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

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$5\lambda$

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(D) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = eV + \phi$,where $\phi = \frac{hc}{\lambda_0}$ is the work function and $\lambda_0$ is the threshold wavelength.For the first case: $\frac{hc}{\lambda} = eV + \phi$  --- $(i)$For the second case: $\frac{hc}{3\lambda} = e(\frac{V}{6}) + \phi$ --- (ii)Multiply equation (ii) by $6$ to eliminate $V$:$6(\frac{hc}{3\lambda}) = eV + 6\phi$$\frac{2hc}{\lambda} = eV + 6\phi$ --- (iii)Subtracting equation $(i)$ from equation (iii):$(\frac{2hc}{\lambda} - \frac{hc}{\lambda}) = (eV + 6\phi) - (eV + \phi)$$\frac{hc}{\lambda} = 5\phi$Since $\phi = \frac{hc}{\lambda_0}$,we have $\frac{hc}{\lambda} = 5(\frac{hc}{\lambda_0})$.Therefore,$\lambda_0 = 5\lambda$.

When light of wavelength $\lambda$ is incident on a photosensitive surface,the stopping potential is $V$. When light of wavelength $3 \lambda$ is incident on the same surface,the stopping potential is $\frac{V}{6}$. The threshold wavelength for the surface is:

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