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(B) 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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(B) 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: $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$  --- $(1)$For the second case: $e(\frac{V}{4}) = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}$ --- $(2)$Dividing equation $(1)$ by equation $(2)$:$4 = \frac{\frac{hc}{\lambda} - \frac{hc}{\lambda_0}}{\frac{hc}{2\lambda} - \frac{hc}{\lambda_0}}$$4 = \frac{\frac{1}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{2\lambda} - \frac{1}{\lambda_0}}$$4(\frac{1}{2\lambda} - \frac{1}{\lambda_0}) = \frac{1}{\lambda} - \frac{1}{\lambda_0}$$\frac{2}{\lambda} - \frac{4}{\lambda_0} = \frac{1}{\lambda} - \frac{1}{\lambda_0}$$\frac{2}{\lambda} - \frac{1}{\lambda} = \frac{4}{\lambda_0} - \frac{1}{\lambda_0}$$\frac{1}{\lambda} = \frac{3}{\lambda_0}$$\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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