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.

De Broglie wavelength associated with He atom = 0.7268 x 10^-10 m

Room temperature, T = 27°C = 27 + 273 = 300 K

Atmospheric pressure, P = 1 atm = 1.01 x 10⁵ Pa

Atomic weight of a He atom = 4

Avogadro’s number, Na = 6.023 x 10²³

Boltzmann constant, k= 1.38 x 10^-23 J mol^-1 K^-1

Average energy of a gas at temperature T, is given as:

E= 3/2 kT
De Broglie wavelength is given by the relation:

λ = h/√[2mE]

Where,

m = Mass of a He atom

= Atomic weight /Na = 4/(6.023 x 10²³)

=6.64 x 10⁻²⁴ g = 6.64 x 10⁻²⁷

Therefore , λ = h/√[3mkT]

= (6.6 x 10⁻³⁴) /√[3 x (6.64 x 10⁻²⁷) x (1.38 x 10^-23) x  300]

= 0.7268 x 10⁻¹⁰ m

We have the ideal gas formula:

PV = RT

PV = kNT

V/N = kT/P_

Where,

V = Volume of the gas

N = Number of moles of the gas

Mean separation between two atoms of the gas is given by the relation:

r³ = (V/N) = (kT/P) = [(1.38 x 10^-23) x (300) /( 1.01 x 10⁵)]

r = 3.35 x 10⁻⁹ m

Hence, the mean separation between the atoms is much greater than the de Broglie wavelength.