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At what temperature does the average translational kinetic energy of a molecule in a gas becomes equal to kinetic energy of an electron accelerated from rest through potential difference of ' $\mathrm{V}$ ' volt?
( $\mathrm{N}=$ Avogadro number, $\mathrm{R}$ = gas constant, $\mathrm{e}=$ electronic charge $)$
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( $\mathrm{N}=$ Avogadro number, $\mathrm{R}$ = gas constant, $\mathrm{e}=$ electronic charge $)$
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Verified Answer
The correct answer is:
$\frac{2 \mathrm{eVN}}{3 \mathrm{R}}$
The average translational kinetic energy of a gas molecule $=\frac{3}{2} \mathrm{kT}=\frac{3}{2} \frac{\mathrm{R}}{\mathrm{N}} \mathrm{T}$
The kinetic energy of a electrons accelerated by a p.d. of V volts
$$
\begin{aligned}
& =\mathrm{eV} \\
& \therefore \frac{3}{2} \frac{\mathrm{RT}}{\mathrm{N}}=\mathrm{eV} \\
& \therefore \mathrm{T}=\frac{2 \mathrm{eVN}}{3 \mathrm{R}}
\end{aligned}
$$
The kinetic energy of a electrons accelerated by a p.d. of V volts
$$
\begin{aligned}
& =\mathrm{eV} \\
& \therefore \frac{3}{2} \frac{\mathrm{RT}}{\mathrm{N}}=\mathrm{eV} \\
& \therefore \mathrm{T}=\frac{2 \mathrm{eVN}}{3 \mathrm{R}}
\end{aligned}
$$
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