Physics - Modern Physics Question with Solution | TestHub

We have $\mathrm{h}v={\mathrm{W}}_{\mathrm{o}}+\mathrm{E}$, where $\mathrm{E}$ is the energy of emitted photoelectron $\Rightarrow \frac{\mathrm{hc}}{\lambda }={\mathrm{W}}_{\mathrm{o}}+\mathrm{E}$ As he and ${\mathrm{W}}_{\mathrm{o}}$ are constant, $\mathrm{E}\propto \frac{1}{\lambda }$ Therefore, as $\lambda$ is doubled, $E$ will become half.