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

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$3\lambda$

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