A He^{+}ion is in its first excited state. Its ionization energy is:

Question:

$\mathrm{AHe}^{+}$ion is in its first excited state. Its ionization energy is:

  1. (1) $48.36 \mathrm{eV}$

  2. (2) $54.40 \mathrm{eV}$

  3. (3) $13.60 \mathrm{eV}$

  4. (4) $6.04 \mathrm{eV}$


Correct Option: , 3

Solution:

(3) $E_{n}=-13.6 \frac{Z^{2}}{n^{2}}$

For $\mathrm{He}^{+}, E_{2}=\frac{-13.6(2)^{2}}{2^{2}}=-13.60 \mathrm{eV}$

Ionization energy $=0-E_{2}=13.60 \mathrm{eV}$

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