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Question

A metallic surface is illuminated with radiation of wavelength $\lambda$, the stopping potential is $V_{0}$. If the same surface is illuminated with radiation of wavelength $2 \lambda$. the stopping potential becomes $\frac{V_{o}}{4}$. The threshold wavelength for this metallic surface will be

$3 \lambda$
$4 \lambda$
$\frac{3}{2} \lambda$
$\frac{\lambda}{4}$

Solution

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