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What temperature does the average translational K.E. of a molecule in a gas becomes equal to K.E. of an electron accelerated from through potential difference of 5 volt? `(K=1.38xx10^(-23))`

A

`38.65xx10^(3)K`

B

`0.3865xx10^(3)K`

C

`3.865xx10^(3)K`

D

`38.65K`

Text Solution

Verified by Experts

The correct Answer is:
A
`K.E.=ev`
`(3)/(2)KT=1.6xx10^(-19)xx5`
`therefore T=(3.2xx5xx10^(-19))/(3xx8.314xx10^(3))`
`=38.65xx10^(3)K`
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