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(B) According to Einstein's photoelectric equation: $eV_s = \frac{hc}{\lambda} - \Phi$, where $\Phi$ is the work function.For the first case: $e(3V_0)

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$4$

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(B) According to Einstein's photoelectric equation: $eV_s = \frac{hc}{\lambda} - \Phi$, where $\Phi$ is the work function.For the first case: $e(3V_0) = \frac{hc}{\lambda} - \Phi$  --- $(1)$For the second case: $e(V_0) = \frac{hc}{2\lambda} - \Phi$  --- $(2)$Subtracting equation $(2)$ from equation $(1)$:$3eV_0 - eV_0 = (\frac{hc}{\lambda} - \Phi) - (\frac{hc}{2\lambda} - \Phi)$$2eV_0 = \frac{hc}{\lambda} - \frac{hc}{2\lambda} = \frac{hc}{2\lambda}$$eV_0 = \frac{hc}{4\lambda}$Substitute $eV_0$ into equation $(2)$:$\frac{hc}{4\lambda} = \frac{hc}{2\lambda} - \Phi$$\Phi = \frac{hc}{2\lambda} - \frac{hc}{4\lambda} = \frac{hc}{4\lambda}$The threshold wavelength $\lambda_0$ is given by $\lambda_0 = \frac{hc}{\Phi}$.Substituting $\Phi = \frac{hc}{4\lambda}$:$\lambda_0 = \frac{hc}{hc / 4\lambda} = 4\lambda$.Comparing this with $\alpha\lambda$, we get $\alpha = 4$.

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