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

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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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A photon of energy 3.4 eV is incident on a metal having work function 2 eV. The maximum K.E. of photo-electrons is equal to

1.4 eV

1.7 eV

5.4 eV

6.8 eV

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Explanation

(a) $K_{\max} = (E - W_{0}) = (3.4 - 2) eV = 1.4 eV$

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The photoelectric threshold wavelength for a metal surface is 6600 Å. The work function for this is

1.87 V

1.87 eV

18.7 eV

0.18 eV

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Explanation

(b) $W_{0} = \frac{12375}{6600} = 1.87 eV$

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