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(B) According to Einstein's photoelectric equation,the stopping potential $V_0$ is given by: $eV_0 = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{hc

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$4\lambda$

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(B) According to Einstein's photoelectric equation,the stopping potential $V_0$ is given by: $eV_0 = \frac{hc}{\lambda} - \phi$,where $\phi = \frac{hc}{\lambda_0}$ is the work function.Thus,$V_0 = \frac{hc}{e} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)$.For the first case: $4.8 = \frac{hc}{e} \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)$ --- $(i)$For the second case: $1.6 = \frac{hc}{e} \left( \frac{1}{2\lambda} - \frac{1}{\lambda_0} \right)$ --- $(ii)$Dividing equation $(i)$ by $(ii)$:$\frac{4.8}{1.6} = \frac{\frac{1}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{2\lambda} - \frac{1}{\lambda_0}}$$3 = \frac{\frac{\lambda_0 - \lambda}{\lambda \lambda_0}}{\frac{\lambda_0 - 2\lambda}{2\lambda \lambda_0}}$$3 = \frac{2(\lambda_0 - \lambda)}{\lambda_0 - 2\lambda}$$3\lambda_0 - 6\lambda = 2\lambda_0 - 2\lambda$$\lambda_0 = 4\lambda$.

When radiation of wavelength $\lambda$ is incident on a metal,the stopping potential of photoelectrons is $4.8 \ V$. When radiation of wavelength $2\lambda$ is incident on the same metal,the stopping potential is $1.6 \ V$. What is the threshold wavelength of the metal?

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