The average kinetic energy of a monatomic molecule is 0 . 414eV at...

For monatomic molecule, degree of freedom $= 3$.

The formula for the average kinetic energy is given by

${K}_{avg} = \dfrac{1}{2}f{K}_{B}T . . . \left(1\right)$

From equation (1), it follows that

$0 . 414eV \times \dfrac{1 . 6 \times {10}^{- 19}J}{1eV} = \dfrac{3}{2} \times 1 . 38 \times {10}^{- 23}J{mol}^{- 1}{K}^{- 1} \times T ⇒ T = \dfrac{0 . 414eV \times \dfrac{1 . 6 \times {10}^{- 19}J}{1eV} \times 2}{3 \times 1 . 38 \times {10}^{- 23}J{mol}^{- 1}{K}^{- 1}} = 3200K$