Question

Consider a balloon filled with helium gas at room temperature and atmospheric pressure. Calculate:
(a) the average de Broglie wavelength of the helium atoms.
(b) the average distance between atoms under these conditions. The average kinetic energy of an atom is equal to $(3/2)kT$, where $k$ is the Boltzmann constant.
(c) Can the atoms be treated as particles under these conditions? Explain.

Answer 1

To calculate the de Broglie wavelength of the helium atom, we need the Planck’s constant and the momentum of the helium atom. We have been given the temperature of the gas and we know the value of the Boltzmann constant, so we can find the average kinetic energy of the helium atom and then find the momentum of the atom using the kinetic energy and the mass of the atom. Let’s see the detailed solution.

Formula Used:
${{p}_{avg}}=\sqrt{2m{{K}_{avg}}}$, ${{K}_{avg}}=\text{ }\dfrac{\text{3kT}}{2}$, $\lambda =\dfrac{h}{{{p}_{avg}}}$, ${{d}_{avg}}=\sqrt[3]{\dfrac{kT}{{{p}_{avg}}}}$

Complete step by step solution:
Average kinetic energy of the helium atom ${{K}_{avg}}=\text{ }\dfrac{\text{3kT}}{2}$
We know that momentum of the helium atom can be related to the kinetic energy as ${{p}_{avg}}=\sqrt{2m{{K}_{avg}}}$ where $m$ is the mass of a helium atom and ${{K}_{avg}}$ is the average kinetic energy
Substituting the value of the average kinetic energy, we get

& {{p}_{avg}}=\sqrt{2m\times \dfrac{\text{3kT}}{2}} \\\ & \Rightarrow {{p}_{avg}}=\sqrt{3mkT} \\\ \end{aligned}$$ The temperature specified in the question is room temperature, that is $$27{}^\circ C$$ or $$300K$$ (because we always consider the absolute temperature of a gas) Substituting the values of temperature and Boltzmann constant in the expression for momentum, we get $$\begin{aligned} & {{p}_{avg}}=\sqrt{3\times 4\times 1.38\times {{10}^{-23}}\times 300}\left[ \because mas{{s}_{helium}}=4 \right] \\\ & \Rightarrow {{p}_{avg}}=\sqrt{496.8}\times {{10}^{-11}} \\\ & \Rightarrow {{p}_{avg}}=22.3\times {{10}^{-11}}kg.m/s \\\ \end{aligned}$$ We know that de Broglie wavelength of an atom is $$\lambda =\dfrac{h}{{{p}_{avg}}}$$ Substituting the value of momentum in the expression for de Broglie wavelength, we get $$\begin{aligned} & \lambda =\dfrac{6.62\times {{10}^{-34}}}{22.3\times {{10}^{-11}}} \\\ & \Rightarrow \lambda =0.3\times {{10}^{-23}}m \\\ & \Rightarrow \lambda =3.0\times {{10}^{-24}}m \\\ \end{aligned}$$ Now we know that the average distance between two atoms is given as $${{d}_{avg}}=\sqrt[3]{\dfrac{kT}{{{p}_{avg}}}}$$ where the meanings of the symbols have been discussed above Substituting the value of Boltzmann constant, the temperature and the average momentum, we get $$\begin{aligned} & {{d}_{avg}}=\sqrt[3]{\dfrac{1.38\times {{10}^{-23}}\times 300}{22.3\times {{10}^{-11}}}} \\\ & \Rightarrow {{d}_{avg}}=\sqrt[3]{18.56}\times {{10}^{-4}}m \\\ & \Rightarrow {{d}_{avg}}=2.62\times {{10}^{-4}}m \\\ \end{aligned}$$ If the de Broglie wavelength of the atoms is much less than the interatomic distance, the atoms can be treated as particles. We can thus say that in this case, we can treat the helium atoms as particles since $$\lambda <<{{d}_{avg}}$$ **Note:** Usually we apply rules of quantum mechanics to atoms and molecules, but in such cases as described in the solution, when the interatomic distance is much greater than the de Broglie wavelength, we can apply rules of classical mechanics and our solutions will be justified. Some students get confused between the symbols for kinetic energy and the Boltzmann constant so you should keep an eye out for that.