The average translational kinetic energy of $N$ molecules in a gas is $E_1$. The kinetic energy of the…
The average translational kinetic energy of $N$ molecules in a gas is $E_1$. The kinetic energy of the electron $(e)$ accelerated from rest through potential difference $V$ volt is $E_2$. The temperature at which $E_1=E_2$ possible is ( $R=$ gas constant, $N=$ number of molecules)
$\frac{3 V N e}{2 R}$
$\frac{V N e}{2 R}$
$\frac{V N e}{3 R}$
$\frac{2 V N e}{3 R}$
Solution
The average translational KE of a gas molecule is $\frac{3}{2}\left(\frac{R}{N}\right) T=E_1$
The KE of an electron accelerated from rest through a potential difference $V$ is given by,
$e V=E_2$
If $E_1=E_2$
$\begin{aligned}
& \frac{3}{2}\left(\frac{R}{N}\right) T^{\prime}=E_2 \\
& \Rightarrow T^{\prime}=\frac{2 N E_2}{3 R}=\frac{2 N V e}{3 R}
\end{aligned}$