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

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$4 \lambda$

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(B) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by $eV_s = \frac{hc}{\lambda} - \phi$,where $\phi$ is the work function.For the first case: $4.8 = \frac{hc}{e\lambda} - \frac{\phi}{e} \quad \dots(i)$For the second case,where wavelength is $2\lambda$: $1.6 = \frac{hc}{e(2\lambda)} - \frac{\phi}{e} \quad \dots(ii)$Subtracting equation $(ii)$ from equation $(i)$:$4.8 - 1.6 = \left(\frac{hc}{e\lambda} - \frac{\phi}{e}\right) - \left(\frac{hc}{2e\lambda} - \frac{\phi}{e}\right)$$3.2 = \frac{hc}{e\lambda} - \frac{hc}{2e\lambda} = \frac{hc}{2e\lambda}$So,$\frac{hc}{e\lambda} = 6.4$.Substituting this into equation $(i)$:$4.8 = 6.4 - \frac{\phi}{e} \Rightarrow \frac{\phi}{e} = 6.4 - 4.8 = 1.6$.The threshold wavelength $\lambda_0$ is given by $\phi = \frac{hc}{\lambda_0}$,so $\frac{\phi}{e} = \frac{hc}{e\lambda_0} = 1.6$.Since $\frac{hc}{e\lambda} = 6.4$,we have $\frac{hc}{e\lambda_0} = \frac{1}{4} \left(\frac{hc}{e\lambda}\right)$.Therefore,$\lambda_0 = 4\lambda$.

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