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QJEEM21Phdc38: JEE Main 2021 Physics - Class 11 JEEM21Phdc38 Solution: The average translational kinetic energy of N 2 gas molecules…

The average translational kinetic energy of N2 gas molecules at ............. ∘^\circ C becomes equal to the K.E. of an electron accelerated from rest through a potential difference of 0.1 volt. (Given kB = 1.38 ×\times 10 −- 23 J/K) (Fill the nearest integer).

Step-by-Step Verified Solution

Given, the average translational kinetic energy of dinitrogen (N2) = Kinetic energy of an electron .... (i)

Translational kinetic energy of dinitrogen (N2)

KE=32KBTKE = {3 \over 2}{K_B}T

Here, T = temperature of the gas,

and KB = Boltzmann constant.

Kinetic energy of an electron = eV

Given, the potential differential of an electron, V = 0.1 V

Substituting the values in the Eq. (i), we get

32KBT=eV{3 \over 2}{K_B}T = eV

⇒32×1.38×10−23×T=1.6×10−19×(0.1)\Rightarrow {3 \over 2} \times 1.38 \times {10^{ - 23}} \times T = 1.6 \times {10^{ - 19}} \times (0.1)

T=773K=773−273∘C=500∘CT = 773K = 773 - 273^\circ C = 500^\circ C

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