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

Sol: Maximum energy is liberated for transition E_{n} → 1 and minimum energy for E_{n} → E_{n-1}

Hence, $\large \frac{E_1}{n^2} – E_1 = 52.224 eV $ …(i)

And , $\large \frac{E_1}{n^2} – \frac{E_1}{(n-1)^2} = 1.224 eV $ …(ii)

Solving above equations simultaneously, we get

E_{1} = -54.4 eV and n = 5 ; Now E_{1}=-(13.6 Z^{2})/1^{2} = -54.4 eV

Hence, Z = 2 i.e., gas in helium originally excited to n = 5 energy state.