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Physics MCQs for NEET — Practice Questions with Answers

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The cathode of a photoelectric cell is changed such that the work function changes from $W_{1}$ to $W_{2}$ $(W_{2} > W_{1})$ . If the current before and after the change are $I_{1}$ and $I_{2}$ , all other conditions remaining unchanged, then (assuming $hν > W_{2}$ ) :

$I_{1}$ = $I_{2}$

$I_{1}$ < $I_{2}$

$I_{1}$ > $I_{2}$

$I_{1} < I_{2} < 2 I_{1}$

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Explanation

(a) The work function has no effect on current so long as $hν > W_{0}$ . The photoelectric current is proportional to the intensity of light. Since there is no change in the intensity of light, therefore $I_{1}$ = $I_{2}$ .

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A beam of light of wavelength $λ$ and with illumination L falls on a clean surface of sodium. If N photoelectrons are emitted each with kinetic energy E, then

N ∝ L and E ∝ L

N ∝ L and $E \propto \frac{1}{λ}$

$N \propto λ$ and E ∝ L

$N \propto \frac{1}{λ}$ and $E \propto \frac{1}{L}$

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Explanation

(b) Number of photons emitted is proportional to the intensity. Also $\frac{hc}{λ} = W_{0} + E$

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

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

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