Question

A hydrogen atom in ground state absorbs a photon of ultraviolet radition 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

Here $\lambda$ = 50 nm

Work function = Energy required to remove the electron from $n_1$ = 1 to $n_2=\infty$

E = $13.6(\frac{1}{1}-\frac{1}{\infty })=13.6$

$\frac{hc}{\lambda }-13.6=KE$

$\frac{1242}{50}-13.6=KE$

$\Rightarrow KE=24.84-13.6$

= 11.24 e V