Calculating the energy of the incident photon 1 Identifying the metals
Reason
The energy of a photon of incident radiation is given by
$\mathit{E}=\frac{\mathit{h}\mathit{c}}{\mathit{\lambda }}$
∴ $\mathit{E}=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{\left(412.5\times {10}^{-9}\right)\times \left(1.6\times {10}^{-19}\right)}\mathrm{eV}$
$=3.01\mathrm{eV}$
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.