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(C) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = \phi + eV_s$,where $\phi = \frac{hc}{\lambda_{th}}$ is the work function and

Vedclass · Accepted Answer

$4\lambda$

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(C) According to Einstein's photoelectric equation: $\frac{hc}{\lambda} = \phi + eV_s$,where $\phi = \frac{hc}{\lambda_{th}}$ is the work function and $V_s$ is the stopping potential.For the first case: $\frac{hc}{\lambda} = \frac{hc}{\lambda_{th}} + e(3V_0)$  --- $(1)$For the second case: $\frac{hc}{2\lambda} = \frac{hc}{\lambda_{th}} + eV_0$  --- $(2)$Multiply equation $(2)$ by $3$: $\frac{3hc}{2\lambda} = \frac{3hc}{\lambda_{th}} + 3eV_0$  --- $(3)$Subtract equation $(1)$ from equation $(3)$:$\frac{3hc}{2\lambda} - \frac{hc}{\lambda} = \frac{3hc}{\lambda_{th}} - \frac{hc}{\lambda_{th}}$$\frac{hc}{2\lambda} = \frac{2hc}{\lambda_{th}}$$\frac{1}{2\lambda} = \frac{2}{\lambda_{th}}$$\lambda_{th} = 4\lambda$.

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