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

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Answer & Solution

Correct Answer

(A) \(500\)

Step-by-step Solution

Detailed explanation

Given

Translation K.E. of \({N}_{2}={KE}\). of electron

\(\frac{3}{2} {kT}={eV}\)

\(\frac{3}{2} \times 1.38 \times 10^{-23} {T}=1.6 \times 10^{-19} \times 0.1\)

\(\Rightarrow {T}=773 {k}\)

\({T}=773-273=500^{\circ} {C}\)

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