$\text{A°/2}$  atoms of X(g) are converted into ${\text{X}}_{\left(g\right)}^{+}$ by absorbing energy $E_1$ . $\text{A°/2}$  ions of  ${\text{X}}_{\left(g\right)}^{+}$ are converted into ${\text{X}}_{\left(g\right)}^{-}$ with release of energy $E_2$ . Hence ionization energy and electron affinity of X(g) are:

$X\left(g\right)+E_1\to X_{\left(g\right)}^{+}+e^{-}$

$\left(\frac{A_0}{2}atoms\right)$

$\frac{A_0}{2}{\text{\hspace{0.17em}Atoms absorbs E}}_1\text{\hspace{0.17em}energy}$

So, 1 atom absorbs  $\frac{2E_1}{A_0}$  energy

Hence, ionization energy of X(g) is  $\frac{2E_1}{A_0}$

$X\left(g\right)+\frac{2E_1}{A_0}\to X_{\left(g\right)}^{+}+e^{-}\mathrm{........}eq\left(1\right)$

$\underset{\left(\frac{A_0}{2}uons\right)}{X_{\left(g\right)}^{+}}+2e^{-}\to X_{\left(g\right)}^{-}+E_2$

$\frac{A_0}{2}$  ions release  $E_2$  energy

So, 1 ion releases $\frac{2E_2}{A_0}$ energy

Hence, $\underset{\left(1ion\right)}{X_{\left(g\right)}^{+}}+2e^{-}\to X_{\left(g\right)}^{-}+\frac{2E_2}{A_0}\mathrm{........}eq\left(2\right)$

Equation for electron affinity of X(g) is, $S\left(g\right)+e^{-}\to X_{\left(g\right)}^{-}+EAofX\left(g\right)$

We can get this equation by adding equation (1) and (2).

$X\left(g\right)+e^{-}\to X_{\left(g\right)}^{-}+\frac{2(E_2-E_1)}{A_0}$

Hence, electron affinity of X(g) is $\frac{2(E_2-E_1)}{A_0}$