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(A) Einstein's photoelectric equation is given by $\frac{hc}{\lambda} = eV_s + W$,where $V_s$ is the stopping potential and $W$ is the work function.F

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$\frac{8}{3}\lambda$

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(A) Einstein's photoelectric equation is given by $\frac{hc}{\lambda} = eV_s + W$,where $V_s$ is the stopping potential and $W$ is the work function.For the first case: $\frac{hc}{\lambda} = e(5V_0) + W$  --- $(1)$For the second case: $\frac{hc}{2\lambda} = eV_0 + W$  --- $(2)$Multiply equation $(2)$ by $5$: $\frac{5hc}{2\lambda} = 5eV_0 + 5W$  --- $(3)$Subtract equation $(1)$ from equation $(3)$:$\frac{5hc}{2\lambda} - \frac{hc}{\lambda} = (5eV_0 + 5W) - (5eV_0 + W)$$\frac{3hc}{2\lambda} = 4W$$W = \frac{3hc}{8\lambda}$Since the work function $W = \frac{hc}{\lambda_0}$,where $\lambda_0$ is the threshold wavelength:$\frac{hc}{\lambda_0} = \frac{3hc}{8\lambda}$$\lambda_0 = \frac{8}{3}\lambda$

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