Question 11.19: What is the de Broglie wavelength of a nitrogen molecule in air at 300 K? Assume that the molecule is moving with the root-mean
square speed of molecules at this temperature. (Atomic mass of nitrogen = 14.0076 u)

Hint: $$\lambda=\frac{h}{v_{rms}}$$

Step 1: Find the mass of nitrogen molecules.
The temperature of the nitrogen molecule, T = 300 K
The atomic mass of nitrogen = 14.0076 u
Hence, mass of the nitrogen molecule, m = 2 x 14.0076 = 28.0152 u
But 1 u= 1.66 x ${10}^{-27}$kg
So, m= 28.0152 x1.66 x ${10}^{-27}$ kg
Step 2: Find root mean square speed.
Planck's constant, h = 6.63 x ${10}^{-34}$ Js
Boltzmann constant, k = 1.38 x
We have the expression that relates the mean kinetic energy kT)of the nitrogen molecule with the root mean square speed as:

$\begin{array}{l}\frac{1}{2}m{v}_{\mathrm{rms}}^{2}=\frac{3}{2}kT\\ {v}_{r\mathrm{ms}}=\sqrt{\frac{3kT}{m}}\end{array}$

Step 3:
Hence, the de Broglie wavelength of the nitrogen molecule is given as:
$\begin{array}{l}\lambda =\frac{h}{m{v}_{rms}}=\frac{h}{\sqrt{3mkT}}\\ =\frac{6.63×{10}^{-34}}{\sqrt{3×28.0152×1.66×{10}^{-27}×1.38×{10}^{-23}×300}}\\ =0.028×{10}^{-9}m\\ =0.028nm\end{array}$
Therefore, the de Broglie wavelength of the nitrogen molecule is 0.028 nm.