When a centimeter thick 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 the same surface is illuminated by light of is
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a
4λ/3
b
4λ
c
6λ
d
8λ/3
answer is B.
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Detailed Solution
eV=hc1λ−1λ0 eV3=hc12λ−1λ0 Dividing eq. (1) by eq. (2), we get 3=1λ−1λ012λ−1λ0 or 312λ−1λ0=1λ−1λ0 32λ−1λ=3λ0−1λ0 or 12λ=2λ0 or λ0=4λ