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  • Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 °C) and 1 atm pressure; and compare it with the mean separation between two atoms under these conditions.

Q: 11.35  Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27\hspace{1mm}^{\circ}C) and 1 atm pressure, and compare it with the mean separation between two atoms under these conditions. 

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The kinetic energy K of a He atom is given by

K=\frac{3}{2}kT

mHe i.e. mass of one atom of He can be calculated as follows

\\m_{He}=\frac{4\times 10^{-3}}{N_{A}}\\ =\frac{4\times 10^{-3}}{6.023\times 10^{23}}\\ m_{He}=6.64\times 10^{-27}\ kg                             (NA is the Avogadro's Number)

De Broglie wavelength is given by

\\\lambda =\frac{h}{p}\\ \lambda =\frac{h}{\sqrt{2m_{He}K}}\\ \lambda =\frac{h}{\sqrt{3m_{He}kT}}\\ \lambda =\frac{6.62\times 10^{-34}}{\sqrt{3\times 6.64\times 10^{-27}\times 1.38\times 10^{-23}\times 300}}\\ \lambda =7.27\times 10^{-11}\ m

The mean separation between two atoms is given by the relation

\\d=\left ( \frac{V}{N} \right )^{\frac{1}{3}}\\

From the ideal gas equation we have

\\PV=nRT\\ PV=\frac{NRT}{N_{A}}\\ \frac{V}{N}=\frac{RT}{PN_{A}}

The mean separation is therefore

\\d=\left ( \frac{RT}{PN_{A}} \right )^{\frac{1}{3}}\\ d=\left ( \frac{kT}{P} \right )^{\frac{1}{3}}\\ d=\left ( \frac{1.38\times 10^{-23}\times 300}{1.01\times 10^{5}} \right )^{\frac{1}{3}}\\ d=3.35\times 10^{-9}\ m

The mean separation is greater than the de Broglie wavelength.

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Sayak

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