Laws of Motion

JEE Main level, numerical answer, medium

157 of 210

Question 157 of 210

The average translational kinetic energy of N2{N}_{2} gas molecules at _________C^{\circ} becomes equal to the K.E.K.E. of an electron accelerated from rest through a potential difference of 0.1volt.0.1volt.
(Given kB=1.38×10−23JK−1)\left.{k}_{B}=1.38\times {10}^{-23}J{K}^{-1}\right) (Fill the nearest integer).

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Answer: 500

Given
Translation K.E.K.E. of N2=K.E{N}_{2}=K.E of electron
32kT=eV\dfrac{3}{2}kT=eV
32×1.38×10−23T=1.6×10−19×0.1⇒T=773K\dfrac{3}{2}\times 1.38\times {10}^{-23}T=1.6\times {10}^{-19}\times 0.1\Rightarrow T=773K
T=773−273=500∘CT=773-273=500^{\circ}C

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