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

Vedclass · Accepted Answer

$5 \lambda_1$

Vedclass · Answer

(C) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by: $eV_s = \frac{hc}{\lambda} - \Phi$,where $\Phi = \frac{hc}{\lambda_0}$ is the work function and $\lambda_0$ is the threshold wavelength.For the first case: $eV_1 = \frac{hc}{\lambda_1} - \frac{hc}{\lambda_0} = hc \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_0} \right)$  --- $(1)$For the second case: $e \left( \frac{V_1}{6} \right) = \frac{hc}{3\lambda_1} - \frac{hc}{\lambda_0} = hc \left( \frac{1}{3\lambda_1} - \frac{1}{\lambda_0} \right)$ --- $(2)$Dividing equation $(1)$ by equation $(2)$:$6 = \frac{\frac{1}{\lambda_1} - \frac{1}{\lambda_0}}{\frac{1}{3\lambda_1} - \frac{1}{\lambda_0}}$$6 \left( \frac{1}{3\lambda_1} - \frac{1}{\lambda_0} \right) = \frac{1}{\lambda_1} - \frac{1}{\lambda_0}$$\frac{2}{\lambda_1} - \frac{6}{\lambda_0} = \frac{1}{\lambda_1} - \frac{1}{\lambda_0}$$\frac{2}{\lambda_1} - \frac{1}{\lambda_1} = \frac{6}{\lambda_0} - \frac{1}{\lambda_0}$$\frac{1}{\lambda_1} = \frac{5}{\lambda_0}$$\lambda_0 = 5\lambda_1$. Thus,the correct option is $C$.

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