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(D) The photoelectric equation is given by $eV_0 = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)$,where $\lambda_0$ is the threshold wavel

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$5 \lambda$

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(D) The photoelectric equation is given by $eV_0 = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)$,where $\lambda_0$ is the threshold wavelength.For the first case: $eV = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) \dots (i)$For the second case: $e \left( \frac{V}{6} \right) = hc \left( \frac{1}{3\lambda} - \frac{1}{\lambda_0} \right) \dots (ii)$Dividing equation $(i)$ by equation $(ii)$:$6 = \frac{\frac{1}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{3\lambda} - \frac{1}{\lambda_0}} = \frac{\frac{\lambda_0 - \lambda}{\lambda \lambda_0}}{\frac{\lambda_0 - 3\lambda}{3\lambda \lambda_0}} = \frac{3(\lambda_0 - \lambda)}{\lambda_0 - 3\lambda}$$6(\lambda_0 - 3\lambda) = 3\lambda_0 - 3\lambda$$6\lambda_0 - 18\lambda = 3\lambda_0 - 3\lambda$$3\lambda_0 = 15\lambda$$\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 $\frac{V}{6}$. Then the threshold wavelength for the surface is:

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