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When a metallic surface is illuminated with radiation of wavelength $λ$ , the stopping potential is V. If the same surface is illuminated with radiation of wavelength $2 λ$ , the stopping potential is $V / 4$ . The threshold wavelength for the metallic surface is

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$4 λ$

$5 λ$

$\frac{5 λ}{2}$

$3 λ$

answer is D.

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Detailed Solution

$\frac{hc}{λ} = \frac{hc}{λ_{0}} + eV . . . . . . . . . . . . (1)$

$\frac{hc}{2 λ} = \frac{hc}{λ_{0}} + \frac{eV}{4} . . . . . . . . . . . . (2)$

solving (1) and (2),

$λ_{0} = 3 λ$

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