Question

The energy of $H{e}^{+}$ ion in the first excited state will be

• A. $-13.6eV$
• B. $-14.4eV$
• C. $-6.8eV$
• D. $-27.2eV$

Answer 1

Answer: (A)

$E=-\left(13.6\right)\frac{n^2}{Z^2}eV$
$E=-13.6\frac{2^2}{2^2}=-13.6eV$