MHT CET · Physics · Kinetic Theory of Gases
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)
- A \(\frac{3 V N e}{2 R}\)
- B \(\frac{V N e}{2 R}\)
- C \(\frac{V N e}{3 R}\)
- D \(\frac{2 V N e}{3 R}\)
Answer & Solution
Correct Answer
(D) \(\frac{2 V N e}{3 R}\)
Step-by-step Solution
Detailed explanation
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 \(E_1=E_2\)
\(\begin{aligned}
& \frac{3}{2}\left(\frac{R}{N}\right) T^{\prime}=E_2 \\
& \Rightarrow T^{\prime}=\frac{2 N E_2}{3 R}=\frac{2 N V e}{3 R}
\end{aligned}\)
The KE of an electron accelerated from rest through a potential difference \(V\) is given by,
\(e V=E_2\)
If \(E_1=E_2\)
\(\begin{aligned}
& \frac{3}{2}\left(\frac{R}{N}\right) T^{\prime}=E_2 \\
& \Rightarrow T^{\prime}=\frac{2 N E_2}{3 R}=\frac{2 N V e}{3 R}
\end{aligned}\)
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