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

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answer is 500.

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Detailed Solution

Given, translational kinetic energy of N₂ = kinetic energy of electron

$\frac{3}{2} k_{B} T = eV$
$\begin{array}{l} \Rightarrow \frac{3}{2} \times 1 .38 \times 10^{- 23} \times T = 1 .6 \times 10^{- 19} \times 0 .1 \\ \Rightarrow T = \frac{2 \times 1 .6 \times 10^{- 20}}{3 \times 1 .38 \times 10^{- 23}} or T = 773 K = (773 - 273)^{\circ} C \\ ∴ T = 500^{\circ} C \end{array}$