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(A) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by: $eV_s = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$,where

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$4$

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(A) According to Einstein's photoelectric equation,the stopping potential $V_s$ is given by: $eV_s = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$,where $\lambda_0$ is the threshold wavelength.For the first case: $e(4.8) = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$  --- $(1)$For the second case: $e(1.6) = \frac{hc}{2\lambda} - \frac{hc}{\lambda_0}$ --- $(2)$Subtracting equation $(2)$ from $(1)$:$e(4.8 - 1.6) = \frac{hc}{\lambda} - \frac{hc}{2\lambda}$$3.2e = \frac{hc}{2\lambda}$ --- $(3)$From equation $(2)$,substitute $\frac{hc}{2\lambda} = 3.2e$:$1.6e = 3.2e - \frac{hc}{\lambda_0}$$\frac{hc}{\lambda_0} = 1.6e$Now,divide equation $(3)$ by this result:$\frac{hc/2\lambda}{hc/\lambda_0} = \frac{3.2e}{1.6e}$$\frac{\lambda_0}{2\lambda} = 2$$\lambda_0 = 4\lambda$.Thus,the threshold wavelength is $4\lambda$.

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