If you are given Avogadro’s number of atoms of a gas ‘X’. If half of the atoms are converted into X⁺_(g) by energy $∆\mathrm{H}$. The IE of ‘X’ is

Given number of atoms=Avogadro's number of atoms =N_A=6.023×10²³
Given that $\frac{{\mathrm{N}}_{\mathrm{A}}}{2}$ atoms are ionised, 

i.e., Ionsation energy of N_A/2 atoms of gas X=Δ H
∴ Ionisation energy of one atom of gas X= $\frac{\mathrm{\Delta H}}{({\mathrm{N}}_{\mathrm{A}}/2)}=\frac{2\times \mathrm{\Delta H}}{{\mathrm{N}}_{\mathrm{A}}}$