$\mathrm{A} \mathrm{He}^{+}$ion is in its first excited state. Its ionization energy is :
Select the correct option:
A
13.60 eV
B
6.04 eV
C
48.36 eV
D
54.40 eV
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$$
E_n: \frac{-E_0 z^2}{n^2}=\frac{-E_0 \times 4}{4}=-E_0
$$
To ionise it $\mathrm{E}_0$ energy must be supplied.
$$
\therefore \mathrm{E}_0=13.6 \mathrm{eV} .
$$
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