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