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.

Solution

Mass of He atom,

m = \frac{Atomic mass of He}{Avogadro' s number}

= \frac{4}{6 \times 10^{23}} g

\Rightarrow

m = 6.67 \times 10^{- 27} kg

De- broglie wavelength,

λ = \frac{h}{p} = \frac{h}{\sqrt{3 mkT}} m

= \frac{6.63 \times 10^{- 34}}{\sqrt{3 \times 6.67 \times 10^{- 27} \times 1.38 \times 10^{- 23} \times 300}} m

i.e.,

λ = 7.3 \times 10^{- 11} m

Now, using the kinetic gas equation for one mole of a gas,

PV = RT = kNT

\Rightarrow

\frac{V}{N} = \frac{KT}{P}

Mean separation,

r_{0} = {(\frac{Molar volume}{Avogadro' s number})}^{1 / 3} = {(\frac{V}{N})}^{1 / 3} = {(\frac{kT}{P})}^{1 / 3}

\Rightarrow

r_{0} = {(\frac{1.38 \times 10^{- 23} \times 300}{1.01 \times 10^{5}})}^{1 / 3} m

i.e.,

r_{0} = 3.4 \times 10^{- 9} m

From the above calculations, we can say that the mean separation between two atoms is much larger than the de-Broglie wavelength.

For Daily Free Study Material Join wiredfaculty