Question

Consider a hydrogen like atom whose energy in $n^{th}$ excited state is given by $E_n=-\frac{13.6}{n^2}{Z}^2$. Whenthis excited atom makes a transition from an excited state to ground state. The most energetic photons have energy $E_{max}=52.224eV$ and the least energetic photons have energy $E_{min}=1.224eV$. Find the atomic number of atom.

Answer 1

Maximum energy is liberated for transition $E_n\to E_1$ and minimum energy for $E_n\to E_{n-1}$
Hence, $\frac{E_1}{n^2}-\frac{E_1}{12}=52.224eV$ and $\frac{E_1}{n^2}-\frac{E_1}{(n-1{)}^2}=1.224eV$
Solving we get, $E_1=-54.4eV$ and $n=5$ hence, $E_1=-\frac{13.6Z^2}{12}=-54.4Z=2$