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.
Given:
Pressure, P = 1 atm = 101325 Pa
Temperature, T = 27° = 300K
De-Broglie wavelength is given by,
…(1)
Where,
h = Planck’s constant = 6.6 × 10-34 Js
m = mass of He
E = total energy
Mass of helium, m = Atomic mass/ number of atoms
Let there be one mole of He,
mass of 1 mol of He = 4 g
Then,
m = 4/ 6.023 × 1023
m = 6.64 × 10-24 = 6.64 × 10-27Kg …(2)
We know that average energy at temperature T is given by,
E = 3/2 kT …(3)
Where,
k = Boltzmann constant = 1.38 × 10-23 Jmol-1K-1
T = absolute temperature
Using the equation (3), we can rewrite equation (1) as,
…(4)
Putting the values in equation (4) we get,
λ = 0.72 × 10-10m
Using ideal gas equation we have,
PV = RT
PV = kNT
Where,
V = volume
N = number of moles
P = pressure
k = Boltzmann constant
T = absolute temperature
Mean distance between, r is given by,
r = 3.3 × 10-9m
The mean separation between the atoms is greater than the De- Broglie wavelength.