Question

The average $KE$ of molecules in a gas at temperature $T$ is $\frac{3}{2}kT$. Find the temperature at which the average KE of molecules equal to binding energy of its atoms.

• A. $1.05\times {10}^{5}K$
• B. $1.05\times {10}^{4}K$
• C. $1.05\times {10}^{3}K$
• D. none of these

Answer 1

The correct option is A $1.05\times {10}^{5}K$

Binding energy of atoms $=13.6eV=2.1789\times {10}^{-18}J$

Now since this energy is equal to $\frac{3}{2}kT$

$T=\frac{2}{3}\frac{2.1789\times {10}^{-18}}{k}=1.05\times {10}^{5}K$