Question:
The average translational kinetic energy of \(\mathrm{N_2}\) gas molecules at ............. \({}^\circ\mathrm{C}\) becomes equal to the K.E. of an electron accelerated from rest through a potential difference of 0.1 volt. (Given \(k_B = 1.38\times10^{-23}\) J/K) (Fill the nearest integer).
The average translational kinetic energy of \(\mathrm{N_2}\) gas molecules at ............. \({}^\circ\mathrm{C}\) becomes equal to the K.E. of an electron accelerated from rest through a potential difference of 0.1 volt. (Given \(k_B = 1.38\times10^{-23}\) J/K) (Fill the nearest integer).
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Equate \(\tfrac32 k_BT\) to \(eV = 0.1\) eV \(= 1.6\times10^{-20}\) J, solve for \(T\) in kelvin and subtract 273.
Updated On: Nov 4, 2026
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Correct Answer: 500
Solution and Explanation
Average translational KE per molecule \(= \dfrac32 k_BT\) (independent of the gas).
KE of electron \(= eV = 1.6\times10^{-19}\times0.1 = 1.6\times10^{-20}\) J.
\(\dfrac32(1.38\times10^{-23})T = 1.6\times10^{-20}\) gives \(T = \dfrac{3.2\times10^{-20}}{4.14\times10^{-23}} \approx 772.9\) K.
\(t = 772.9 - 273 \approx 499.9 \approx 500^\circ\)C.
Answer: 500
KE of electron \(= eV = 1.6\times10^{-19}\times0.1 = 1.6\times10^{-20}\) J.
\(\dfrac32(1.38\times10^{-23})T = 1.6\times10^{-20}\) gives \(T = \dfrac{3.2\times10^{-20}}{4.14\times10^{-23}} \approx 772.9\) K.
\(t = 772.9 - 273 \approx 499.9 \approx 500^\circ\)C.
Answer: 500
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