Obtain an order-of-magnitude value for the temperature at which helium might begin to exhibit quantum/ superfluid behaviour. See equation (9.43). (Helium's specific gravity is about $\mathbf{0}\mathbf{.}\mathbf{12}$.)

$\frac{\mathbf{N}}{\mathbf{V}}\frac{{\mathbf{n}}^{\mathbf{3}}}{{\mathbf{\left(}{\mathbf{mk}}_{\mathbf{B}}\mathbf{T}\mathbf{\right)}}^{\frac{\mathbf{3}}{\mathbf{2}}}}\mathbf{<}\mathbf{<}\mathbf{1}$

Where,

$\mathbf{N}\mathbf{\to }$Number of particles

$\mathbf{V}\mathbf{\to }$Volume

$\mathbf{\hslash }\mathbf{\to }$Planck's reduced constant

$\mathbf{m}\mathbf{\to }$Mass of the object

${\mathbf{k}}_{\mathbf{B}}\mathbf{\to }$Boltzmann' constant

$\mathbf{T}\mathbf{\to }$Temperature.

Converting mass of helium from u to kg unit

$\begin{array}{c}{\mathrm{m}}_{\mathrm{A}}=4\mathrm{u}\\ =\left(4\mathrm{u}\right)\left(\frac{1.66\times {10}^{-27}\text{kg}}{1\mathrm{u}}\right)\\ =6.64\times {10}^{-27}\text{kg}\end{array}$

$\begin{array}{c}\frac{\mathrm{N}}{\mathrm{V}}\frac{{\mathrm{\hslash }}^3}{{\left({\mathrm{mk}}_{\mathrm{B}}\mathrm{T}\right)}^{\frac{3}{2}}}<<1\\ \frac{\mathrm{N}}{\mathrm{V}}{\mathrm{\hslash }}^3<<{\left({\mathrm{mk}}_{\mathrm{B}}\mathrm{T}\right)}^{\frac{3}{2}}\\ {\left(\frac{\mathrm{N}}{\mathrm{V}}\right)}^{\frac{3}{2}}{\mathrm{\hslash }}^2<<{\mathrm{mk}}_{\mathrm{B}}\mathrm{T}\\ \mathrm{T}>>\frac{{\mathrm{\hslash }}^2}{{\mathrm{mk}}_{\mathrm{R}}}{\left(\frac{\mathrm{N}}{\mathrm{V}}\right)}^{\frac{2}{3}}\end{array}$

The $N/V$ can be rewritten as the mass per unit volume over the mass per atom (or just the mass of an atom), or the bulk density $D$over the atomic mass $m$.

$\begin{array}{c}\mathrm{T}>>\frac{{\mathrm{\hslash }}^2}{{\mathrm{mk}}_{\mathrm{B}}}{\left(\frac{\mathrm{N}}{\mathrm{V}}\right)}^{\frac{2}{3}}\\ >>\frac{{\mathrm{\hslash }}^2}{{\mathrm{mk}}_{\mathrm{R}}}{\left(\frac{\mathrm{N}}{\mathrm{V}}\right)}^{\frac{2}{3}}\end{array}$

The specific gravity of helium is 0.12, so the density of the helium is 0.12 times that of density of the air, which has an approximate density of $1.2\text{kg}/{\text{m}}^3$. Therefore, its density of helium is

$\mathrm{D}=0.12\left({\mathrm{D}}_{\text{air}}\right)$

Substitute $1.2\text{kg}/{\text{m}}^3$for density of the air ${\mathrm{D}}_{\text{air}}$.

$\begin{array}{c}\mathrm{D}=\left(0.12\right)(1.2\text{kg}/{\text{m}}^3)\\ =0.144\text{kg}/{\text{m}}^3\end{array}$

Substitute $0.144\text{kg}/{\text{m}}^3$ for $\mathrm{D},1.38\times {10}^{-23}\text{J}/\text{K}$ for ${\mathrm{k}}_{\mathrm{B}},1.055\times {10}^{-34}\text{J}\cdot \text{s}$ and ${\mathrm{m}}_{\mathrm{A}}$for 6.64 $\times {10}^{-27}\text{kg}$

$\begin{array}{c}\frac{{\mathrm{\hslash }}^2}{{\mathrm{mk}}_{\mathrm{B}}}{\left(\frac{\mathrm{D}}{\mathrm{m}}\right)}^{\frac{2}{3}}=\frac{{(1.055\times {10}^{-34}\text{J}\cdot \text{s})}^2}{(6.64\times {10}^{-27}\text{kg})(1.38\times {10}^{-23}\text{J}/\mathrm{K})}{\left[\frac{(0.144\text{kg}/{\text{m}}^3)}{(6.64\times {10}^{-27}\text{kg})}\right]}^{\frac{2}{3}}\\ =0.0094\text{K}\end{array}$,