MHT CET · Physics · Kinetic Theory of Gases
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 \()\)
- A \(\frac{2 \mathrm{eVN}}{3 \mathrm{R}}\)
- B \(\frac{\mathrm{eVN}}{\mathrm{R}}\)
- C \(\frac{\mathrm{eVN}}{4 \mathrm{R}}\)
- D \(\frac{3 \mathrm{eVN}}{2 \mathrm{R}}\)
Answer & Solution
Correct Answer
(A) \(\frac{2 \mathrm{eVN}}{3 \mathrm{R}}\)
Step-by-step Solution
Detailed explanation
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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