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Dual Nature of Matter and Radiation MCQs for NEET — Physics Questions with Answers

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Which of the following statements is correct

The current in a photocell increases with increasing frequency of light

The photocurrent is proportional to applied voltage

The photocurrent increases with increasing intensity of light

The stopping potential increases with increasing intensity of incident light

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Explanation

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For intensity I of a light of wavelength 5000Å the photoelectron saturation current is 0.40 $μA$ and stopping potential is 1.36 V, the work function of metal is

2.47 eV

1.36 eV

1.10 eV

0.43 eV

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Explanation

$E = \frac{12375}{5000} = 2.475 eV and K_{\max} = {eV}_{0} = 1.36 eV So 2.475 = W_{0} + 1.36 \Rightarrow \Rightarrow W_{0} = 1.1 eV$

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The work functions of metals A and B are in the ratio 1 : 2. If light of frequencies f and 2f are incident on the surfaces of A and B respectively, the ratio of the maximum kinetic energies of photoelectrons emitted is (f is greater than threshold frequency of A, 2f is greater than threshold frequency of B)

1 : 1

1 : 2

1 : 3

1 : 4

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Explanation

(b)

$E = W_{0} + K_{\max} . . . . (i) \Rightarrow hf = W_{A} + K_{A} . . . (ii)$

and $2 hf = W_{B} + K_{B} = 2 W_{A} + K_{B} (∵ \frac{W_{A}}{W_{B}} = \frac{1}{2})$

Dividing equation (i) and (ii)

$\frac{1}{2} = \frac{W_{A} + K_{A}}{2 W_{A} + K_{B}} \Rightarrow \frac{K_{A}}{K_{B}} = \frac{1}{2}$

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Light of frequency v is incident on a substance of threshold frequency $v_{0} (v_{0} < v)$ . The energy of the emitted photo-electron will be

$h (v - v_{0})$

h/v

$he (v - v_{0})$

$h / v_{0}$

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Explanation

(a) Work function, $W = {hv}_{0}$

Energy of incident light = hv

Energy of emitted electrons = E- ${hv}_{0}$

= $h (v - v_{0})$

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4 eV is the energy of the incident photon and the work function in 2eV. What is the stopping potential ?

$2 \sqrt{2} V$

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Explanation

(a) $E = W_{0} + {eV}_{0} \Rightarrow 4 eV = 2 eV + {eV}_{0} \Rightarrow V_{0} = 2 volt$

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The number of photons of wavelength 540 nm emitted per second by an electric bulb of power 100W is (taking h = $6 \times 10^{- 34}$ J-sec)

100

1000

$3 \times 10^{20}$

$3 \times 10^{18}$

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Explanation

(c) $p = \frac{nhc}{λt} \Rightarrow 100 = \frac{n \times 6 \times 10^{- 34} \times 3 \times 10^{8}}{540 \times 10^{- 9} \times 1} \Rightarrow n = 3 \times 10^{20}$

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Light of frequency 4 $v_{0}$ is incident on the metal of the threshold frequency $v_{0}$ . The maximum kinetic energy of the emitted photoelectrons is

$3 {hv}_{0}$

$2 {hv}_{0}$

$\frac{3}{2} {hv}_{0}$

$\frac{1}{2} {hv}_{0}$

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Explanation

(a) $E = {hv}_{0} + K_{\max} \Rightarrow h (4 v_{0}) = {hv}_{0} + K_{\max} \Rightarrow K_{\max} = 3 {hv}_{0}$

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Two identical photo-cathodes receive light of frequencies $f_{1}$ and $f_{2}$ . If the velocities of the photo electrons (of mass m) coming out are respectively $v_{1}$ and $v_{2}$ , then

$v_{1} - v_{2} = {[\frac{2 h}{m} (f_{1} - f_{2})]}^{1 / 2}$

$v_{1}^{2} - v_{2}^{2} = \frac{2 h}{m} (f_{1} - f_{2})$

$v_{1} + v_{2} = {[\frac{2 h}{m} (f_{1} + f_{2})]}^{1 / 2}$

$v_{1}^{2} + v_{2}^{2} = \frac{2 h}{m} (f_{1} + f_{2})$

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Explanation

(b) Using Einstein photoelectric equation

$E = W_{0} + K_{\max}$

${hf}_{1} = W_{0} + \frac{1}{2} {mv}_{1}^{2} . . . (i) {hf}_{2} = W_{0} + \frac{1}{2} {mv}_{2}^{2} . . . (ii) \Rightarrow h (f_{1} - f_{2}) = \frac{1}{2} m (v_{1}^{2} - v_{2}^{2}) \Rightarrow (v_{1}^{2} - v_{2}^{2}) = \frac{2 h}{m} (f_{1} - f_{2})$

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When radiation of wavelength $λ$ is incident on a metallic surface, the stopping potential is 4.8 volts. If the same surface is illuminated with radiation of double the wavelength, then the stopping potential becomes 1.6 volts. Then the threshold wavelength for the surface is

$2 λ$

$4 λ$

$6 λ$

$8 λ$

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Explanation

(b) By using

$\frac{hc}{e} (\frac{1}{λ} - \frac{1}{λ_{0}}) = V_{0}$

$\Rightarrow \frac{hc}{e} (\frac{1}{λ} - \frac{1}{λ_{0}}) = 4.8$ .....(i)

and $\Rightarrow \frac{hc}{e} (\frac{1}{2 λ} - \frac{1}{λ_{0}}) = 1.6$ .....(ii)

From equation (i) and (ii)

$\frac{(\frac{1}{λ} - \frac{1}{λ_{0}})}{(\frac{1}{2 λ} - \frac{1}{λ_{0}})} = \frac{4.8}{1.6} \Rightarrow λ_{0} = 4 λ$

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If the energy of the photon is increased by a factor of 4, then its momentum

Does not change

Decreases by a factor of 4

Increases by a factor of 4

Decreases by a factor of 2

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Explanation

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Yes. You can attempt every Dual Nature of Matter and Radiation question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.