Question

The ionisation energy of $H$ - atom is $13.6$ eV. What will be the ionisation energy (in eV) of $L{i}^{2+}$ ion

Answer 1

Given,
The ionisation energy of hydrogen atom = 13.6 eV
We know,
Ionisation energy = $13.6\frac{Z^2}{n^2}$ eV
[Since ionisation energy is defined for ground state only, thus $n=1$]
Thus,
$\text{Ionisation enrgy (I.E)}\propto Z^2$
$\therefore \frac{I.E_H}{I.E_{L{i}^{2+}}}=\frac{Z_H^2}{Z_{L{i}^{2+}}^2}$
$\Rightarrow I.E_{L{i}^{2+}}=(I.E{)}_H\frac{Z_{L{i}^{2+}}^2}{Z_H^2}$
Therefore, the ionisation energy of $L{i}^{2+}$ ion
= $13.6\times \frac{3^2}{1^2}eV=122.4eV$