#### Question

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)

#### Solution

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 × 10^{−23} J K^{−1}

We have the expression that relates mean kinetic energy `(3/2 kT)` of the nitrogen molecule with the root mean square speed `(V_(nms))` as:

`1/2 mv_"rms"^2 = 3/2 kT`

`v_"rms" = sqrt((3kT)/m)`

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

`lambda = h/(mv_"rms") = h/sqrt(3mkT)`

`= (6.63 xx 10^(-34))/sqrt(3xx28.0152 xx 1.66 xx 10^(-27) xx 1.38 xx 20^(-23) xx 300 )`

` = 0.028 xx 10^(-9) m`

= 0.028 nm

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