Question

Light from a discharge tube containing hydrogen atoms falls on the surface of a piece of sodium. The kinetic energy of the fastest photo electrons emitted from sodium is 0.73 eV. The work function for sodium is 1.82 eV. Find the quantum numbers of the two levels involved in the emission of these photons.

Answer 1

$E_k=0.73eV,W=1.82eV$
Ionization energy of H atom $=13.6eV$
The electronic energy levels of H atoms are given by:
$E_n=\frac{Rhc}{n^2}=-\frac{13.6}{n^2}eV$
For $n=1,E_1=-13.6eV$
For $n=2,E_2=-3.4eV$
For $n=3,E_3=-1.51eV$
For $n=4,E_4=-0.85eV$
Clearly, $E_4-E_2=-0.85eV-(-3.4eV)=2.55eV$
i.e., Quantum numbers involved in the photons of energy 2.55 eV are 2 and 4. The transition is specified by
$n_1=4\to n_2=2$