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(D) According to Einstein's photoelectric equation: $eV = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{hc}{\lambda_0}$ is the work function.Case $I$

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$4\lambda$

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(D) According to Einstein's photoelectric equation: $eV = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{hc}{\lambda_0}$ is the work function.Case $I$: $eV = \frac{hc}{\lambda} - \phi$  --- $(i)$Case $II$: $e(V/3) = \frac{hc}{2\lambda} - \phi$  --- $(ii)$Multiply equation $(ii)$ by $3$: $eV = \frac{3hc}{2\lambda} - 3\phi$ --- $(iii)$Equating $(i)$ and $(iii)$: $\frac{hc}{\lambda} - \phi = \frac{3hc}{2\lambda} - 3\phi$$2\phi = \frac{3hc}{2\lambda} - \frac{hc}{\lambda} = \frac{hc}{2\lambda}$$\phi = \frac{hc}{4\lambda}$Since $\phi = \frac{hc}{\lambda_0}$,we get $\lambda_0 = 4\lambda$.

When a photosensitive 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 $V/3$. The threshold wavelength for the surface is:

Similar Questions

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