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Assuming photoemission to take place, the factor by which the maximum velocity of the emitted photoelectrons changes when the wavelength of the incident radiation is increased four times, is

$\frac{1}{4}$

$\frac{1}{2}$

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Explanation

(d) $\frac{hc}{λ} = W_{0} + \frac{1}{2} {mv}_{\max}^{2}$

Assuming $W_{0}$ to be negligible in comparison to $\frac{hc}{λ}$

i.e. $v_{\max}^{2} \propto \frac{1}{λ} \Rightarrow v_{\max} \propto \frac{1}{\sqrt{λ}}$

(On increasing wavelength $λ$ to 4 $λ$ , $v_{\max}$ becomes half).

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If the work function of a metal is $' ϕ'$ and the frequency of the incident light is ' $ν$ ', there is no emission of photoelectron if

$v < \frac{ϕ}{h}$

$v = \frac{ϕ}{h}$

$v > \frac{ϕ}{h}$

$v > = \frac{ϕ}{h}$

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Explanation

(a) For no emission of photoelectron, energy of incident light < Work function

$\Rightarrow hv < ϕ \Rightarrow v < \frac{ϕ}{h}$

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Light of wavelength $λ$ strikes a photo-sensitive surface and electrons are ejected with kinetic energy E. If the kinetic energy is to be increased to 2E, the wavelength must be changed to $λ'$ where

$λ' = \frac{λ}{2}$

$λ' = 2 λ$

$\frac{λ}{2} < λ' < λ$

$λ' > λ$

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Explanation

(c)

$E = \frac{hc}{λ} - W_{0} and 2 E = \frac{hc}{λ'} - W_{0} \Rightarrow \frac{λ'}{λ} = \frac{E + W_{0}}{2 E + W_{0}} \Rightarrow λ' = λ (\frac{1 + W_{0} / E}{2 + W_{0} / E}) Since (\frac{1 + W_{0} / E}{2 + W_{0} / E}) > \frac{1}{2} so λ' > \frac{λ}{2}$

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If in a photoelectric experiment, the wavelength of incident radiation is reduced from 6000 Å to 4000 Å then

Stopping potential will decrease

Stopping potential will increase

Kinetic energy of emitted electrons will decrease

The value of work function will decrease

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Explanation

(b) Stopping potential $V_{0} = \frac{hc}{e} [\frac{1}{λ} - \frac{1}{λ_{0}}]$ . As $λ$ decreases so $V_{0}$ increases.

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The photoelectric work function for a metal surface is 4.125 eV. The cut-off wavelength for this surface is

4125 Å

2062.5 Å

3000 Å

6000 Å

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Explanation

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As the intensity of incident light increases

Photoelectric current increases

Photoelectric current decreases

Kinetic energy of emitted photoelectrons increases

Kinetic energy of emitted photoelectrons decreases

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Explanation

(a) Intensity increases means more photons of same energy will emit more electrons of same energy, hence only photoelectric current increases.

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Which of the following is dependent on the intensity of incident radiation in a photoelectric experiment

Work function of the surface

Amount of photoelectric current

Stopping potential will be reduced

Maximum kinetic energy of photoelectrons

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Explanation

Amount of photo electric current depends on the intensity of incident radiation.

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The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately

540 nm

400 nm

310 nm

220 nm

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Explanation

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The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential in volts is

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Explanation

(b) $K_{\max} = (|V_{s}|) eV \Rightarrow |V_{s}| = 4 V$

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Work function of a metal is 2.1 eV. Which of the waves of the following wavelengths will be able to emit photoelectrons from its surface ?

4000 Å, 7500 Å

5500 Å, 6000 Å

4000 Å, 6000 Å

None of these

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Explanation

(d)

Cutoff -wavelength $λ_{0} = \frac{12375}{2.1} = 5892.8 \overset{0}{A}$

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