Question

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

• A. $1.243\times {10}^{22}$ atoms
  
• B. $1.243\times {10}^{23}$ atoms
  
• C. $2.243\times {10}^{22}$ atoms
  
• D. $2.243\times {10}^{23}$ atoms
  

Answer 1

The correct option is B $1.243\times {10}^{23}$ atoms


Energy released in conversion of =$6\times {10}^{23}$ atoms to $C{l}^{-}$ ions.
$=6\times {10}^{23}\times \text{electron affinity}=6\times {10}^{23}\times 3.61=2.166\times {10}^{24}eV$
Let x, chlorine atoms are conversed to $C{l}^{+}$ ion.
Energy absorbed $=N\times \text{ionization energy}=N\times 17.422eV$
$\therefore N\times 17.422eV=2.166\times {10}^{24}$
$N=1.243\times {10}^{23}$ atoms
hence, number of chlorine atom = $1.243\times {10}^{23}$ atoms