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Question

When a surface 1 cm thick is illuminated with light of wave length λ the stopping potential is V0 ,but when the same surface is illuminated by light of wavelength 3λ , the stopping potential is V06. The threshold wavelength for metallic surface is

A
4λ
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B
5λ
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C
3λ
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D
2λ
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Solution

The correct option is C 5λ
V0 = hce(1λ1λ0) ( l )
V06=hce(13λ1λ0) (ll)
Now equation ( I ) ÷ equation ( l l )
6= 3(λ0λ)λ03λ
6λ018λ =3λ03λ
λ0=5λ
So, the answer is option (B).

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