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

(N/A) The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{3mkT}}$.
Given: $T = 27^{\circ}C = 300 \; K$,$P = 1.01 \times 10^5 \; Pa$,$m = \frac{4 \times 10^{-3} \; kg/mol}{6.023 \times 10^{23} \; mol^{-1}} \approx 6.64 \times 10^{-27} \; kg$.
Using $h = 6.63 \times 10^{-34} \; J \cdot s$ and $k = 1.38 \times 10^{-23} \; J/K$:
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{3 \times 6.64 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300}} \approx 0.73 \times 10^{-10} \; m$.
The mean separation $r$ between atoms is given by $r = (V/N)^{1/3} = (kT/P)^{1/3}$.
$r = \left( \frac{1.38 \times 10^{-23} \times 300}{1.01 \times 10^5} \right)^{1/3} \approx 3.4 \times 10^{-9} \; m$.
Comparing the two,the mean separation $r$ is much larger than the de Broglie wavelength $\lambda$ $(r \approx 46 \lambda)$.