In a photoelectric experiment, the stopping potential for a metal surface is $V_s=2.0\text{ V}$ when light of wavelength $\lambda =300\text{ nm}$ is used. What is the work function $\Phi$ of the metal?

Solution

1. The energy of the incident photon is $$\begin{array}{rl}E& =hc/\lambda \\ & =1240\text{ eV nm}/300\text{ nm}\approx 4.13\text{ eV}\end{array}$$.
2. The maximum kinetic energy of the emitted electrons is $$\begin{array}{rl}{K}_{max}& =e{V}_s\\ & =2.0\text{ eV}\end{array}$$.
3. From Einstein's equation, $K_{max}=E-\Phi$, so $\Phi =E-K_{max}$.
4. $$\begin{array}{rl}\Phi & =4.13\text{ eV}-2.0\text{ eV}\\ & =2.13\text{ eV}\end{array}$$.
5. Hence the answer is (B).