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. $\frac{2E_1}{N_0}\text{,}\frac{2\left(E_1-E_2\right)}{N_0}$
• B. $\frac{2E_1}{N_0}\text{,}\frac{2E_2}{N_0}$
• C. $\frac{\left(E_1-E_2\right)}{N_0}\text{,}\frac{2E_2}{N_0}$
• 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 ${\left(\text{X}\right)}^{+}\left(g\right)$ .
$\text{X }\left(\text{g}\right)\to {\left(\text{X}\right)}^{+}\left(\text{g}\right)+{\left(\text{e}\right)}^{-}$
If I is the ionisation potential, then
$\frac{N_0}{2}\left(\text{I}\right)=E_1$
$\therefore$ $\text{I}=\frac{2E_1}{N_0}$
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
$\text{X }\left(\text{g}\right)+{\left(\text{e}\right)}^{-}\to {\left(\text{X}\right)}^{-}\left(\text{g}\right)$
$\frac{N_0}{2}\left(\text{E}\right)=E_2$
$\therefore$ $\text{E}=\frac{2E_2}{N_0}$