Question

The electric field at a point associated with a light wave is
$E=\left(100{\mathrm{Vm}}^{-1}\right)\sin \left[\right(3.0\times {10}^{15}{s}^{-1}\left)t\right]\sin \left[\right(6.0\times {10}^{15}{s}^{-1}\left)t\right].$
If this light falls on a metal surface with a work function of 2.0 eV, what will be the maximum kinetic energy of the photoelectrons?

Answer 1

Given:
$$\begin{gathered} E=100\sin \left[\right(3\times {10}^{-15}{s}^{-1}\left)t\right]\sin \left[\right(6\times {10}^{-15}{s}^{-1}\left)t\right] \\ =100\times \frac{1}{2}\cos \left[\right(9\times {10}^{15}{s}^{-1}\left)t\right]-\cos \left[\right(3\times {10}^{15}{s}^{-1}\left)t\right] \end{gathered}$$
The values of angular frequency $\omega$ are 9 × 10¹⁵ and 3 × 10¹⁵ .
Work function of the metal surface, $\varphi$ = 2 eV
Maximum frequency,
$v=\frac{{\omega }_{max}}{2\mathrm{\pi }}$=$\frac{9\times {10}^{15}}{2\mathrm{\pi }}$ Hz
From Einstein's photoelectric equation, kinetic energy,
K = hv $-$ $\varphi$
$\Rightarrow$K = $6.63\times {10}^{-34}\times \frac{9\times {10}^{15}}{2\mathrm{\pi }}\times \frac{1}{1.6\times {10}^{-19}}$ $-$ 2 eV
$\Rightarrow$K = 3.938 eV