Question

How many $Cl$ atoms can you ionize in the process $Cl\to C{l}^{+}+e^{-}$ by the energy liberated from the process $Cl+e^{-}\to C{l}^{-}$ for Avogadro number of atoms?

Given, $IP=13.0eV$ and $EA=4.29eV.$

• A. $1\times {10}^{23}$
• B. $1.5\times {10}^{23}$
• C. $2\times {10}^{23}$
• D. $2.5\times {10}^{23}$

Answer 1

The correct option is C $2\times {10}^{23}$

Energy liberated $=4.29\times 6.02\times {10}^{23}$

Now, by 13 eV energy number of atoms that get ionized = 1

So, by $4.29\times 6.02\times {10}^{23}eV$, number of atoms ionize $=\frac{1}{13}\times 4.29\times 6.02\times {10}^{23}\approx 2\times {10}^{23}$

Hence, the correct option is $\text{C}$