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Physics MCQs for NEET — Practice Questions with Answers

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The photo-electrons emitted from a surface of sodium metal are such that

They all are of the same frequency

They have the same kinetic energy

They have the same de Broglie wavelength

They have their speeds varying from zero to a certain maximum

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Explanation

The photo-electrons have kinetic energy in varying proportion.

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In a photo cell, the photo-electrons emission takes place

After $10^{- 1}$ sec on incident of light rays

After $10^{- 3}$ sec on incident of light rays

After $10^{- 6}$ sec on incident of light rays

After $10^{- 8}$ sec on incident of light rays

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Explanation

(d)

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When light falls on a metal surface, the maximum kinetic energy of the emitted photo-electrons depends upon

The time for which light falls on the metal

Frequency of the incident light

Intensity of the incident light

Velocity of the incident light

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Explanation

(b) $K_{\max} = (hv - W_{0})$ ; v = frequency of incident light.

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The electrons are emitted in the photoelectric effect from a metal surface

Only if the frequency of the incident radiation is above a certain threshold value

Only if the temperature of the surface is high

At a rate that is independent of the nature of the metal

With a maximum velocity proportional to the frequency of the incident radiation

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Explanation

(a) Refer to threshold frequency

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The work function of a metal is 4.2 eV, its threshold wavelength will be

4000 Å

3500 Å

2955 Å

2500 Å

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Explanation

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The number of photo-electrons emitted per second from a metal surface increases when

The energy of incident photons increases

The frequency of incident light increases

The wavelength of the incident light increases

The intensity of the incident light increases

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Explanation

(d) Intensity ∝ (No. of photons) ∝ (No. of photoelectrons)

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The work function of metal is 1 eV. Light of wavelength 3000 Å is incident on this metal surface. The velocity of emitted photo-electrons will be

10 m/sec

$1 \times 10^{3}$ m/sec

$1 \times 10^{4}$ m/sec

$1 \times 10^{6}$ m/sec

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Explanation

(d)

$E = W_{0} + K_{\max}; E = \frac{12375}{3000} = 4.125 eV \Rightarrow K_{\max} = E - W_{0} = 4.125 eV - 1 eV = 3.125 eV \Rightarrow \frac{1}{2} {mv}_{\max}^{2} = 3.125 \times 1.6 \times 10^{- 19} J \Rightarrow v_{\max} = \sqrt{\frac{2 \times 3.125 \times 1.6 \times 10^{- 19}}{9.1 \times 10^{- 31}}} = 1 \times 10^{6} m / s$

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The work function of a metal is $1.6 \times 10^{- 19}$ J. When the metal surface is illuminated by the light of wavelength 6400 Å, then the maximum kinetic energy of emitted photo-electrons will be
(Planck's constant = $6.4 \times 10^{- 34} Js$ )

$14 \times 10^{- 19} J$

$2.8 \times 10^{- 19} J$

$1.4 \times 10^{- 19} J$

$1.4 \times 10^{- 19} eV$

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Explanation

(c) $K_{\max} = \frac{hc}{λ} - W_{0} = \frac{6.4 \times 10^{- 34} \times 3 \times 10^{8}}{6400 \times 10^{- 10}} - 1.6 \times 10^{- 19} = 1.4 \times 10^{- 19} J$

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Ultraviolet radiations of 6.2 eV falls on an aluminium surface (work function 4.2 eV ). The kinetic energy in joules of the fastest electron emitted is approximately

$3.2 \times 10^{- 21}$

$3.2 \times 10^{- 19}$

$3.2 \times 10^{- 17}$

$3.2 \times 10^{- 15}$

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Explanation

(b) $K_{\max} (eV) = E (eV) - W_{0} (eV) = 6.2 - 4.2 = 2 eV ∴ K_{\max} (Joules) = 2 \times 1.6 \times 10^{- 19} J = 3.2 \times 10^{- 19} J$

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The work function for tungsten and sodium are 4.5 eV and 2.3 eV respectively. If the threshold wavelength $λ$ for sodium is 5460 Å, the value of $λ$ for tungsten is

5893 Å

10683 Å

2791 Å

528 Å

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Explanation

$W_{0} = \frac{hc}{λ_{0}}; ∴ \frac{{(W_{0})}_{T}}{{(W_{0})}_{Na}} = \frac{λ_{Na}}{λ_{T}} or λ_{T} = \frac{λ_{Na} \times {(W_{0})}_{Na}}{{(W_{0})}_{T}} = \frac{5460 \times 2.3}{4.5} = 2791 Å$

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