Question

The average thermal energy of a oxygen atom at room temperature $\left({27}^{o}C\right)$

• A. $4.5\times {10}^{-21}J$
• B. $6.2\times {10}^{-21}J$
• C. $3.4\times {10}^{-21}J$
• D. $1.8\times {10}^{-21}J$

Answer 1

The correct option is B $6.2\times {10}^{-21}J$

Boltzman's constant is given as $k=1.38\times {10}^{-23}J/k$

Average kinetic energy will be $KE=\frac{FkT}{2}$

Where $F$ is degree of freedom which will be $3$ because $O$ is a mono-atomic molecule.

Temperature, $T=27+273=300K$

So energy=$\frac{3\times 1.38\times {10}^{-23}\times 300}{2}=6.2\times {10}^{-21}Joule$

Option B is correct.