Ncert Solution for Class 12 Physics Chapter 11 Dual Nature of Radiation and Matter

Dual Nature of Radiation and Matter ncert solutions: Ncert Class 12 Physics Chapter 11 Exercise Solutions

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Question 11.1: Find the
(a) maximum frequency, and
(b) minimum wavelength of X-rays produced by 30 kV electrons.

Solution 11.1:

• Electron potential, V = 30 kV = 3 × 10⁴ V
• Hence, electron energy, E = 3 × 10⁴ eV
• Where, e = Charge on one electron = 1.6 × 10^-19 C

(a) Maximum frequency by the X-rays = ν

The energy of the electrons: E = hν

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Where,

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

Therefore,

\(v = \frac{E}{h}\)

= \(\frac{1.6\times 10^{-19}\times 3 \times 10^{4}}{6.626\times 10^{-34}}\)

= 7.24 x 10¹⁸ Hz

Hence,  7.24 x 10¹⁸ Hz is the maximum frequency of the X-rays.

(b) The minimum wavelength produced = \(\lambda =\frac{c}{v}\)

= \(\frac{3\times 10^{8}}{7.24\times 10^{18}}\)

= 4.14 x 10^-11 m = 0.0414 nm

Hence, the minimum wavelength of X-rays produced is 0.0414 nm.

Question 11.2: The work function of caesium metal is 2.14 eV. When light of frequency 6 ×10¹⁴Hz 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?

Solution 11.2: Work function of caesium, \(\Phi _{o}\) = 2.14eV

Frequency of light, v = 6.0 x 10¹⁴ Hz

(a) The maximum kinetic energy of the emitted electrons:

K = hν – \(\Phi _{o}\)

Where,

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

Therefore,

K = \(\frac{6.626\times 10^{-34}\times 6\times 10^{14}}{1.6\times 10^{-19}}\) – 2.14

= 2.485 – 2.140 = 0.345 eV

Hence, 0.345 eV is the maximum kinetic energy of the emitted electrons.

(b) For stopping potential V_o, we can write the equation for kinetic energy as:

K = eV_o

Therefore, V_o = \(\frac{K}{e}\)

= \(\frac{0.345\times 1.6\times 10^{-19}}{1.6\times 10^{-19}}\)

= 0.345 V

Hence, 0.345 V is the stopping potential of the material.

(c) Maximum speed of photoelectrons emitted = ν

Following is the kinetic energy relation:

K = \(\frac{1}{2}mv^{2}\)

Where,

m = mass of electron = 9.1 x 10^-31 Kg

\(v^{2}=\frac{2K}{m}\)

= \(\frac{2\times 0.345\times 1.6\times 10^{-19}}{9.1\times 10^{-31}}\)

= 0.1104 x 10¹²

Therefore, ν = 3.323 x 10⁵ m/s = 332.3 km/s

Hence, 332.3 km/s is the maximum speed of the emitted photoelectrons.

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?

Solution 11.3: The maximum kinetic energy of the emitted photoelectrons can be determined using the photoelectric equation, which relates the kinetic energy of the photoelectrons to the stopping (cut-off) voltage.

The kinetic energy K_max of the photoelectrons is given by:

K_max = eV_cutoff

where:

• e = 1.602 × 10^-19 C (elementary charge),
• eV_cutoff = 1.5 V (cut-off voltage, as given).

Substituting the values:

K_max = 1.602 × 10^-19 C × 1.5 V = 2.403 × 10^-19 J

Thus, the maximum kinetic energy of the photoelectrons is:

K_max = 2.4 × 10^-19 J

So, the maximum kinetic energy of the emitted photoelectrons is 2.4 × 10^-19 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? (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?

Solution 11.4: Monochromatic light having a wavelength, λ = 632.8 nm = 632.8 × 10^-9 m

Given that the laser emits the power of, P = 9.42 mW = 9.42 × 10^-3 W

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

Speed of light, c = 3 × 10⁸ m/s

Mass of a hydrogen atom, m = 1.66 × 10^-27 kg

(a) The energy of each photon in the light beam:

E = \(\frac{hc}{\lambda }\)

= \(\frac{6.626\times 10^{-34}\times 3\times 10^{8}}{632.8\times 10^{-9}}\)

= 3.141 x 10^-19 J.

