Question

The energy of a hydrogen atom in the ground state is $-13.6\, eV$. The energy of a $He^+$ ion in the first excited state will be

• A. -13.6 eV
• B. -27.2 eV
• C. -54.4e V
• D. -6.8eV

Answer 1

Answer: (A)

-13.6 eV

Answer 2

Energy of an hydrogen like atom like He$^+$ in an n$^{th}$ orbit is given by $E_n =-\frac{13.6Z^2}{n^2}eV$ For hydrogen atom, $Z = 1$ $\therefore \, \, \, E_n =-\frac{13.6}{n^2}eV$ For ground state, $n = 1$ $\therefore \, \, E_1=-\frac{13.6}{1^2}eV=-13.6\,eV$ For $He^{+} ion , Z=2$ $E_n=-\frac{4(13.6)}{n^2}eV$ For first excited state, n = 2 $\therefore \, \, \, E_2=-\frac{4(13.6)}{(2)^2}eV=-13.6\,eV$ Hence, the energy in $He^+$ ion in first excited state is same that of energy of the hydrogen atom in ground state i.e.$- 13.6 \,eV$