If the average translational kinetic energy of molecule in a gas is equal to the kinetic energy of an electron acceleration from rest through 10 V , then the temperature of the gas molecule is
(Boltzmann constant `= 1 . 38 xx 10^(-23) JK ^(-1)` )

Options

A. `7 . 73 xx 10^(23) K`
B. `730 K`
C. 737 K
D. `77.3 xx 10^(3) K`

Answer

D

Explanation

Given,
V = 10 V
`k = 13.8 xx 10^(-23 JK^(-1)`
According to question the transiation KE of a
molecule of gas ` = (3)/(2) kT`
According to the question,
`(3)/(2) kT = eV` `= (xx 16 xx 10^(-19) xx 10)/(3 xx 1.38 xx 10^(-23))`
`= 7.73 xx 10^(4)`
`77.3 xx 10^(3) K`