Login/Sign Up
Home » Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature and 1 atm pressure, and compare it with the mean separation between two atoms under these conditions.
Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature \left(27^{\circ} \mathrm{C}\right) and 1 atm pressure, and compare it with the mean separation between two atoms under these conditions.
CBSE | Class 12 | Dual Nature of Radiation and Matter | NCERT | Physics
Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature \left(27^{\circ} \mathrm{C}\right) and 1 atm pressure, and compare it with the mean separation between two atoms under these conditions.

Room temperature is given as T=27^{\circ} \mathrm{C}=27+273=300 \mathrm{~K}

Atmospheric pressure is given as P=1 \mathrm{~atm}=1.01 \times 10^{5} \mathrm{~Pa}

Atomic weight of He atom as we know is 4

Avogadro’s number is N_{A}=6.023 \times 10^{23}

Boltzmann’s constant is k=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{mol} / \mathrm{K}

The expression for De Broglie wavelength is given as,

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

E=(3 / 2) \mathrm{kT}

m= mass of the He atom

= Atomic weight / \mathrm{N}_{\mathrm{A}}

=4 /\left(6.023 \times 10^{23}\right)

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

m=6.64 \times 10^{-27} \mathrm{~kg}

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

\lambda=\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

P V=R T

P V=k N T

\mathrm{V} / \mathrm{N}=\mathrm{kT} / \mathrm{P}

Here,

V is the volume of the gas

\mathrm{N} is the number of moles of the gas

Mean separation between the two atoms of the gas is given as

r=\left[\frac{V}{N}\right]^{1 / 3}=\left[\frac{k T}{P}\right]^{1 / 3}

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

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

The mean separation between the atom is greater than the de Broglie wavelength.

i

More Material