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When a metal surface is illuminated by two different wavelengths $λ$ and $2 λ$ , the stopping potentials in the two cases were observed to be 5.6 V and 1.4 V respectively. The threshold wavelength of the metal is

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$λ$

$4 λ$

$5 λ$

$3 λ$

answer is D.

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Detailed Solution

$\frac{h c [\frac{1}{λ} - \frac{1}{λ_{0}}]}{h c [\frac{1}{2 λ} - \frac{1}{λ_{0}}]} = \frac{5.6}{1.4} = \frac{4}{1} \Rightarrow λ_{0} = 3 λ$

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