The average translational kinetic energy of

Question:
The average translational kinetic energy of $\mathrm{N}_{2}$ gas molecules at $.{ }^{\circ} \mathrm{C}$ becomes equal to the K.E. of an electron accelerated from rest through a potential difference of $0.1$ volt.

$\left(\right.$ Given $\left.\mathrm{k}_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}\right)$

(Fill the nearest integer).

Solution:

Given

Translation K.E. of $\mathrm{N}_{2}=\mathrm{K}$.E. of electron

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

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

$\Rightarrow \mathrm{T}=773 \mathrm{k}$

$\mathrm{T}=773-273=500^{\circ} \mathrm{C}$

The average translational kinetic energy of

Leave a comment

All Study Material