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(D) According to Einstein's photoelectric equation,the stopping potential $V$ is given by $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$,where $\lam

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$3\lambda$

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(D) According to Einstein's photoelectric equation,the stopping potential $V$ is given by $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$,where $\lambda_0$ is the threshold wavelength. For the first case: $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$  --- $(i)$ For the second case: $e(\frac{V}{4}) = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}$ --- $(ii)$ Multiplying equation $(ii)$ by $4$,we get: $eV = \frac{2hc}{\lambda} - \frac{4hc}{\lambda_0}$ --- $(iii)$ Equating $(i)$ and $(iii)$: $\frac{hc}{\lambda} - \frac{hc}{\lambda_0} = \frac{2hc}{\lambda} - \frac{4hc}{\lambda_0}$ Rearranging the terms: $\frac{3hc}{\lambda_0} = \frac{hc}{\lambda}$ Therefore,$\lambda_0 = 3\lambda$.

When a metallic surface is illuminated with radiation of wavelength $\lambda$,the stopping potential is $V$. If the same surface is illuminated with radiation of wavelength $2\lambda$,the stopping potential is $\frac{V}{4}$. The threshold wavelength for the metallic surface is:

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