Question

Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27°C) 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 = 0.7268 × 10⁻¹⁰ m

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 gas at temperature T, is given as:

`"E" = 3/2 "kT"`

De Broglie wavelength is given by the relation:

`lambda = "h"/(sqrt(2"mE"))`

Where,

m = Mass of a He atom

= `"Atomic weight"/"N"_"A"`

= `4/(6.023 xx 10^23)`

= 6.64 × 10⁻²⁴ g

= 6.64 × 10⁻²⁷ kg

∴ `lambda = "h"/sqrt(3"mkT")`

= `(6.6 xx 10^(-34))/sqrt(3xx6.64 xx10^(-27)xx 1.38 xx 10^(-23) xx 300)`

= 0.7268 × 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:

`"r" = ("V"/"N")^(1/3) = ("kT"/"P")^(1/3)`

= `[(1.38 xx 10^(-23) xx 300)/(1.01 xx 10^5)]^(1/3)`

= 3.35 × 10⁻⁹ m

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