Q: Consider a hydrogen-like atom whose energy in nth 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 Emax = 52.224 eV and least energetic photons have energy Emin = 52.224 eV. Find the atomic number of atom the state of excitation.
Sol: Maximum energy is liberated for transition En → 1 and minimum energy for En → En-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
E1 = -54.4 eV and n = 5 ; Now E1=-(13.6 Z2)/12 = -54.4 eV
Hence, Z = 2 i.e., gas in helium originally excited to n = 5 energy state.