Question

How many chlorine atoms can you ionize in the process $Cl\to C{l}^{+}+e^{-}$, by the energy liberated from the following process? $Cl+e^{-}\to C{l}^{-}$ for 6 $\times {10}^{23}$ atoms. Given electron affinity of Cl = 3.61 eV/atom, and IP of Cl= 17.422 eV/atom

• 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

The correct option is A

$1.24\times {10}^{23}$atoms

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

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

Let $x$ Cl atoms be converted to $C{l}^{+}$ ion.

Energy absorbed = $x$ $\times$ ionization energy

$\Rightarrow x\times 17.422=2.166\times {10}^{24}$

$\Rightarrow x=1.243\times {10}^{23}$atoms