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)

• A. \(\frac{2VNe}{3R}\)
• B. \(\frac{VNe}{2R}\)
• C. \(\frac{3NeV}{2R}\)
• D. \(\frac{5NeV}{3R}\)

Hint:

Average translational KE is (3/2) kT per molecule, with k = R/N.

Answer 1

The Correct Option is A

Step 1: Use Boltzmann constant
Write $k = R/N$. Setting $eV = \frac32\frac RN T$ and solving for $T$ directly gives $T = \frac{2eVN}{3R}$. Option (A).

Final Answer:
2VNe/(3R). \[ \boxed{\text{(A)}\ T=\frac{2VNe}{3R}} \]