Q.The threshold wavelength for photoelectric emission from a certain metal is $500\ \text{nm}$. Calculate the work function of the metal in electron-volts.

Setting up. At the threshold wavelength, the photon's energy exactly equals the work function: $\phi_0=\dfrac{hc}{\lambda_0}$.

Calculation. $\phi_0=\frac{6.63\times10^{-34}\times3\times10^{8}}{500\times10^{-9}}\approx3.98\times10^{-19}\ \text{J}=\frac{3.98\times10^{-19}}{1.6\times10^{-19}}\approx2.49\ \text{eV}$

This value lies close to sodium's known work function, consistent with sodium's threshold wavelength lying in the visible range.

Apply $\phi_0=hc/\lambda_0$ directly (since at threshold, $K_{max}=0$, so the whole photon energy equals the work function), then convert the result from joules to eV.

• Forgetting the final conversion from joules to electron-volts, leaving the answer as an unfamiliar power-of-ten joule value instead of the requested eV figure.
• Using the wrong wavelength-to-frequency conversion (mixing up $c=\nu\lambda$ with $\nu=c\lambda$).