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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 .

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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.

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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 ...

Veda
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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