91Ó°ÊÓ

Electron affinity is the measure of the energy produced when an electron is added to a neutral atom to create an anion.

The Energy input=Energy Output

Energy input+liberated Energy by Chlorine=Ionisation Energy of Sodium

Thus, Energy Input=Ionisation Energy of Sodium-Liberated Energy by chlorine..

$5.14\text{ e³Õ}−3.61\text{ e³Õ}=1.53\text{ e³Õ}$

Thus The Energy input is, $1.53\text{ e³Õ}$.

The potential energy between the Sodium ion and the chlorine ion is given by

$\mathrm{U}=−\frac{{\mathrm{e}}^2}{4\mathrm{π}{∈}_0\mathrm{a}}$

To, Calculate the Separation in order to release $1.53\text{ e³Õ}$

$\mathrm{a}=−\frac{{\mathrm{e}}^2}{4\mathrm{π}{∈}_0\mathrm{U}}$

Substitute,$\mathrm{e}=1.6×{10}^{−19}$ ,${∈}_0=8.85×{10}^{−12}{\text{ C}}^{\text{2}}{\text{/N.m}}^2$, and$\mathrm{U}=1.53\text{ e³Õ}$in the above equation.

Hence, Ion need to be at least as close at 0.94 nm so that it can release as much energy as it spends when forming Ions.

$\mathrm{U}=−\frac{{\mathrm{e}}^2}{4\mathrm{π}{∈}_0\mathrm{a}}$

Substitute,$\mathrm{e}=1.6×{10}^{−19}$,${∈}_0=8.85×{10}^{−12}{\text{ C}}^{\text{2}}{\text{/N.m}}^2$, and$\mathrm{a}=0.24\text{ n³¾}$in the above equation.

Thus Sodium Energy’s is lowered by

$(5.99\text{ e³Õ}−1.53\text{ e³Õ})=4.46\text{ e³Õ}$

Hence, The molecule Sodium Chlorine’s energy is lowered by $4.46\text{ e³Õ}$..