91影视

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..

$5.14\text{鈥塭痴}鈭�3.61\text{鈥塭痴}=1.53\text{鈥塭痴}$

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

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

$\mathrm{U}=鈭�\frac{{\mathrm{e}}^2}{4\mathrm{蟺}{鈭�}_0\mathrm{a}}$

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

$\mathrm{a}=鈭�\frac{{\mathrm{e}}^2}{4\mathrm{蟺}{鈭�}_0\mathrm{U}}$

Substitute,$\mathrm{e}=1.6脳{10}^{鈭�19}$ ,${鈭�}_0=8.85脳{10}^{鈭�12}{\text{鈥塁}}^{\text{2}}{\text{/N.m}}^2$, and$\mathrm{U}=1.53\text{鈥塭痴}$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.

$\mathrm{U}=鈭�\frac{{\mathrm{e}}^2}{4\mathrm{蟺}{鈭�}_0\mathrm{a}}$

Substitute,$\mathrm{e}=1.6脳{10}^{鈭�19}$,${鈭�}_0=8.85脳{10}^{鈭�12}{\text{鈥塁}}^{\text{2}}{\text{/N.m}}^2$, and$\mathrm{a}=0.24\text{鈥塶尘}$in the above equation.

Thus Sodium Energy鈥檚 is lowered by

$(5.99\text{鈥塭痴}鈭�1.53\text{鈥塭痴})=4.46\text{鈥塭痴}$

Hence, The molecule Sodium Chlorine鈥檚 energy is lowered by $4.46\text{鈥塭痴}$..