Question

Electron affinity of chlorine is -3.61 eV. $IP$ value of the same element is 12.967 eV. $3\times {10}^{23}$ atoms of chlorine are converted to $C{l}^{-}$ ions and the energy liberated is used to ionize chlorine atoms to $C{l}^{+}$ ions. How many chlorine atoms could be ionized in the above process?

• A. $5.45\times {10}^{21}$
• B. $6.11\times {10}^{18}$
• C. $8.35\times {10}^{22}$
• D. $7.11\times {10}^{28}$

Answer 1

The correct option is C $8.35\times {10}^{22}$

Energy released for the conversion of $Cl$ to $C{l}^{-}=-3.61$ eV per atom
Energy released for the $3\times {10}^{23}Cl$ atoms $=3\times {10}^{23}\times 3.61$ eV
Energy required for the conversion of $Cl$ to $C{l}^{+}=12.97$ eV per atom
No. of atoms ionized $=x$
Energy required $=x\times 12.97$
$x\times 12.97=3\times {10}^{23}\times 3.61$
$x=\frac{3\times {10}^{23}\times 3.61}{12.97}=8.35\times {10}^{22}$