$A_{0} / 2$ atoms of $X (g)$ are converted into $X^{+} (g)$ by energy $E_{1} \cdot A_{0} / 2$ atoms of $X (g)$ are converted into $X^{-} (g)$ of energy $E_{2}$ . Hence, ionization energy and electron affinity of $X (g)$ are :

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$\frac{2 E_{1}}{A_{0}}, \frac{2 (E_{1} - E_{2})}{A_{0}}$

none of these

$\frac{(E_{1} - E_{2})}{A_{0}}, \frac{2 E_{2}}{A_{0}}$

$\frac{2 E_{1}}{A_{0}}, \frac{2 E_{2}}{A_{0}}$

answer is B.

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Detailed Solution

The ionization energy and electron affinity of $X (g)$ are, ionization energy and electron affinity of $X (g)$ are $\frac{2 E_{1}}{A_{0}}, \frac{2 E_{2}}{A_{0}}$ .

The ionization energy for the formula is found as

$X ⟶ X^{+}; {IE}_{1}$

$\begin{array}{r} \frac{N_{0}}{2} I = E_{1} \end{array}$

$I = \frac{2 E_{1}}{N_{0}}$

The electron affinity for the formula is found as

$X ⟶ X^{-}; {EA}_{1}$

$\begin{aligned} \frac{N_{0} E}{2} = E_{2} \end{aligned}$

$E = \frac{2 E_{2}}{N_{\circ}}$

Hence, the ionization energy and electron affinity of $X (g)$ are ionization energy and electron affinity of $X (g)$ are $\frac{2 E_{1}}{A_{0}}, \frac{2 E_{2}}{A_{0}}$