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Question

When a centimeter thick surface is illuminated with ligh of wavelength λ, the stopping potential is V. When the same surface is illuminated by light of wavelengh2λ, the stopping potential is V/3. The threshold wavelength for he surface is

A
4λ3
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B
4λ
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C
6λ
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D
8λ3
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Solution

The correct option is B 4λ
According to question,
ev=h0(1λ1λ0)
(i) ev3=hc(12λ1λ0)
(ii) Dividing equ (1) by eq (2)
We get 3=(1λ1λ0)(12λ1λ0)
or 32λ1λ=3λ01λ0
12λ=2λ0
Threshold wavelength for metallic surface
λ0=4λ

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