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Q.

A metallic surface is illuminated with radiation of wave length $λ$ , the stopping potential is $V_{O}$ . If the same surface is illuminated with radiation of wavelength $2 λ$ , the stopping potential becomes $\frac{V_{O}}{4}$ . The threshold wavelength for this metallic surface will be

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a

$\frac{λ}{4}$

b

$4 λ$

c

$3 λ$

d

$\frac{3}{2} λ$

answer is C.

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Detailed Solution

$\begin{aligned} e V_{o} = \frac{h c}{λ} - \frac{h c}{λ_{0}} e (\frac{V_{0}}{4}) = \frac{h c}{2 λ} - \frac{h c}{λ_{o}} \frac{1}{4} [\frac{h c}{λ} - \frac{h c}{λ_{o}}] = \frac{h c}{2 λ} - \frac{h c}{λ_{o}} \\ \frac{1}{4 λ} - \frac{1}{4 λ_{o}} = \frac{1}{2 λ} - \frac{1}{λ_{o}} \frac{1}{λ_{o}} - \frac{1}{4 λ_{o}} = \frac{1}{2 λ} - \frac{1}{4 λ} \\ \Rightarrow \frac{3}{4 λ_{o}} = \frac{1}{4 λ} λ_{o} = 3 λ \end{aligned}$

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