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

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$9 \lambda$

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(D) According to Einstein's photoelectric equation,the stopping potential $V$ is given by $eV = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{hc}{\lambda_0}$ is the work function and $\lambda_0$ is the threshold wavelength.Initially,$eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} \quad ...(1)$Finally,$e(V/4) = \frac{hc}{3\lambda} - \frac{hc}{\lambda_0} \quad ...(2)$Multiply equation $(2)$ by $4$,we get $eV = \frac{4hc}{3\lambda} - \frac{4hc}{\lambda_0} \quad ...(3)$Equating $(1)$ and $(3)$,we have $\frac{hc}{\lambda} - \frac{hc}{\lambda_0} = \frac{4hc}{3\lambda} - \frac{4hc}{\lambda_0}$.Rearranging the terms,$\frac{4hc}{\lambda_0} - \frac{hc}{\lambda_0} = \frac{4hc}{3\lambda} - \frac{hc}{\lambda}$.$\frac{3hc}{\lambda_0} = \frac{hc}{3\lambda}$.Therefore,$\lambda_0 = 9\lambda$.

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 $3 \lambda$,the stopping potential is $V/4$. The threshold wavelength for the metallic surface is.

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