Calculate the temperature at which the average K.E. of a molecule of a gas will be the same as that of an electron accelerated through 1 volt. Boltzmann constant `= 1.4 xx 10^(-23) J "molecule"^(-1) K^(-1)`, charge of an electron `= 1.6 xx 10^(-19)C`.

Answer

`7619 K`

Explanation

Energy of an electron accelerated through 1 volt
`=1 eV = 1.6 xx 10^(-19) C xx 1 V = 1.6 xx 10^(-19) J`
From kinetic theory of gases, average K.E. Of a gas molecule is `3/2 kT = 1.6 xx 10^(-19) J`
or `T= (2 xx 1.6 xx 10^(-19))/(3K) = (2 xx 1.6 xx 10^(-19))/(3 xx 1.4 xx 10^(-23))`
`= 7619 K`