91Ó°ÊÓ

The given data can be listed below as,

• The average kinetic energy of an atom is,${\left(KE\right)}_{avg}=\left(\frac{3}{2}\right)kT$ .
• The value of room temperature is,$T=300\text{K}$ .

The value of the average kinetic energy of a moving atom can be obtained with the help of its absolute temperature and the Boltzmann constant. The relation between the average kinetic energy and temperature is directly linear.

The expression to calculate the average de Broglie wavelength of the helium atoms is expressed as,

$\begin{array}{rcl}{\mathrm{λ}}_{avg}& =& \frac{h}{p_{avg}}\\ & =& \frac{h}{\sqrt{2m_{He}{\left(KE\right)}_{avg}}}\\ & =& \frac{h}{\sqrt{2m_{He}\left(\frac{3}{2}kT\right)}}\\ & =& \frac{h}{\sqrt{3m_{He}kT}}\end{array}$

Here, ${\mathrm{λ}}_{avg}$is the average de Broglie wavelength of the helium atoms,k is the Boltzmann constant whose value is $1.38×{10}^{-23}\text{J/K}$ , his the Plank’s constant whose value is$6.63×{10}^{-34}\text{J}·\text{s}$, $m_{He}$is the mass a helium atom whose value is$m_{He}=4.00\text{u}$and $p_{avg}$is the average momentum.

Substitute all the known values in the above equation.

$\begin{array}{rcl}{\mathrm{λ}}_{avg}& =& \frac{\left(6.63×{10}^{-34}\text{J}·\text{s}\right)\left(\frac{1\text{N}·\text{m}·\text{s}}{1\text{J}·\text{s}}\right)}{\sqrt{3\left(4.0\text{u}\right)\left(\frac{1.67×{10}^{-27}\text{kg}}{1\text{u}}\right)\left(1.38×{10}^{-23}\text{J/K}\right)\left(300\text{K}\right)\left(\frac{1{\text{N}}^{\text{2}}·{\text{s}}^{\text{2}}}{1\text{kg}·\text{J}}\right)}}\\ & ≈& 7.28×{10}^{-11}\text{m}\\ & ≈& \left(7.28×{10}^{-11}\text{m}\right)\left(\frac{{10}^{12}\text{pm}}{1\text{m}}\right)\\ & ≈& 72.8\text{pm}\end{array}$

Thus, the average de Broglie wavelength of the helium atoms is $72.8\text{pm}$.

The expression to calculate the average momentum of helium atoms is expressed as,

$\begin{array}{rcl}{p}_{avg}& =& \sqrt{2m_{He}{\left(KE\right)}_{avg}}\\ & =& \sqrt{2m_{He}\left(\frac{3}{2}kT\right)}\\ & =& \sqrt{3m_{He}kT}\end{array}$

Here, $p_{avg}$is the average momentum of helium atoms.

Substitute all the known values in the above equation.

$\begin{array}{rcl}{p}_{avg}& =& \sqrt{3\left(4.0\text{u}\right)\left(\frac{1.67×{10}^{-27}\text{kg}}{1\text{u}}\right)\left(1.38×{10}^{-23}\text{J/K}\right)\left(300\text{K}\right)\left(\frac{1{\text{kg}}^{\text{2}}·{\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}}{1\text{kg}·\text{J}}\right)}\\ & ≈& 9.11×{10}^{-24}\text{kg}·\text{m/s}\end{array}$

The expression to calculate the average distance between atoms is expressed as,

$d_{avg}=\sqrt[3]{\frac{kT}{p_{avg}}}$

Here, $d_{avg}$is the average distance between atoms.

Substitute all the known values in the above equation.

$\begin{array}{rcl}{d}_{avg}& =& \sqrt[3]{\frac{\left(1.38×{10}^{-23}\text{J/K}\right)\left(300\text{K}\right)}{9.11×{10}^{-24}\text{kg}·\text{m/s}}\left(\frac{1{\text{m}}^{\text{2}}}{1\text{J}·\text{s/kg}·\text{m}}\right)}\\ & ≈& 7.69\text{m}\end{array}$

Thus, the average distance between atoms is $7.69\text{m}$.

Since the value of the average de Broglie wavelength of the helium atoms $\left(7.28×{10}^{-11}\text{m}\right)$is less than the average distance between atoms $\left(7.69\text{m}\right)$. So, the helium atoms can be treated as particles.

Thus, the helium atoms can be treated as particles.