Question

How many chlorine atoms can you ionize in the process?
$Cl\to {Cl}^{+}+e^{-}$ by the energy liberated from the following process:
$Cl+e\to {Cl}^{-}$ for $6\times {10}^{23}$ atoms
given that electron affinity of chlorine is $3.61eV$ and ionization energy of chlorine is $17.422eV$.

Answer 1

Total energy liberated per mole when one electron is added in $6.023\times {10}^3$ in Cl atoms

$=3.61\times 1.6\times {10}^{-19}\times 6.023\times {10}^3=34.78\times {10}^{4}J/mole\to \left(1\right)$

Let the number of Chlorine atoms ionized be $n$

Total energy liberate when one electron is added in $n$ Cl atoms

$=17.422\times 1.6\times {10}^{-19}\times n\to \left(2\right)$

Energy of $\left(1\right)$ is used in $\left(2\right)$

$\Rightarrow \left(1\right)=\left(2\right)34.78\times {10}^4=17.422\times 1.6\times {10}^{-19}\times n\Rightarrow n=1.247\times {10}^{23}atoms$