Question:
The average transnational kinetic energy of a molecule 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 (mass of molecule and electron is same) (N = number of molecules, V = Velocity, R = gas constant)
The average transnational kinetic energy of a molecule 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 (mass of molecule and electron is same) (N = number of molecules, V = Velocity, R = gas constant)
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Average translational KE is (3/2) kT per molecule, with k = R/N.
Updated On: Oct 1, 2026
- \(\frac{2VNe}{3R}\)
- \(\frac{VNe}{2R}\)
- \(\frac{3NeV}{2R}\)
- \(\frac{5NeV}{3R}\)
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The Correct Option is A
Solution and Explanation
Step 1: Gas molecule
\(E_1 = \frac32kT = \frac32\frac{R}{N}T\), where \(k = \frac RN\) is the Boltzmann constant.
Step 2: Electron
The electron gains \(E_2 = eV\).
Step 3: Equate
\[ \frac{3RT}{2N} = eV \Rightarrow T = \frac{2NeV}{3R} \]
Option (A).
Final Answer:
The temperature is 2VNe/(3R). \[ \boxed{\text{(A)}\ T=\frac{2VNe}{3R}} \]
\(E_1 = \frac32kT = \frac32\frac{R}{N}T\), where \(k = \frac RN\) is the Boltzmann constant.
Step 2: Electron
The electron gains \(E_2 = eV\).
Step 3: Equate
\[ \frac{3RT}{2N} = eV \Rightarrow T = \frac{2NeV}{3R} \]
Option (A).
Final Answer:
The temperature is 2VNe/(3R). \[ \boxed{\text{(A)}\ T=\frac{2VNe}{3R}} \]
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