When radiation of wavelength λ is incident on a metallic surface, the stopping potential is 4.8 volts. If the same surface is illuminated with radiation of double the wavelength, then the stopping potential becomes 1.6 volts. Then the threshold wavelength for the surface is
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a
2λ
b
4λ
c
6λ
d
8λ
answer is B.
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Detailed Solution
By usinghce1λ−1λo=Vo⇒hce1λ−1λo=4.8 . . . . iand hce12λ−1λo=1.6 . . . . iiFrom equation i by ii,1λ−1λo12λ−1λo=4.81.6⇒λo=4λ