Find the typical de Broglie wavelength associated

Solution:

De Broglie wavelength associated with $\mathrm{He}$ atom $=0.7268 \times 10^{-10} \mathrm{~m}$

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, 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:

$E=\frac{3}{2} k T$

De Broglie wavelength is given by the relation:

$\lambda=\frac{h}{\sqrt{2 m E}}$

Where,

 

m = Mass of a He atom

$=\frac{\text { Atomic weight }}{\mathrm{N}_{\mathrm{A}}}$

$=\frac{4}{6.023 \times 10^{23}}$

$=6.64 \times 10^{-24} \mathrm{~g}=6.64 \times 10^{-27} \mathrm{~kg}$

$\therefore \lambda=\frac{h}{\sqrt{3 m k T}}$

$=\frac{6.6 \times 10^{-34}}{\sqrt{3 \times 6.64 \times 10^{-27} \times 1.38 \times 10^{-23} \times 300}}$

$=0.7268 \times 10^{-10} \mathrm{~m}$

We have the ideal gas formula:

PV = RT

 

PV = kNT

$\frac{V}{N}=\frac{k T}{P}$

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:

$r=\left(\frac{V}{N}\right)^{\frac{1}{3}}=\left(\frac{k T}{P}\right)^{\frac{1}{3}}$

$=\left[\frac{1.38 \times 10^{-23} \times 300}{1.01 \times 10^{5}}\right]^{\frac{1}{3}}$

$=3.35 \times 10^{-9} \mathrm{~m}$

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