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Q. Ionization energy of $He^{+}$ ion at minimum energy position is

NTA AbhyasNTA Abhyas 2020Atoms

Solution:

The energy of helium ions.
$E_{n}=-\frac{13 .6 \, Z^{2}}{n^{2}} \, eV$
In minimum position, $n=1$ .
For $ \, He^{+}, \, Z=2$ .
$E=\frac{-13.6 \times(2)^{2}}{1} eV$
$E=-\text{54.4} \, eV$
Ionization Energy = $-E=54.4eV$