Let us first find mass ‘m’ of each helium atom.
m =

Atomic‌weight‌of‌Helium

Avogadro′s‌number

m =

4

6×1023

g =

2

3

×10−23 g =

2

3

×10−26 kg
Absolute temperature, T = 273 + 27 = 300 K
Average K.E. of a He atom at absolute temperature T
K =

1

2

mv2 =

3

2

kT
m2v2 = p2 = 3 mkT
Momentum, p = √3mkT
Wavelength of the wave associated with He atom at room temperature
λ =

h

p

=

h

√3mkT

=
λ = 0.73 × 10−10 m ... (i)
Now let us find mean separation between He atoms.
Mean separation, r = [

Molar‌volume

Avagadro′s‌number

]1∕3
Here for 1 mole, PV = RT
PV = NkT or

V

N

=

kT

P

So, mean separation r = [

kT

P

]1∕3
Here, k = Boltzman’s constant = 1.38 × 10–23JK–1
T = Absolute temperature = 300 K
P = Atmospheric pressure = 1.01 × 105 Pa
r = [

1.38×10−23×300

1.01×1058

]1∕3 m = 3.4 × 10−9 m ... (ii)
Comparing the wavelength ‘λ’ with mean separation ‘r’, it can be observed that separation is larger than wavelength, i.e. (r ≫ λ)