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.

I found an answer from content.wolfram.com
The Physics of Subatomic Particles
that is, its atomic number times the average mass of the proton and neutron. The nuclear reaction in question would be written as follows : 9 Be-ece. + particle .
For more information, see The Physics of Subatomic Particles
I found an answer from en.wikipedia.org
Thermal de Broglie wavelength - Wikipedia
In physics, the thermal de Broglie wavelength is roughly the average de Broglie wavelength of the gas particles in an ideal gas at the specified temperature.
For more information, see Thermal de Broglie wavelength - Wikipedia
I found an answer from www.khanacademy.org
De Broglie wavelength (video) | Khan Academy
Bohr model energy levels (derivation using physics) ... This means that a microscope using electron "matter waves" instead of photon light waves can see much ...
For more information, see De Broglie wavelength (video) | Khan Academy
Given that
Room temperature T = 27\degree C = 300 K \because 1\degree C = 273 K
Pressure P = 1 atm = 1.01 * 10^{5} pa
Mass of an He atom m = \frac{\text{ Atomic weight of He }}{\text{ Avogadro number }}
m=\frac{4}{6.022*10^{23}\ \ mol^{-1}}
m = 6.64 * 10^{-24} g
m = 6.64 * 10^{- 27} Kg
Step 1: Ge an expression for the de Broglie wavelength associated with a He
Gaseous particle kinetic energy at a certain temperature K.E = \frac{3}{2} kT ............................(1)
Where, k - Boltzmann constant. = 1.38 * 10^{-23} J/K
de Broglie wavelength \lambda = \frac{h}{p}
Where, Plank constant h = 6.63 * 10^{-34} and p - momentum
\lambda = \frac{h}{\sqrt{2mK.E}} \because K.E = \frac{p^2}{2m}
\lambda = \frac{h}{3mkT} \because \text{ equation (1)}
Step 2: Plug in the known values in the above equation ( \lambda )
\lambda = \frac{6.63* 10^{-34}}{3 * 6.64* 10^{-27} * 1.38*10^{-23} * 300}
\lambda = 0.7269 * 10^{-10} m
Therefore, de Broglie wavelength associated with a He \lambda = 0.7269 * 10^{-10} m
Step 3: Mean separation between the two atoms
Ideal gas equation PV = nRT
Where, P - gas pressure, V - gas volume, n - amount of gas(substance), R - gas constant and T - temperature.
PV = kNT
\frac{V}{N} = k \frac{T}{P}
Mean separation between the two points
(\frac{V}{N})^{\frac{1}{3}} = (\frac{kT}{P})^{\frac{1}{3}}
= \frac{1.38*10^{-23} * 300}{1.01 * 10^{5}}
= 3.35 * 10^{-9} m
Therefore, separation between the two points = 3.35 * 10^{-9} m