1) Let the energy of an emitted photoelectron be E and wavelength of incident light be $\lambda$. What will be the change in E if $\lambda$ is doubled? A) E B) E/2 C) 2E D) E/4 Answer: Option BExplanation:We have hv= W0+E, where E is the energy of emitted photoelectron $\Rightarrow \frac{hc}{\lambda}=W_{0}+E$ As hc and W0 are constant $E\propto \frac{1}{\lambda}$ Therefore, as $\lambda$ is doubled, E will become half
