Question

What is the maximum kinetic energy in eV of electrons ejected from sodium metal by 450-NM EM radiation, given that the binding energy is 2.28 eV

Answer 1

The maximum kinetic energy of electrons ejected from metal

$$K_{\max}=\frac{hc}{\lambda}-\varphi$$

$$K_{\max}=\frac{1239.8 }{\lambda\:(\rm nm)}-\varphi$$

$$K_{\max}=\frac{1239.8 }{450\:(\rm nm)}-2.28=0.475\:\rm eV$$