If the average translational kinetic energy of a molecule in a gas is equal to the kinetic energy of an electron accelerating from rest through 10 ~V , then the temperature of the gas molecule is ( Boltzmann constant =1.38 10⁻²³ JK ⁻¹ )

Step-by-step solution

According to question, the translation KE of a molecule of gas = 3 2 k t According to the question, aligned & 3 2 k t=e V 3 2 k t=10 e & aligned T & = 20 e 3 K_B = 20 1.6 10⁻¹⁹ 3 1.38 10⁻²³ & =77.3 10^3 ~K aligned aligned