Therefore, each photon has a momentum of:

P = \(\frac{h}{\lambda }\)

= \(\frac{6.626\times 10^{-34}}{632.8\times 10^{-9}}\)

= 1.047 x 10^-27 kg m/s.

(b) Number of photons/second arriving at the target illuminated by the beam  = n

Assuming the uniform cross-sectional area of the beam is less than the target area,

The equation for power is written as, P = nE

Therefore, n = \(\frac{P}{E}\)

= \(\frac{9.42\times 10^{-3}}{3.141\times 10^{-19}}\)

= 3 x 10¹⁶ photons/s.

(c) Given that the momentum of the hydrogen atom is equal to the momentum of the photon,

P = 1.047 x 10^-27 kg m/s

Momentum is given as: P = mv

Where,

ν = speed of hydrogen atom

Therefore, ν = \(\frac{p}{m}\)

= \(\frac{1.047\times 10^{-27}}{1.66\times 10^{-27}}\)

= 0.621 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 × 10^–15 V s. Calculate the value of Planck’s constant.

Solution 11.5: The slope of the cut-off voltage versus frequency graph in a photoelectric effect experiment is related to Planck’s constant. The relationship is given by Einstein’s photoelectric equation:

\(eV_{\text{cut-off}} = h\nu – \phi\)

Where:

• e is the charge of an electron ( 1.6 × 10^–19 C ),
• V_cut-off is the cut-off voltage,
• h is Planck’s constant,
• V is the frequency of the incident light,
• \( \phi \) is the work function (which is irrelevant for this specific calculation).

From the equation, the slope of the cut-off voltage versus frequency graph is:

\(\text{slope} = \frac{h}{e}\)

We are given the slope as \( 4.12 \times 10^{-15} \, \text{V} \cdot \text{s} \). Therefore:

\(\frac{h}{e} = 4.12 \times 10^{-15} \, \text{V} \cdot \text{s}\)

To find ( h ), we multiply both sides of the equation by ( e ):

h = (4.12 × 10^-15 × (1.6 × 10^-19) C

Now calculate:

h = 4.12 × 10^-15 × 1.6 × 10^-19 = 6.592 × 10^-34 J . s

Thus, the value of Planck’s constant is approximately:

h = 6.59 × 10^-34 J . s

Question 11.6: 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 cutoff voltage for the photoelectric emission.

Solution 11.6: To predict the cutoff voltage for photoelectric emission, we can use the photoelectric equation derived from Einstein’s photoelectric effect theory:

\(K.E. = h \nu – h \nu_0\)

Where:

• K.E. is the kinetic energy of the emitted electrons.
• h is Planck’s constant ( 6.626 × 10^-34 Js ).
• \( \nu \) is the frequency of the incident light.
• \( \nu_0 \) is the threshold frequency for the metal.

The maximum kinetic energy of the emitted electrons is related to the cutoff voltage V_cutoff by the equation:

K.E. = eV_cutoff

Where:

• e is the charge of the electron ( 1.602 × 10^-19 C ).
• V_cutoff is the stopping or cutoff voltage.

Step 1: Calculate the kinetic energy of the emitted electrons.

Using the formula:

\( K.E. = h (\nu – \nu_0)\)

Given:

• \( \nu = 8.2 \times 10^{14} \, \text{Hz} \)
• \( \nu_0 = 3.3 \times 10^{14} \, \text{Hz} \)
• \( h = 6.626 \times 10^{-34} \, \text{Js} \)

K.E. = 6.626 × 10^-34 × (8.2 × 10¹⁴ – 3.3 × 10¹⁴)

K.E. = 6.626 × 10^-34 × 4.9 × 10¹⁴

K.E. = 3.24674 × 10^-19 J

Step 2: Convert the kinetic energy to the cutoff voltage.

Now, use the relationship \(K.E. = e V_{\text{cutoff}} \):

\(V_{\text{cutoff}} = \frac{K.E.}{e}\)

\(V_{\text{cutoff}} = \frac{3.24674 \times 10^{-19}}{1.602 \times 10^{-19}}\)

\(V_{\text{cutoff}} \approx 2.03 \, \text{V}\)

The cutoff voltage for the photoelectric emission is approximately 2.03 V.

Question 11.7: The work function for a certain metal is 4.2 eV. Will this metal give hotoelectric emission for incident radiation of wavelength 330 nm?

