Question

The energy of $H{e}^{+}$ in the ground state is $-54.4eV$, then the energy of $L{i}^{++}$ in the first excited state will be

• A. $-30.6eV$
• B. $27.2eV$
• C. $-13.6eV$
• D. $-27.2eV$

Answer 1

The correct option is A $-30.6eV$

Energy of electron in $n^{th}$ orbit is, $E_n=-\left(Rch\right)\frac{Z^2}{n^2}=-54.4eV$
For $H{e}^{+}$ is ground state:
$E_1=-\left(Rch\right)\frac{(2{)}^2}{(1{)}^2}=-54.4\Rightarrow Rch=13.6$

For $L{i}^{++}$ in first excited state $(n=2)$:
$E=-13.6\times \frac{(3{)}^2}{(2{)}^2}=-30.6eV$