When radiation of wavelength $λ$ is used to illuminate a metallic surface, the stopping potential is V. When the same surface is illuminated with radiation of wavelength $3 λ$ , the stopping potential is $\frac{V}{4} .$ If the threshold wavelength for the metallic surface is $n λ$ then value of n will be ____.

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answer is 9.

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Detailed Solution

$E = W + K E$

$\frac{h c}{λ} = W + e V \to (1)$

$\frac{h c}{3 λ} = W + \frac{e V}{4} \to (2)$

From (1) – (2)

$\frac{2}{3} \frac{h c}{λ} = \frac{3}{4} e v$

$e v = \frac{8}{9} \frac{h c}{λ}$

$\frac{h c}{λ} = W + \frac{8}{9} \frac{h c}{λ}$

$∴ W = \frac{h c}{λ} - \frac{8}{9} \frac{h c}{λ}$

$\frac{h c}{λ_{0}} = \frac{h c}{9 λ}$

$λ_{0} = 9 λ$

$∴ n = 9$

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