Solution 11.7: Work function of the metal, \(\Phi _{o}\) = 4.2eV

Charge on an electron, e = 1.6 x 10^-19 C

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

Wavelength of the incident radiation, \(\lambda\) = 330nm = 330  × 10^-9 m

Speed of light, c = 3 × 10⁸ m/s

The energy of the incident photon is given as: E = \(\frac{hc}{\lambda }\)

= \(\frac{6.626 \times 10^{-34}\times 3\times 10^{8}}{330\times 10^{-9}}\)

= \(\frac{6.0\times 10^{-19}}{1.6\times 10^{-19}}\)

= 3.76 eV

The energy of the incident radiation is less than the work function of the metal. Hence, there is no photoelectric emission takes place.

Question 11.8: 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?

Solution 11.8: Frequency of the incident photon, ν = 488nm = 488 x 10^-9m

The maximum speed of the electrons, v = 6.0 x 10⁵ m/s

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

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

For threshold frequency v_o, the relation for kinetic energy is written as: \(\frac{1}{2}mv^{2}\) = h(ν – ν_o)

ν_o = ν – \(\frac{mv^{2}}{2h}\)

= 7.21 x 10¹⁴ – \(\frac{(9.1 \times 10^{-31})\times (6\times 10^{5})^{2}}{2 \times (6.626 \times 10^{-34})}\)

= 7.21 x 10¹⁴ – 2.472 x 10¹⁴ 

= 4.738 x 10¹⁴ Hz

Therefore, 4.738 x 10¹⁴ Hz is the threshold frequency for the photoemission of the electrons.

Question 11.9: 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 0.38 V. Find the work function of the material from which the emitter is made.

Solution 11.9: The wavelength of light produced by the argon laser, \(\lambda\) = 488nm = 488 x 10^-9 m

Stopping potential of the photoelectrons, V_o = 0.38 V

1eV = 1.6 x 10^-19 J

Therefore, V_o = \(\frac{0.38}{1.6\times 10^{-19}}\)eV

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

Charge on an electron, e = 1.6 x 10^-19 C

Speed of light, c = 3 x 10⁸ m/s

Using Einstein’s photoelectric effect, the following is the relation for the work function \(\Phi _{o}\) of the material of the emitter as:

eV_o = \(\frac{hc}{\lambda } – \phi _{o}\)

\(\Phi _{o}\) = \(\frac{hc}{\lambda }\)– eV_o

= \(\frac{6.6\times 10^{-34}\times 3\times 10^{8}}{1.6\times 10^{-19}\times 488\times 10^{-9}} – \frac{1.6\times 10^{-19}\times 0.38}{1.6\times 10^{-19}}\)

= 2.54 – 0.38 

= 2.16 eV

Therefore, 2.16 eV is the work function of the material with which the emitter is made.

Question 11.10: 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 ?

Solution 11.10: (a) Mass of the bullet, m = 0.040 Kg

The speed of the bullet, v = 1.0 km/s = 1000 m/s

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

de Broglie wavelength of the bullet is given by the relation: \(\lambda = \frac{h}{mv}\)

= \(\frac{6.6\times 10^{-34}}{0.040\times 1000}\)

= 1.65 x 10^-35 m.

(b) Mass of the ball, m = 0.060 Kg

The speed of the ball, v = 1.0 m/s

de Broglie wavelength of the ball is given by the relation: \(\lambda = \frac{h}{mv}\)

= \(\frac{6.6\times 10^{-34}}{0.060\times 1}\)

= 1.1 x 10^-32 m.

(c) Mass of the dust particle, m = 1 x 10^-9 Kg

The speed of the dust particle, v = 2.2 m/s

de Broglie wavelength of the dust particle is given by the relation: \(\lambda = \frac{h}{mv}\)

= \(\frac{6.6\times 10^{-34}}{2.2\times 1\times 10^{-9}}\)

= 3.0 x 10^-25 m.

Question 11.11: Show that the wavelength of electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

Solution 11.11: The momentum of a photon having energy (hv) is given as: p = \(\frac{hv}{c}\) = \(\frac{h}{\lambda }\)

\(\lambda = \frac{h}{p}\) ………. (i)

Where,

\(\lambda\) = wavelength of the electromagnetic radiation

c = speed of light

h = Planck’s constant

De Broglie wavelength of the photon is given as: \(\lambda = \frac{h}{mv}\)

But, p = mv

Therefore,

\(\lambda = \frac{h}{p}\) ……………(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.