In a photoelectric effect experiment a metallic surface of work function $2.2eV$ is illuminated with light of wavelength $400nm$. Assume that an electron makes two collisions before being emitted and in each collision ten per cent additional energy is lost. Find the kinetic energy of this electron as it comes out of the metal.

Energy of photon,

$E=\frac{hc}{\lambda }=\frac{1.24\times {10}^3}{400}=3.1eV$

Energy lost in first collision $=\left(3.1\right)\times \left(\frac{10}{100}\right)=0.31eV$

Remaining energy $=3.1-0.31=2.79eV$

Energy lost in second collision $=\left(2.79\right)\times \left(\frac{10}{100}\right)=0.279eV$

Total energy lost in two collisions

$=\left(0.31+0.279\right)eV=0.589eV$

So, from conservation of energy, we have

$$\begin{gathered} \frac{hc}{\lambda }=\varphi +K{E}_{max}+energy lost in collision \\ 3.1=2.2+K{E}_{max}+0.589 \end{gathered}$$

or $K{E}_{max}=0.31eV$