What is the de Broglie wavelength of a nitrogen molecule in air at 300 K?

Temperature of the nitrogen molecule, T = 300 K

Atomic mass of nitrogen = 14.0076 u

Hence, mass of the nitrogen molecule, m = 2 × 14.0076 = 28.0152 u

But 1 u = 1.66 × 10−27 kg

∴m = 28.0152 ×1.66 × 10−27 kg

Planck’s constant, h = 6.63 × 10−34 Js

Boltzmann constant, $k=1.38 \times 10^{-23} \mathrm{JK}^{-1}$

We have the expression that relates mean kinetic energy $\left(\frac{3}{2} k T\right)$ of the nitrogen molecule with the root mean square speed $\left(v_{\text {rms }}\right)$ as:

$\frac{1}{2} m v_{\mathrm{rms}}^{2}=\frac{3}{2} k T$

$v_{\text {rms }}=\sqrt{\frac{3 k T}{m}}$

Hence, the de Broglie wavelength of the nitrogen molecule is given as:

$\lambda=\frac{h}{m v_{\mathrm{rms}}}=\frac{h}{\sqrt{3 m k T}}$

$=\frac{6.63 \times 10^{-34}}{\sqrt{3 \times 28.0152 \times 1.66 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300}}$

$=0.028 \times 10^{-9} \mathrm{~m}$

$=0.028 \mathrm{~nm}$

Therefore, the de Broglie wavelength of the nitrogen molecule is 0.028 nm.