• # question_answer $\frac{{{N}_{0}}}{2}$ atoms of $X(g)$ are converted into ${{X}^{+}}(g)$ by energy ${{E}_{1}}\cdot \frac{{{N}_{0}}}{2}$ atoms of $X(g)$ are converted into ${{X}^{-}}(g)$by energy ${{E}_{2}}$. Hence, ionization potential and electron affinity of $X(g)$ are A)  $\frac{2{{E}_{1}}}{{{N}_{0}}},\,\,\frac{2({{E}_{1}}-{{E}_{2}})}{{{N}_{0}}}$    B)  $\frac{2{{E}_{1}}}{{{N}_{0}}},\,\,\frac{2{{E}_{2}}}{{{N}_{0}}}$ C)  $\frac{{{E}_{1}}-{{E}_{2}}}{{{N}_{0}}},\,\,\frac{2{{E}_{1}}}{{{N}_{0}}}$        D)  None of these

$\because \,\,X(g)\,\to \,{{X}^{+}}(g)\,+{{e}^{-}};\,\,\Delta H={{E}_{1}}$ for $\frac{1}{2}$ mol $\therefore$                $IP=\frac{2{{E}_{1}}}{{{N}_{0}}}$ Again $\because$     $X(g)+{{e}^{-}}\to {{X}^{-}}(g),$ $\Delta H=-{{E}_{2}}$ for $\frac{1}{2}$ mole $\therefore$    $EA=\frac{2{{E}_{2}}}{{{N}_{0}}}$