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$V_{1} a n d V_{2}$ are the stopping potentials for the incident radiations of wave lengths
$λ_{1} a n d λ_{2}$ respectively are incident on a metallic surface. $λ_{1} = 3 λ_{2}$ then

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$V_{1} > \frac{V_{2}}{3}$

$V_{1} < \frac{V_{2}}{3}$

$V_{2} = 3 V_{1}$

$V_{1} {>3V}_{2}$

answer is D.

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Detailed Solution

$\frac{h c}{λ_{1}} = w + e V_{1}$

$\frac{h c}{λ_{2}} = w + e V_{2}$

$λ_{1} = 3 λ_{2}$

$\frac{h c}{3 λ_{2}} = w + e V_{1}$

$w + e V_{2} = 3 w + 3 e V_{1}$

$3 e V_{1} = e V_{2} - 2 w$

$V_{1} = \frac{V_{2}}{3} - \frac{2 w}{3 e}$

$V_{1} < \frac{V_{2}}{3}$

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