$\left(\mathbb{E}{)}_1\text{ and }\right(\mathbb{E}{)}_2$ of  Mg(g) are $740,1540\mathrm{kJ}{\mathrm{mol}}^{-1}$. Calculate percentage of ${\mathrm{Mg}}_{\left(\mathrm{g}\right)}^{+}$ and  ${\mathrm{Mg}}^{2+}$ if 1 mole of ${\mathrm{Mg}}_{\left(\mathrm{g}\right)}$ absorbs 1300 kJ of energy.

$Mg+740\mathrm{KJ}\to M^{+}$
                After IP₁ Energy remained = 1300 – 740 = 560 KJ

$\frac{560}{1540}\times 1=0.38\text{ moles of }{\mathrm{Mg}}^{+}$
NO.OF MOLES $M{g}^{+}\text{ionized }=0.38$
$\text{ no of moles of }{\mathrm{Mg}}^{+}\mathrm{Left}=1-0.38=0.62\text{ i.e }62\mathrm{\%}$
$\therefore \text{ no of moles of }M{g}^{+2}\text{ formed }=0.38=38\mathrm{\%}$