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
$$
\begin{aligned}
& \frac{3}{2} \mathrm{kT}=\mathrm{eV} \\
& \frac{3}{2} \times 1.38 \times 10^{-23} \mathrm{~T}=1.6 \times 10^{-19} \times 0.1 \\
& \Rightarrow \mathrm{~T}=773 \mathrm{k} \\
& \mathrm{~T}=773-273=500^{\circ} \mathrm{C}
\end{aligned}
$$
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