Chemistry - Atomic Structure Question with Solution | TestHub

Since, the striking photon has energy equal to $hv\mathrm{v}$ and the minimum energy required to eject the electron is $h{v}_0$ (work function ${\mathrm{W}}_0$), the difference in energy $\left(hv-h{v}_0\right)$ is transferred as the kinetic energy of the photoelectron. $hv=h{v}_0+\mathrm{KE}$ $h{v}_0=$ minimum energy required to remove electron) Energy of a $300\mathrm{nm}$ photon is given by $\begin{array}{rl}& \Rightarrow \mathrm{E}=\frac{hc}{\lambda }\\ & =\frac{6.626\times 10^{-34}\mathrm{Js}\times 3\times 10^8{\mathrm{ms}}^{-1}}{300\times 10^{-9}\mathrm{m}}\\ & =6.626\times 10^{-19}\mathrm{J}\end{array}$ Energy of one mole of photons $\begin{array}{rl}& =6.626\times 10^{-19}\mathrm{J}\times 6.022\times 10^{23}{\mathrm{mol}}^{-1}\\ & =3.99\times 10^5\mathrm{J}{\mathrm{mol}}^{-1}\end{array}$ The minimum energy needed to remove one mole of electrons $=(3.99-1.68)\times 10^5\mathrm{J}{\mathrm{mol}}^{-1}$ $=2.31\times 10^5\mathrm{J}{\mathrm{mol}}^{-1}$