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(B) According to Einstein's photoelectric equation,the stopping potential $V_0$ is given by: $\frac{hc}{e} \left( \frac{1}{\lambda} - \frac{1}{\lambda

Vedclass · Accepted Answer

$4\lambda$

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(B) According to Einstein's photoelectric equation,the stopping potential $V_0$ is given by: $\frac{hc}{e} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) = V_0$ For the first case: $\frac{hc}{e} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) = 4.8 \quad ...(i)$ For the second case,where wavelength is $2\lambda$: $\frac{hc}{e} \left( \frac{1}{2\lambda} - \frac{1}{\lambda_0} \right) = 1.6 \quad ...(ii)$ Dividing equation $(i)$ by $(ii)$: $\frac{\frac{1}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{2\lambda} - \frac{1}{\lambda_0}} = \frac{4.8}{1.6} = 3$ $\frac{1}{\lambda} - \frac{1}{\lambda_0} = 3 \left( \frac{1}{2\lambda} - \frac{1}{\lambda_0} \right)$ $\frac{1}{\lambda} - \frac{1}{\lambda_0} = \frac{3}{2\lambda} - \frac{3}{\lambda_0}$ $\frac{3}{\lambda_0} - \frac{1}{\lambda_0} = \frac{3}{2\lambda} - \frac{1}{\lambda} = \frac{1}{2\lambda}$ $\frac{2}{\lambda_0} = \frac{1}{2\lambda}$ $\lambda_0 = 4\lambda$

When radiation of wavelength $\lambda$ is incident on a metallic surface,the stopping potential is $4.8 \ V$. If the same surface is illuminated with radiation of double the wavelength,then the stopping potential becomes $1.6 \ V$. Then the threshold wavelength for the surface is

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