Question

A hydrogen atom in ground state absorbs a photon of ultraviolet radiation of wavelength 50 nm. Assuming that the entire photon energy is taken up by the electron with what kinetic energy will the electron be ejected?

Answer 1

Given:
Wavelength of ultraviolet radiation, λ = 50 nm = 50$\times$10^$-9$ m

We know that the work function of an atom is the energy required to remove an electron from the surface of the atom. So, we can find the work function by calculating the energy required to remove the electron from n₁ = 1 to n₂ = ∞.

Work function, $$\begin{gathered} W_0=13.6\left(\frac{1}{1}-\frac{1}{\infty }\right) \\ =13.6\mathrm{eV} \end{gathered}$$
Using Einstein's photoelectric equation, we get

$$\begin{gathered} E=W_0+\mathrm{KE} \\ \Rightarrow \frac{hc}{\lambda }-13.6=\mathrm{KE}\left(\because E=\frac{hc}{\lambda }\right) \\ \Rightarrow \frac{1242}{50}-13.6=\mathrm{KE} \\ \Rightarrow \mathrm{KE}=24.84-13.6 \\ =11.24\mathrm{eV} \end{gathered}$$