Calculate the temperature at which the average kinetic energy of a molecule of a gas will be same as that of an electron accelerated through 1 volt. [ Given: \(k_B = 1.4 \times 10^{-23}\) J/K, \(e = 1.6 \times 10^{-19}\) C ]

Q.13 Calculate the temperature at which the average kinetic energy of a molecule of a gas will be same as that of an electron accelerated through 1 volt. [ Given: \(k_B = 1.4 \times 10^{-23}\) J/K, \(e = 1.6 \times 10^{-19}\) C ]

Doubtly Asked 27 Oct 2026 Board exam · March 2026 hsc 12th Save WhatsApp Report

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Answer: \(T \approx 7.62\times10^3\) K (about 7619 K)

Energy gained by an electron accelerated through 1 V: \(E = eV = 1.6\times10^{-19}\) J.

Average K.E. of a gas molecule \(= \dfrac{3}{2}k_BT\). Equating:

\[\frac{3}{2}k_B T = eV \Rightarrow T = \frac{2eV}{3k_B} = \frac{2\times1.6\times10^{-19}\times1}{3\times1.4\times10^{-23}} \approx 7619\ \text{K}\]

Doubtly AI · 27 Oct 2026 AI-verified · Oct 2026 Report

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Calculate the temperature at which the average kinetic energy of a molecule of a gas will be same as that of an electron accelerated through 1 volt. [ Given: \(k_B = 1.4 \times 10^{-23}\) J/K, \(e = 1.6 \times 10^{-19}\) C ]

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