Question

At what temperature the average translational KE of the molecules of a gas will become equal to the KE of an electron accelerated from rest through 1 V potential difference?

• A. ${10}^{4}K$
• B. $2.34\times {10}^{4}K$
• C. $7.73\times {10}^{3}K$
• D. none of these

Answer 1

Answer: (C)

$7.73\times {10}^{3}K$

$\frac{3}{2}KT=1eV=1.6\times {10}^{-19}J$
or
$T=\frac{2\times 1.6\times {10}^{-19}}{3\times 1.38\times {10}^{-23}}=7730K=7.73\times {10}^{3}K$