Chemistry - Atomic Structure Question with Solution | TestHub

Energy of photon $=\frac{hc}{\lambda }\mathrm{J}=\frac{hc}{e\lambda }\mathrm{eV}=\frac{6.625\times 10^{-34}\times 3\times 10^8}{300\times 10^{-9}\times 1.602\times 10^{-19}}=4.14\mathrm{eV}$ For photoelectric effect to occur, energy of incident photons must be greater than work function of metal. Hence, only $\mathrm{Li},\mathrm{Na},\mathrm{K}$ and $\mathrm{Mg}$ have work functions less than $4.14\mathrm{V}$.