Question

If the energy of hydrogen atom in the ground state is $-13.6eV$, then energy of $H{e}^{+}$ ion in first excited state will be

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

Answer 1

The correct option is B $-13.6eV$

$E_{He}=-\frac{Z^2}{n^2}\times 13.6eV$
For $H{e}^{+}$ ion $Z=2$ and for first excited state $n=2$
$\therefore E_{H{e}^{+}}=-\frac{2^2}{2^2}\times 13.6eV$
$=-13.6eV$.