The energy of a hydrogen atom in the ground state is $-13.6 \mathrm{eV}$. The energy of a…
The energy of a hydrogen atom in the ground state is $-13.6 \mathrm{eV}$. The energy of a $\mathrm{He}^{+}$ion in the first excited state will be
$-13.6 \mathrm{eV}$
$-27.2 \mathrm{eV}$
$-54.4 \mathrm{eV}$
$-6.8 \mathrm{eV}$
Solution
Energy E of an atom with principal quantum number $\mathrm{n}$ is given by $E=\frac{-13.6}{\mathrm{n}^2} Z^2$ for first excited state $\mathrm{n}=2$ and for $\mathrm{He}^{+} \mathrm{Z}=2$
$\begin{aligned}
\Rightarrow \quad \mathrm{E} & =\frac{-13.6 \times(2)^2}{(2)^2} \\
& =-13.6 \mathrm{eV}
\end{aligned}$