The ionisation energies of Lithium and Sodium are 520kJ Mol^–1and 495kJ Mol^–1 respectively. The energy required to convert all the atoms present in 7mg of Li vapour and 23 mg of sodium vapour to their respective gaseous cations respectively are

IE of $\mathrm{Li}=520\mathrm{Kj}/\mathrm{mol}$

1 mole of gaseous Li requires $520\mathrm{KJ}$ of energy to ionise each atom. $1 \mathrm{mol}=6.02\times {10}^{23}$atoms

So, 1 atom $=\frac{1}{6.02\text{ times }{10}^{23}} \mathrm{mol}$

$520\mathrm{KJ}=520000 \mathrm{J}=5.2\times {10}^5 \mathrm{J}$

$1 \mathrm{mol}\to 5.2\times {10}^5 \mathrm{J}$

$=\frac{7\times {10}^{-3}}{7} \mathrm{mol}\to 5.2\times {10}^5\times {10}^{-3} \mathrm{J}$

$=5.2\times {10}^2 \mathrm{J}$

$=520 \mathrm{J}$

IE of $\mathrm{Na}=495\mathrm{KJ}/\mathrm{mol}$

$1 \mathrm{mol}⟶495\mathrm{KJ}$ energy

$=\frac{23\times {10}^{-3}}{23} \mathrm{mol}\to 495\times {10}^{-3}\mathrm{KJ}$

$=495 \mathrm{J}$

So, energy required to convert $\mathrm{Li}$and $\mathrm{N}$ a to their respective gaseous cations are $520 \mathrm{J}$ and $495 \mathrm{J}$.

Therefore, Option B is the correct answer.