Room temperature is given as $T=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}$

Atmospheric pressure is given as $P=1 \mathrm{~atm}=1.01 \times 10^{5} \mathrm{~Pa}$

Atomic weight of He atom as we know is $4$

Avogadro’s number is $N_{A}=6.023 \times 10^{23}$

Boltzmann’s constant is $k=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{mol} / \mathrm{K}$

The expression for De Broglie wavelength is given as,

$\lambda=\frac{h}{\sqrt{2 m E}}$

$E=(3 / 2) \mathrm{kT}$

$m=$ mass of the He atom

$=$ Atomic weight $/ \mathrm{N}_{\mathrm{A}}$

$=4 /\left(6.023 \times 10^{23}\right)$

$=6.64 \times 10^{-24} \mathrm{~g}$

$m=6.64 \times 10^{-27} \mathrm{~kg}$

$\lambda=\frac{h}{\sqrt{3 m k T}}$

$\lambda=\frac{6.6 \times 10^{-34}}{\sqrt{3 \times 6.64 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300}}$

$=0.7268 \times 10^{-10} \mathrm{~m}$

We have the ideal gas formula

$P V=R T$

$P V=k N T$

$\mathrm{V} / \mathrm{N}=\mathrm{kT} / \mathrm{P}$

Here,

$V$ is the volume of the gas

$\mathrm{N}$ is the number of moles of the gas

Mean separation between the two atoms of the gas is given as

$r=\left[\frac{V}{N}\right]^{1 / 3}=\left[\frac{k T}{P}\right]^{1 / 3}$

$r=\left[\frac{1.38 \times 10^{-23} \times 300}{1.01 \times 10^{5}}\right]^{1 / 3}$

$=3.35 \times 10^{-9} \mathrm{~m}$

The mean separation between the atom is greater than the de Broglie wavelength.