mv²/r

r = mv/eB = v/(e/m)B

= (5.20 x 10⁶)/ (1.76 x 10¹¹) x (1.30 x 10^-4) = 0.227 m = 22.7 cm

Therefore, the radius of the circular path is 22.7 cm.

(b) Energy of the electron beam, E = 20 Mev = 20 x 10⁶ x 1.6 x 10^-19 J

The energy of the electron beam, E = (1/2) mv²

⇒

\(\begin{array}{l}v =\sqrt{\frac{2E}{m}}\end{array} \)

\(\begin{array}{l}v =\sqrt{\frac{2\times 20\times 10^{6}\times1.6\times 10^{-19} }{9.1\times 10^{-31}}} = 2.652 \times 10^{9}m/s\end{array} \)

The result is greater than the speed of light. Therefore, it is wrong. The expression (1/2) mv² for energy should be used in the non -relativistic limit, i.e., v << c.

In the relativistic limits, the total mass is given as

m = m₀ (1- v²/c²)^1/2

m₀ = mass of the particle at rest

Therefore, the radius of the circular path is

r = mv/eB

\(\begin{array}{l}r = \frac{m_{0}v}{eB\sqrt{\frac{c^{2}-v^{2}}{c^{2}}}}\end{array} \)

Q11.22Written answer

An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure (∼10^–2 mm of Hg). A magnetic field of 2.83 × 10^–4 T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method.) Determine e/m from the data.

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Worked solution

Potential of the collector, V = 100 V

Magnetic field experienced by the electron, B = 2.83 × 10^–4 T

Radius of the circular orbit, r = 12 cm = 12.0 x 10^-2 m

Kinetic energy, (1/2)mv² = eV

v² = 2eV/m ——–(1)

The magnetic field that curves the path of the electron provides the centripetal force

evB = mv²/r

eB = mv/r

v = eBr/m ——–(2)

Substituting (2) in (1)

\(\begin{array}{l}\frac{2eV}{m}=\frac{e^{2}B^{2}r^{2}}{m^{2}}\end{array} \)

\(\begin{array}{l}\frac{e}{m}=\frac{2V}{B^{2}r^{2}}\end{array} \)

\(\begin{array}{l}\frac{e}{m}=\frac{2\times 100}{(2.83\times 10^{-4})^{2}(12\times 10^{-2})^{2}} = 1.73 \times 10^{11}Ckg^{-1}\end{array} \)

Therefore, the specific charge ration e/m is 1.73 x 10¹¹ Ckg^-1.

Q11.23Written answer

Question 11. 23: (a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength ending at 0.45 Å. What is the maximum energy of a photon in radiation?
(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?

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Worked solution

(a) Wavelength produced by the X-ray tube, λ = 0.45 Å = 0.45 x 10^-10 m

Speed of light, c = 3 x 10⁸ m/s

Planck’s constant, h = 6.626 x 10^-34 Js

The maximum energy of a photon is given as

E_max = hc/λ_min

= (6.626 x 10^-34)(3 x 10⁸ m/s)/(0.45 x 10^-10 m x 1.6 x 10^-19)

= 19.878 x 10^-26/0.72 x 10^-29

= 27.60 x 10³ eV = 27.6 keV.

(b) To incident electron should have an energy of 27.6 keV to get an X-ray of 27.6 keV. Therefore, the accelerating voltage of the order of 30 keV is required for producing X-rays.

Q11.24Written answer

In an accelerator experiment on high-energy collisions of electrons with positrons, a certain event is interpreted as the annihilation of an electron-positron pair of total energy 10.2 BeV into two γ-rays of equal energy. What is the wavelength associated with each γ-ray? (1BeV = 10⁹ eV)

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Worked solution

Total energy of the electron-positron pair, E = 10.2 BeV = 10.2 x 10⁹ eV = 10.2 x 10⁹ x 1.6 x 10^-19 J

Hence, the energy of each γ-ray , E’ = E/2 = (10.2 x 10⁹ x 1.6 x 10^-19 )/2 = 8.16 x 10^-10 J

Energy and wavelength relation is given as,

E’ = hc/λ

Therefore, λ = hc/E’

Here, h = 6.626 x 10^-34 Js

c = 3 x 10⁸ m/s

\(\begin{array}{l}\lambda = \frac{6.626\times 10^{-34}\times 3\times 10^{8}}{8.16\times 10^{-10}}= 2.436 \times 10^{-16}m\end{array} \)

Therefore, the wavelength associated with each γ-ray is 2.436 x 10^-16 m.

Q11.25Written answer

Estimating the following two numbers should be interesting. The first number will tell you why radio engineers do not need to worry much about photons. The second number tells you why our eye can never ‘count photons’, even in barely detectable light.
(a) The number of photons emitted per second by a medium wave transmitter of 10 kW power, emitting radio waves of wavelength 500 m.
(b) The number of photons entering the pupil of our eye per second corresponding to the minimum intensity of white light that we humans can perceive (∼10^–10 W m^–2). Take the area of the pupil
to be about 0.4 cm², and the average frequency of white light to be about 6 × 10¹⁴ Hz.

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Worked solution

(a) Power of the medium wave transmitter, P = 10 kW = 10⁴ W

The energy emitted by the transmitter per second, E = 10⁴

The wavelength of the radio waves, λ = 500 m

The energy of the wave is given as, E’ = hc/λ

E’ = (6.6 x 10^-34 x 3 x 10⁸)/500

= 3.96 x 10^-28 J

Let n be the number of photons emitted by the transmitter

nE’ = E

n = E/E’

= 10⁴/(3.96 x 10^-28 )

= 0.2525 x 10³²

The energy E’ of the radio photon is very less, but the number of photons emitted is large. The total energy of the radio ways can be considered as continuous, and the existence of the minimum quantum energy can be ignored.

(b) Intensity of the light perceived by the human eye, I = 10^–10 W m^–2

Area of the pupil, A = 0.4 cm² = 0.4 x 10^-4 m²

Frequency of the white light, ν = 6 x 10¹⁴ Hz

h = Planck’s constant = 6.6 x 10^-34 Js

The energy of the emitted photon, E = hν

= 6.6 x 10^-34 x6 x 10¹⁴

= 3.96 x 10^-19 J

Let n be the total number of photons falling per unit area per unit time. The total energy per unit for n photons is

E = n x 3.96 x 10^-19 J/s/m²

The total energy per unit for n photons is equal to the intensity of the light.

E = I

I = n x 3.96 x 10^-19 J/s/m²

n = I/3.96 x 10^-19

= 10^-10/3.96 x 10^-19

= 2.52 x 10⁸ m²/s

The total number of photons entering the pupil is given as,

nA = 2.52 x 10⁸ x 0.4 x 10^-4

= 1.008 x 10⁴ s^-1

The number is large, so the human eye can never count the number of individual photons.

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Chapter 11: Dual Nature of Radiation and Matter

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