Question

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.

Answer 1

De Broglie wavelength associated with He atom =

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

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

Atomic weight of a He atom = 4

Avogadro’s number, N_A = 6.023 × 10²³

Boltzmann constant, k = 1.38 × 10⁻²³ J mol⁻¹ K⁻¹

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

De Broglie wavelength is given by the relation:

Where,

m = Mass of a He atom

We have the ideal gas formula:

PV = RT

PV = kNT

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:

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