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 2λ, stopping potential is V/3. Threshold wavelength for metallic surface is

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4λ3

4λ

6λ

8λ3

answer is B.

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Detailed Solution

eV=hcλ−hcλ0...(1) & eV3=hc12λ−1λ0...(2)From (1 )and (2), we get λ0=4λ

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