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 10-27kg
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 10-23 J K-1
We have the expression that relates the mean kinetic energy kT)of the nitrogen molecule with the root mean square speed as:

12mvrms2=32kTvrms=3kTm

Step 3:
Hence, the de Broglie wavelength of the nitrogen molecule is given as:
λ=hmvrms=h3mkT=6.63×10343×28.0152×1.66×1027×1.38×1023×300=0.028×109m=0.028nm
Therefore, the de Broglie wavelength of the nitrogen molecule is 0.028 nm.