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