Question:
The average translational 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 (N = number of molecules, R = gas constant)
The average translational 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 (N = number of molecules, R = gas constant)
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Average translational energy per molecule is (3/2) kT, with k = R/N.
Updated On: Oct 1, 2026
- \(\frac{eVN}{2R}\)
- \(\frac{2eVN}{3R}\)
- \(\frac{eVN}{R}\)
- \(\frac{3eVN}{4R}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
Average translational kinetic energy of a gas molecule is \(E_1 = \frac32 kT\), with Boltzmann constant \(k = R/N\) (N being the Avogadro number here). An electron accelerated through V volts gains \(E_2 = eV\).
Step 2: Equate
\[ \frac32\cdot\frac{R}{N}T = eV \Rightarrow T = \frac{2eVN}{3R} \]
Option (A) has the factor 2 in the denominator instead of the numerator, and (C) leaves out the 3/2.
Final Answer:
The temperature is \(\dfrac{2eVN}{3R}\), option (B). \[ \boxed{\frac{2eVN}{3R}} \]
Average translational kinetic energy of a gas molecule is \(E_1 = \frac32 kT\), with Boltzmann constant \(k = R/N\) (N being the Avogadro number here). An electron accelerated through V volts gains \(E_2 = eV\).
Step 2: Equate
\[ \frac32\cdot\frac{R}{N}T = eV \Rightarrow T = \frac{2eVN}{3R} \]
Option (A) has the factor 2 in the denominator instead of the numerator, and (C) leaves out the 3/2.
Final Answer:
The temperature is \(\dfrac{2eVN}{3R}\), option (B). \[ \boxed{\frac{2eVN}{3R}} \]
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