Question

What is the de Broglie wavelength of a nitrogen molecule in air at $300K$? Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen $=14.0076u)$

Answer 1

Given, Temperature of the nitrogen molecule,
$T=300K$
Atomic mass of nitrogen = 14.0076 u
Hence, mass of the nitrogen molecule, $m=2\times 14.0076=28.0152u$
But 1 $u=1.66\times {10}^{-27}\text{kg}$
Therefore, $m=28.0152\times 1.66\times {10}^{-27}\text{kg}$
Planck's constant, $h=6.63\times {10}^{-34}Js$
Boltzmann constant, $k=1.38\times {10}^{-23}J/K$
We have the expression that relates mean kinetic energy $\left(\frac{3}{2}kT\right)$ of the nitrogen molecule with the root mean square speed $\left(V_{rms}\right)$ as:
$\frac{1}{2}m{v}_{rms}^2=\left(\frac{3}{2}kT\right)$
$V_{rms}=\sqrt{\frac{3kT}{m}}$
For nitrogen molecule, the de Broglie wavelength is given as:
$\lambda =\frac{h}{m{v}_{rms}}=\frac{h}{\sqrt{3mkT}}$
$=\frac{6.63\times {10}^{-34}}{\sqrt{3\times 28.0152\times 1.66\times {10}^{-27}\times 1.38\times {10}^{-23}\times 300}}$
$=0.028\times {10}^{-9}m$
$=0.028\text{nm}$
Therefore, the de Broglie wavelength of the nitrogen molecule is $0.028\text{nm}$.
Final Answer $0.028\text{nm}$