The average translational kinetic energy of a molecule in a gas becomes equal to (0.69 eV ) at temperature about, [Boltzmann's constant (=138 10⁻²³ ~J ~K ⁻¹ ) ]

Step-by-step solution

Given, average translational kinetic energy (=0.69 eV =0.69 1.6 10⁻¹⁹ ~V ) As we know that, average translational kinetic energy (= 3 2 k T ) ( aligned & 0.69 1.6 10⁻¹⁹= 3 2 1.38 10⁻²³ T & T= 0.69 1.6 10⁻¹⁹ 2 3 1.38 10⁻²³ & T=5333 ~K =5333-273=5060^ C aligned )