Question

A monochromatic light source of intensity 5 mW emits 8 × 10¹⁵ photons per second. This light ejects photoelectrons from a metal surface. The stopping potential for this setup is 2.0 V. Calculate the work function of the metal.

Answer 1

Given:
Intensity of light, I = 5 mW
Number of photons emitted per second, n = 8 × 10¹⁵
Stopping potential, V₀ = 2 V
Energy, E = $hv$ = $\frac{I}{n}=\frac{5\times {10}^{-3}}{8\times {10}^{15}}$
From Einstein's photoelectric equation, work function,
$W_0=hv-e{V}_0$
Here, h = Planck's constant
e = 1.6$\times$10^$-19$ C
On substituting the respective values, we get:
$$\begin{gathered} W_0=\frac{5\times {10}^{-3}}{8\times {10}^{15}}-1.6\times {10}^{-19}\times 2 \\ =6.25\times {10}^{-19}-3.2\times {10}^{-19} \\ =3.05\times {10}^{-19} \\ =\frac{3.05\times {10}^{-19}}{1.6\times {10}^{-15}}=1.906\mathrm{eV} \end{gathered}$$