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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max} = eV_s = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$,where $\lam

Vedclass · Accepted Answer

$4\lambda$

Vedclass · Answer

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max} = eV_s = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$,where $\lambda_0$ is the threshold wavelength.For the first case,wavelength is $\lambda$ and stopping potential is $3V_0$:$\frac{hc}{\lambda} - \frac{hc}{\lambda_0} = e(3V_0)$  --- $(1)$For the second case,wavelength is $2\lambda$ and stopping potential is $V_0$:$\frac{hc}{2\lambda} - \frac{hc}{\lambda_0} = eV_0$  --- $(2)$Dividing equation $(1)$ by equation $(2)$:$\frac{\frac{hc}{\lambda} - \frac{hc}{\lambda_0}}{\frac{hc}{2\lambda} - \frac{hc}{\lambda_0}} = \frac{3}{1}$$\frac{hc}{\lambda} - \frac{hc}{\lambda_0} = 3 \left( \frac{hc}{2\lambda} - \frac{hc}{\lambda_0} \right)$$\frac{hc}{\lambda} - \frac{hc}{\lambda_0} = \frac{3hc}{2\lambda} - \frac{3hc}{\lambda_0}$$\frac{3hc}{\lambda_0} - \frac{hc}{\lambda_0} = \frac{3hc}{2\lambda} - \frac{hc}{\lambda}$$\frac{2hc}{\lambda_0} = \frac{hc}{2\lambda}$$\frac{2}{\lambda_0} = \frac{1}{2\lambda} \implies \lambda_0 = 4\lambda$.

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