Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 ºC) and 1 atm pressure; and compare it with the mean separation between two atoms under these conditions.
De Broglie wavelength associated with $\mathrm{He}$ atom $=0.7268 \times 10^{-10} \mathrm{~m}$
Room temperature, T = 27°C = 27 + 273 = 300 K
Atmospheric pressure, P = 1 atm = 1.01 × 105 Pa
Atomic weight of a He atom = 4
Avogadro’s number, NA = 6.023 × 1023
Boltzmann constant, k = 1.38 × 10−23 J mol−1 K−1
Average energy of a gas at temperature T,is given as:
$E=\frac{3}{2} k T$
De Broglie wavelength is given by the relation:
$\lambda=\frac{h}{\sqrt{2 m E}}$
Where,
m = Mass of a He atom
$=\frac{\text { Atomic weight }}{\mathrm{N}_{\mathrm{A}}}$
$=\frac{4}{6.023 \times 10^{23}}$
$=6.64 \times 10^{-24} \mathrm{~g}=6.64 \times 10^{-27} \mathrm{~kg}$
$\therefore \lambda=\frac{h}{\sqrt{3 m k T}}$
$=\frac{6.6 \times 10^{-34}}{\sqrt{3 \times 6.64 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300}}$
$=0.7268 \times 10^{-10} \mathrm{~m}$
We have the ideal gas formula:
PV = RT
PV = kNT
$\frac{V}{N}=\frac{k T}{P}$
Where,
V = Volume of the gas
N = Number of moles of the gas
Mean separation between two atoms of the gas is given by the relation:
$r=\left(\frac{V}{N}\right)^{\frac{1}{3}}=\left(\frac{k T}{P}\right)^{\frac{1}{3}}$
$=\left[\frac{1.38 \times 10^{-23} \times 300}{1.01 \times 10^{5}}\right]^{\frac{1}{3}}$
$=3.35 \times 10^{-9} \mathrm{~m}$
Hence, the mean separation between the atoms is much greater than the de Broglie wavelength.