Physics - Modern Physics Question with Solution | TestHub

$$\mathrm{h}v=4.2\mathrm{eV},\varphi =2.4\mathrm{eV},(\mathrm{KE}{)}_{\max }=4.2-2.4=1.8\mathrm{eV}$$ The electrons will not be able to escape from the surface when its potential becomes $1.8\mathrm{V}$. $$\begin{array}{rl}& \mathrm{V}=\frac{1}{4\pi {ϵ}_0}\frac{\mathrm{q}}{\mathrm{r}}\\ & 1.8=9\times 10^9\times \frac{\mathrm{q}}{0.1}\\ & \mathrm{q}=2\times 10^{-11}\mathrm{C}\end{array}$$ The number of photoelectrons emitted before the emission is stopped is, $$\mathrm{n}=\frac{\mathrm{q}}{\mathrm{e}}=\frac{2\times 10^{-11}}{1.6\times 10^{-19}}=1.25\times 10^8$$