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Physics MCQs for NEET — Practice Questions with Answers

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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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Which of one is correct

$E^{2} = p^{2} c^{2}$

$E^{2} = p^{2} c$

$E^{2} = {pc}^{2}$

$E^{2} = p^{2} / c^{2}$

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Explanation

(a) Momentum $p = \frac{E}{c} \Rightarrow E^{2} = p^{2} c^{2}$

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The work function for metals A, B and C are respectively 1.92 eV, 2.0 eV and 5 eV. According to Einstein’s equation, the metals which will emit photo electrons for a radiation of wavelength 4100 Å is/are

None of these

A only

A and B only

All the three metals

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Explanation

$= \frac{12375}{4100} = 3.01 eV$
Work function of metal A and B are less than 3.01 eV , so A and B will emit photo electrons.

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The magnitude of saturation photoelectric current depends upon

Frequency

Intensity

Work function

Stopping potential

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Explanation

(b) The value of saturation current depends on intensity. It is independent of stopping potential

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The light rays having photons of energy 1.8 eV are falling on a metal surface having a work function 1.2 eV. What is the stopping potential to be applied to stop the emitting electrons

3 eV

1.2 eV

0.6 eV

1.4 eV

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Explanation

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A photon and an electron have equal energy E. $λ_{photon} / λ_{electron}$ is proportional to

$\sqrt{E}$

$1 / \sqrt{E}$

1/E

Does not depend upon E

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Explanation

(b) $λ_{photon} = \frac{hc}{E} and λ_{photon} = \frac{h}{\sqrt{2 mE}} \Rightarrow \frac{λ_{photon}}{λ_{electron}} = c \sqrt{\frac{2 m}{E}} \Rightarrow \frac{λ_{photon}}{λ_{electron}} \propto \frac{1}{\sqrt{E}}$

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An image of the sun is formed by a lens of focal length of 30 cm on the metal surface of a photoelectric cell and a photoelectric current I is produced. The lens forming the image is then replaced by another of the same diameter but of focal length 15 cm. The photoelectric current in this case is

$\frac{I}{2}$

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Explanation

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A photon of $1.7 \times 10^{- 13}$ Joules is absorbed by a material under special circumstances. The correct statement is:

Electrons of the atom of absorbed material will go the higher energy states

Electron and positron pair will be created

Only positron will be produced

Photoelectric effect will occur and electron will be produced

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Explanation

(b) For electron and positron pair production, minimum energy is 1.02 MeV.
Energy of photon is given $1.7 \times 10^{- 3}$ J

= $\frac{1.7 \times 10^{- 13}}{1.6 \times 10^{- 19}}$
= 1.06 MeV.
Since energy of photon is greater than 1.02 MeV,
So electron, positron pair will be created.

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The maximum velocity of an electron emitted by light of wavelength $λ$ incident on the surface of a metal of work function $ϕ$ is

${[\frac{2 (hc + λϕ)}{mλ}]}^{1 / 2}$

${[\frac{2 (hc - λϕ)}{mλ}]}^{1 / 2}$

${[\frac{2 (hλ - ϕ)}{m}]}^{1 / 2}$

Where h = Planck's constant, m = mass of electron and c = speed of light.

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Explanation

$\frac{hc}{λ} = ϕ + \frac{1}{2} {mv}^{2} \Rightarrow v = {[\frac{2 (hc - λϕ)}{mλ}]}^{1 / 2}$

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