Calculating the energy of the incident photon 1 Identifying the metals Reason The energy of a photon of incident radiation is given by $\;E=\;\frac{h c}{\lambda}$ ∴ $\;E=\;\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{(412.5 \times 10^{-9}) \times (1.6 \times 10^{-19})}\ \mathrm{eV}$ $\;=3.01\ \mathrm{eV}$ Hence, only $\mathrm{Na}$ and $\mathrm{K}$ will show photoelectric emission Reason: The energy of the incident photon is more than the work function of only these two metals.