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, NA = 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:

We have the ideal gas formula:
PV = RT
PV = kNT

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