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The average translational kinetic energy of $N_2 $ ....

The average translational kinetic energy of $N_2 $ gas molecules at _ _ _ _ $^\circ 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 \times 10^{-23} J/K $) (Fill the nearest integer)

Solution

K.E. of electron =  Translational K.E. of $N_2 $ molecules

$\therefore eV = \frac {3}{2} kT $

$\therefore 2 \times 1.6 \times 10^{-19} \times 0.1 = 3 \times 1.38 \times 10^{-23} \times T $

$\therefore T = \frac {1.6 \times 10^3 }{3 \times 0.69 } \approx 773 K = 500 ^\circ C $

Answer: 500