Q. Consider a hydrogen like atom whose energy in $n^{th}$ excited state is given by $E_{n} = - \frac{13.6Z^{2}}{n^{2}}$. 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) 5
C) 4
D) None of these
Explanation
(a) Maximum energy is liberated for transition $E_{n} \rightarrow 1$ and minimum energy for $E_{n} \rightarrow E_{n - 1}$
Hence $\frac{E_{1}}{n^{2}} - E_{1} = 52.224eV$ …(i)
and $\frac{E_{1}}{n^{2}} - \frac{E_{1}}{(n - 1)^{2}} = 1.224eV$ …(ii)
Solving equations (i) and (ii), we get
$E_{1} = - 54.4\ eV$ and $n = 5$
Now $E_{1} = - \frac{13.6Z^{2}}{1^{2}} = - 54.4\ eV$. Hence $Z = 2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.