At what temperature does the average translational K.E. of a molecule in a gas becomes equal to K.E. of an electron accelerated from rest through potential difference of V volt ? All symbols have their usual meaning.

Options

A. `(2eVN)/(3R)`
B. `(3R)/(2eVN)`
C. `(NeV)/R`
D. `(2NeV)/R`

Answer

A

Explanation

The average translational K.E. of the molecules of a gas is given by `1/2mv^(2)=3/2KT=3/2(RT)/N" "...(1)`
and the K.E. acquired by an electron accelerated from rest through a P.D., (V) is given by `1/2mv^(2)=eV" "...(2)`
`therefore` from (1) and (2), `3/2(RT)/N=eV`
`thereforeT=(2eVN)/(3R)`