Question

It takes $E_{1}$ units of energy to convert $N_{0}/2$ atoms of $\text{X}\left(\text{g}\right)$ into $\left(\text{X}\right)^{+}\left(\text{g}\right)$ and $E_{2}$ units of energy to convert $N_{0}/2$ atoms of $\text{X}\left(\text{g}\right)$ into $\left(\text{X}\right)^{-}\left(\text{g}\right)$ . What is the ionisation potential and electron affinity of $\text{X}\left(\text{g}\right)$ ?

• A. $\frac{2 E_{1}}{N_{0}}\text{,}\frac{2 \left(E_{1} - E_{2}\right)}{N_{0}}$
• B. $\frac{2 E_{1}}{N_{0}}\text{,}\frac{2 E_{2}}{N_{0}}$
• C. $\frac{\left(E_{1} - E_{2}\right)}{N_{0}}\text{,}\frac{2 E_{2}}{N_{0}}$
• D. None is correct.

Answer 1

Answer: (B)

Ionisation potential is the amount of energy required to convert one atom of $\text{X }\left(\text{g}\right)$ into $\left(\text{X}\right)^{+} \, \left(g\right)$ .
$\text{X }\left(\text{g}\right) \rightarrow \left(\text{X}\right)^{+}\left(\text{g}\right)+\left(\text{e}\right)^{-}$
If I is the ionisation potential, then
$\frac{N_{0}}{2}\left(\text{I}\right)=E_{1}$
$\therefore $ $\text{I}=\frac{2 E_{1}}{N_{0}}$
Electron affinity is the amount of energy required to convert one atom of $\text{X }\left(\text{g}\right)$ into $\left(\text{X}\right)^{-} \, \left(g\right)$ .
If E is electron-affinity, then
$\text{X }\left(\text{g}\right)+\left(\text{e}\right)^{-} \rightarrow \left(\text{X}\right)^{-}\left(\text{g}\right)$
$\frac{N_{0}}{2}\left(\text{E}\right)=E_{2}$
$\therefore $ $\text{E}=\frac{2 E_{2}}{N_{0}}$