Question

At what temperature does the average translational K.E. of a molecule in a gas becomes equal to the K.E. of an electron accelerated through a potential density of 3 V?

• A. 232 K
• B. 2320 K
• C. 23,200 K
• D. 2,32,000 K

Answer 1

Answer: (C)

23,200 K

Answer 2

The average translational kinetic energy of a gas molecule is given by $KE_{avg} = \frac{3}{2}kT$, where $k$ is the Boltzmann constant and $T$ is the temperature. The kinetic energy of an electron accelerated through a potential difference $V$ is $KE_{electron} = eV$, where $e$ is the elementary charge. Setting $KE_{avg} = KE_{electron}$: $\frac{3}{2}kT = eV$ Solving for $T$: $T = \frac{2eV}{3k}$ Substituting the values $e \approx 1.602 \times 10^{-19}$ C, $V = 3$ V, and $k \approx 1.381 \times 10^{-23}$ J/K: $T = \frac{2 \times (1.602 \times 10^{-19}) \times 3}{3 \times (1.381 \times 10^{-23})} \approx 2.319 \times 10^4 \text{ K} \approx 23190 \text{ K}$ This is closest to 23,200 K.