# Q: 11.19  What is the de Broglie wavelength of a nitrogen molecule in air at $\dpi{100} 300\hspace{1mm}K$? Assume that the molecule is moving with the root-mean-square speed of molecules at this temperature. (Atomic mass of nitrogen $\dpi{100} =14.0076\hspace{1mm} \mu$)

S Sayak

Since the molecule is moving with the root-mean-square speed the kinetic energy K will be given by

K=3/2 kT where k is the Boltzmann's constant and T is the absolute Temperature

In the given case Kinetic Energy of a Nitrogen molecule will be

$\\K=\frac{3}{2}\times 1.38\times 10^{-23}\times 300\\ K=6.21\times 10^{-21}J$

Mass of Nitrogen molecule = 2$\times$14.0076$\times$1.66$\times$10-27=4.65$\times$10-26kg

The momentum of the molecule is

$\\p=\sqrt{2mK}\\ p=\sqrt{2\times 4.65\times 10^{-26}\times 6.21\times 10^{-21}}\\ p=2.4\times 10^{-23}kg\ m\ s^{-1}$

Associated De Broglie wavelength is

$\\\lambda =\frac{h}{p}\\ \lambda =\frac{6.62\times 10^{-34}}{2.4\times 10^{-23}}\\ \lambda= 2.75\times 10^{-11}\ m$

The nitrogen molecule will have a De Broglie wavelength of 0.0275 nm.

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