Question

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.

Answer 1

Mass of the He-atom,

m = \(\frac{\textit{atomic weight of He}}{\textit{Avogadro Number}}\) = \(\frac{4}{6.02\times 10^{23}}\)

= 6.645 × 10^-24 g = 6645 × 10^-27 kg
find the de-Broglie wavelength of the He-atom at 27°C.
It can be obtained that de-Broglie wavelength of the He-atom,
λ = 0.73 x 10^-10 m = 0.73 Å
If V is volume of the gas and dj the mean separation between two atoms in the gas, then V = d³ N,
Where N is Avogadro’s number
From the perfect gas equation

i.e., the separation between the two He-atoms in the gas is much larger than the Broglie wavelength of the atom.