Question

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

$\begin{array}{cc}\text{Metal}& \text{Work Function (eV)}\\ \text{Na}& 1.92\\ \text{K}& 2.15\\ \text{Ca}& 3.20\\ \text{Mo}& 4.17\end{array}$

Answer 1

Wavelength of incident light,
$\lambda =412.5$ nm

= $412.5\times {10}^{-9}$m
$\therefore$ Energy of incident light

$E=\frac{hc}{\lambda }=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{412.5\times {10}^{-9}}$

= $4.82\times {10}^{-19}J=3.01$ eV

Photoelectric emission from a metal surface will take place when the incident energy on it is greater than its work function.

Clearly, 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.