Q. The temperature at which the average translational kinetic energy of a molecule is equal to the energy gained by an electron in accelerating from rest through a potential difference of $1\ volt$ is
A) $4.6 \times 10^{3}K$
B) $11.6 \times 10^{3}K$
C) $23.2 \times 10^{3}K$
D) $7.7 \times 10^{3}K$
Explanation
(d) $\frac{3}{2}kT = 1\ eV \Rightarrow T = \frac{2}{3}\frac{eV}{k} = \frac{\frac{2}{3} \times 1.6 \times 10^{- 19}}{1.38 \times 10^{- 23}} = 7.7 \times 10^{3}K$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.