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A metallic surface is illuminated with radiation of wavelength $λ$ , 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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$4 λ$

$\frac{λ}{4}$

$\frac{3}{2} λ$

$3 λ$

answer is C.

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Detailed Solution

${eV}_{0} = \frac{hc}{λ} - \frac{hc}{λ_{o}} . . . . . . . . . . (i)$

$e (\frac{V_{0}}{4}) = \frac{hc}{2 λ} - \frac{hc}{λ_{o}} . . . . . . . . (ii)$

$\Rightarrow \frac{1}{4} [\frac{hc}{λ} - \frac{hc}{λ_{o}}] = \frac{hc}{2 λ} - \frac{hc}{λ_{o}}$

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

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

$\Rightarrow \frac{3}{4 λ_{o}} = \frac{1}{4 λ}$

$\Rightarrow λ_{o} = 3 λ$

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