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.


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