Question

The electric field associated with a monochromatic beam is 1.2 × 10¹⁵ times per second. Find the maximum kinetic energy of the photoelectrons when this light falls on a metal surface whose work function is 2.0 eV.

Answer 1

Given:
Electric field of the monochromatic beam, E = 1.2 × 10¹⁵ times per second
Frequency, v = $\frac{1.2\times {10}^{15}}{2}=0.6\times {10}^{15}\mathrm{Hz}$
Work function of the metal surface, $\varphi$ = 2.0 eV
From Einstein's photoelectric equation, kinetic energy,
$$\begin{gathered} K=hv-{\mathrm{\varphi }}_0 \\ \Rightarrow K=\frac{6.63\times {10}^{-34}\times 0.6\times {10}^{15}}{1.6\times {10}^{-19}}-2 \\ =0.486\mathrm{eV} \end{gathered}$$
Thus, the maximum kinetic energy of a photon is 0.486 eV.