The ionisation energy of a hydrogen atom is $13.6\,\text{eV}$. The energy required to remove the only electron of a $\text{He}^{+}$ ion from its first excited state ($n=2$) is
A$3.4\,\text{eV}$
B$54.4\,\text{eV}$
C$27.2\,\text{eV}$
D$13.6\,\text{eV}$
Answer & Solution
Correct answer: D. $13.6\,\text{eV}$
For a hydrogen-like ion, $E_n = -13.6 \cdot \tfrac{Z^2}{n^2}\,\text{eV}$. For $\text{He}^+$, $Z=2$. In the first excited state ($n=2$): $E_2 = -13.6 \cdot \tfrac{4}{4} = -13.6\,\text{eV}$. The energy required to free the electron (take it to $n=\infty$ where $E=0$) is $13.6\,\text{eV}$.
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