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When a metal surface is illuminated by light of wavelength $λ$ . The stopping potential is 8V. When the same surface is illuminated by light of wavelength $3 λ$ stopping potential is 2V. The threshold wavelength for this surface is

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$9 λ$

$5 λ$

$4.5 λ$

$3 λ$

answer is A.

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Detailed Solution

$e v = \frac{h c}{λ} - \frac{h c}{λ_{0}}$ ; $K E_{\max} = h v - ϕ$

$e (8 v) = \frac{h c}{λ} - \frac{h c}{λ_{0}}$ .......... (1)

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

$\frac{(1)}{(2)} \Rightarrow \frac{e (8 v)}{e (2 v)} = \frac{\frac{1}{λ} - \frac{1}{λ_{0}}}{\frac{1}{3 λ} - \frac{1}{λ_{0}}} \Rightarrow 4 = \frac{\frac{λ_{0} - λ}{λ λ_{0}}}{\frac{λ_{0} - 3 λ}{λ λ_{0}}}$

$\Rightarrow λ_{0} = 9 λ$

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