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)

Temperature of the nitrogen molecule, T = 300 K

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

Therefore, m = 28.0152 x 1.66 x 10⁻²⁷ kg

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

Boltzmann constant, k = 1.38 x 10⁻²³J K⁻¹

We have the expression that relates mean kinetic energy (3KT/2 )of the nitrogen molecule with the root mean square speed (v_rms) as:

1/2   x m (v_rms)² = 3/2 kT

v_rms= √(3KT/m)

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

λ = h/(mv_rms)= h/√(3mKT)

=(6.63 x 10⁻³⁴) /√[3 x 28.0152 x (1.66 x 10⁻²⁷) x (1.38 x 10⁻²³) x 300

= 0.028 x 10⁻⁹ m

= 0.028 nm

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