When a metallic surface is illuminated with light of wavelength $^{'} λ^{'},$ stopping potential is V, when the same surface is illuminated by light of wavelength $2 λ,$ stopping potential is $\frac{V}{3} .$ Threshold wavelength for metallic surface is

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$\frac{4 λ}{3}$

$4 λ$

$6 λ$

$\frac{8 λ}{3}$

answer is B.

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Detailed Solution

$e v = \frac{h c}{λ} - \frac{h c}{λ_{0}}$ …….(1)

$\frac{e v}{3} = (\frac{h c}{2 λ} - \frac{h c}{λ_{0}})$ ………(2)

From (1) and (2), We get $λ_{0} = 4 λ$

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