Question

$\frac{N_0}{2}$ atoms of $X\left(g\right)$ are converted into $X^{+}\left(g\right)$ by energy $E_1$. $\frac{N_0}{2}$ atoms of $X\left(g\right)$ are converted into $X^{-}\left(g\right)$ by energy $E_2$. Hence, ionisation potential and electron affinity of $X\left(g\right)$ are:

• A. $\frac{2E_1}{N_0},\frac{2(E_1-E_2)}{N_0}$
• B. $\frac{2E_1}{N_0},\frac{2E_2}{N_0}$
• C. $\frac{(E_1-E_2)}{N_0},\frac{2E_2}{N_0}$
• D. none of the above

Answer 1

The correct option is B $\frac{2E_1}{N_0},\frac{2E_2}{N_0}$

IP is the energy require to remove valence electron from atom.
similarly, electron affinity is energy releases when one atom accepts one electron.
Let $IP=I$
$EA=E$
$X\left(g\right)\to X^{+}\left(g\right)+e^{-}$

$\frac{N_0}{2}I=E_1$

$I=\frac{2E_1}{N_0}$

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

$\frac{N_{0}E}{2}=E_2$

$E=\frac{2E_2}{N_0}$