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Q.

When radiation of wave length $λ$ is incident on a metallic surface, the stopping potential is 4.8Volt. If the same surface is illuminated with radiation of double the wavelength, then the stopping potential becomes 1.6V. Then the threshold wave length for the surface is

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a

$8 λ$

b

$4 λ$

c

$6 λ$

d

$2 λ$

answer is B.

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Detailed Solution

$\frac{h c}{λ} = e v + \frac{h c}{λ_{0}}$

$\frac{h c}{2 λ} = e v_{2} + \frac{h c}{λ_{0}}$

$e v_{1} = \frac{h c}{λ} - \frac{h c}{λ_{0}}$

$e v_{2} = \frac{h c}{2 λ} - \frac{h_{1}}{λ_{0}}$

$\frac{v_{1}}{v_{2}} = \frac{\frac{1}{λ} - \frac{1}{λ_{0}}}{\frac{1}{2 λ} - \frac{1}{λ_{0}}} = \frac{4.8}{1.6} = 3$

$\frac{1}{λ} - \frac{1}{λ_{0}} = \frac{3}{2 λ} - \frac{3}{λ_{0}}$

$\frac{2}{λ_{0}} = \frac{1}{2 λ} \Rightarrow λ_{0} = 4 λ$

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When radiation of wave lengthλ is incident on a metallic surface, the stopping potential is 4.8Volt. If the same surface is illuminated with radiation of double the wavelength, then the stopping potential becomes 1.6V. Then the threshold wave length for the surface is