The average translational... | Kinetic Theory JEE Main 2021

JEE MAIN 2021

01-09-21 S2

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 $\mathrm{k}_{\mathrm{B}}=1.38 \times \left.10^{-23} \mathrm{~J} / \mathrm{K}\right)$ (Fill the nearest integer)

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Solution

Given Translation K.E. of $\mathrm{N}_2=$ K.E. of electron $$ \begin{aligned} & \frac{3}{2} \mathrm{kT}=\mathrm{eV} \\ & \frac{3}{2} \times 1.38 \times 10^{-23} \mathrm{~T}=1.6 \times 10^{-19} \times 0.1 \\ & \Rightarrow \mathrm{~T}=773 \mathrm{k} \\ & \mathrm{~T}=773-273=500^{\circ} \mathrm{C} \end{aligned} $$

Question Tags

JEE Main

Physics

Medium

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