If light of wavelength 412.5 nm is incident on each of the metals given below, which ones will show photoelectric emission and why?

Wavelength of incident light,
$\begin{array}{l}\mathrm{\lambda }=412.5\mathrm{nm}\\ =412.5\times {10}^{-9}\mathrm{n}\end{array}$
$\therefore \mathrm{Energy} \mathrm{of} \mathrm{incident} \mathrm{light}$
$\begin{array}{l}\mathbf{E}=\frac{\mathrm{hc}}{\mathrm{\lambda }}=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{412.5\times {10}^{-9}}\\ =4.82\times {10}^{-19}\mathrm{J}\end{array}$
Since, the energy of incident radiation is greater than the work function of sodium and potassium, but less than that of calcium and molybdenum, therefore, photoelectric emission will take place in sodium and potassium.