Question

The energy of a hydrogen atom in the ground state is $-13.6eV$. The energy of a $H{e}^{+}$ 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)

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