Question

How many chlorine atoms can you ionize in the process $Cl \rightarrow Cl^{+} + e^{-},$ by the energy liberated from the following process :

$\mathbf{Cl +}\mathbf{e}^{\mathbf{-}}\mathbf{\rightarrow C}\mathbf{l}^{\mathbf{-}}$for$6 \times 10^{23}$ atoms

Given electron affinity of $Cl = 3.61eV,$ and $IP$ of

$Cl = 17.422eV$

• A. 1.24$\times 10^{23}$ atoms
• B. $9.82 \times 10^{20}$ atoms
• C. $2.02 \times 10^{15}$atoms
• D. None of these

Answer 1

Answer: (A)

1.24$\times 10^{23}$ atoms

Answer 2

Energy released in conversion of $6 \times 10^{23}$ atoms of $Cl^{-}$ ions = $6 \times 10^{23}$× electron affinity

= 6×$10^{23} \times 3.61 = 2.166 \times 10^{24}$eV.

Let x Cl atoms are converted to $Cl^{+}$ ion

Energy absorbed $= x \times$ ionization energy

$x \times 17.422 = 2.166 \times 10^{24}$; $x = 1.243 \times 10^{23}$ atoms