Q.The work function of caesium is $2.14\ \text{eV}$. Calculate

Given. $\phi_0=2.14\ \text{eV}=2.14\times1.6\times10^{-19}\ \text{J}=3.424\times10^{-19}\ \text{J}$.

Threshold frequency. $\nu_0=\frac{\phi_0}{h}=\frac{3.424\times10^{-19}}{6.63\times10^{-34}}\approx5.16\times10^{14}\ \text{Hz}$

Threshold wavelength. $\lambda_0=\frac{c}{\nu_0}=\frac{3\times10^{8}}{5.16\times10^{14}}\approx5.81\times10^{-7}\ \text{m}=581\ \text{nm}$

This falls in the visible range, close to yellow-green light -- consistent with caesium's well-known usefulness in photocells designed to respond to ordinary visible light.

Convert $\phi_0$ from eV to joules, apply $\nu_0=\phi_0/h$, then find $\lambda_0=c/\nu_0$.

• Forgetting to convert the work function from eV to joules before dividing by $h$ (which is in J s) -- using the eV value directly gives a wildly wrong frequency.
• Mixing up $\nu_0=\phi_0/h$ (correct) with $\nu_0=h/\phi_0$ (incorrect, wrong dimensions).