Question

Consider a hydrogen like atom whose energy in nth excited state is given by $ {{E}_{n}}=-\frac{13.6{{Z}^{2}}}{{{n}^{2}}}, $ when this excited atom makes a 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 }}=1.224\,eV. $ The atomic number of atom is

• A. 2
• B. 4
• C. 5
• D. None of the above

Answer 1

Answer: (A)

Maximum energy is liberated for transition
$ {{E}_{n}}\to 1 $ and minimum energy for
$ {{E}_{n}}\to {{E}_{n-1}} $
Hence, $ \frac{{{E}_{1}}}{{{n}^{2}}}-{{E}_{1}}=52.224\,eV $ ..(i)
and $ \frac{{{E}_{1}}}{{{n}^{2}}}-\frac{E}{{{(n-1)}^{2}}}=1.224\,eV $ ..(ii)
Solving Eqs. (i) and (ii),
we get $ {{E}_{1}}=-54.4\,eV $ and $ n=5 $
But $ {{E}_{1}}=-\frac{13.6{{Z}^{2}}}{{{1}^{2}}} $
$ \therefore $ $ -54.4=\frac{13.6}{{{1}^{2}}}{{Z}^{2}} $
$ \Rightarrow $ $ Z=2 $