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(B) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by $eV_s = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$,where $

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$4 \lambda$

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(B) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by $eV_s = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$,where $\lambda_0$ is the threshold wavelength.For the first case: $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$  --- $(1)$For the second case: $e(\frac{V}{3}) = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}$  --- $(2)$Multiply equation $(2)$ by $3$: $eV = \frac{3hc}{2\lambda} - \frac{3hc}{\lambda_0}$  --- $(3)$Equating $(1)$ and $(3)$: $\frac{hc}{\lambda} - \frac{hc}{\lambda_0} = \frac{3hc}{2\lambda} - \frac{3hc}{\lambda_0}$Rearranging the terms: $\frac{3hc}{\lambda_0} - \frac{hc}{\lambda_0} = \frac{3hc}{2\lambda} - \frac{hc}{\lambda}$$\frac{2hc}{\lambda_0} = \frac{hc}{2\lambda}$$\frac{2}{\lambda_0} = \frac{1}{2\lambda}$$\lambda_0 = 4\lambda$.

When a metallic surface is illuminated with light of wavelength $\lambda$,the stopping potential is $V$. When the same surface is illuminated by light of wavelength $2 \lambda$,the stopping potential is $\frac{V}{3}$. The threshold wavelength for the metallic surface is:

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