Question 4 of 7: Fermi Energy, Fermi Temperature and Occupation Probability of Liquid He-3
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — Phys-A6: Solid State Physics — May 2013 (3 hours; closed book; useful equations and physical constants annexed to the paper). Any FIVE of the SEVEN questions constitute a complete exam paper (first five as they appear in the answer book are marked); all seven are solved below as a complete study resource.
Reference texts: C. Kittel, Introduction to Solid State Physics, 8th ed. (Ch. 1–3, 4, 6, 8, 14, 18); N. W. Ashcroft & N. D. Mermin, Solid State Physics, 1st ed. (Ch. 2, 4–7, 22, 28, 31–32).
Question 4: Fermi Energy, Fermi Temperature and Occupation Probability of Liquid He-3 (20 marks)
Given. He³ treated as a free-fermion gas; atomic mass $=3\ \text{amu}=3\times1.66053\times10^{-24}\ \text{g}=4.982\times10^{-24}\ \text{g}$; mass density $\rho=0.081\ \text{g/cm}^3$; occupation level $\epsilon=20\epsilon_F$; temperature $T=45\ \text{K}$; chemical potential $\mu\approx\epsilon_F$.
Find. (a) $\epsilon_F$; (b) $T_F=\epsilon_F/k_B$; (c) the Fermi-Dirac occupation probability $f(\epsilon)$ at $\epsilon=20\epsilon_F,\,T=45\ \text{K}$.
Fig. 4 — Spherical Fermi surface of radius $k_F$ in $k$-space (Fig. P4); all states with $|\mathbf{k}| < k_F$ are filled at $T=0$.
Approach. Compute the number density $N/V$ from the given mass density, apply the free-fermion-gas formula eq. (13) for $\epsilon_F$; divide by $k_B$ for $T_F$; substitute into the Fermi-Dirac distribution eq. (12) for the occupation probability.
Part (a) — Fermi energy. Number density of He³ atoms:
$$\frac{N}{V}=\frac{\rho}{m_{\text{He3}}}=\frac{0.081\ \text{g/cm}^3}{4.982\times10^{-24}\ \text{g}}=1.626\times10^{22}\ \text{cm}^{-3}$$
Applying eq. (13), $\epsilon_F=\dfrac{\hbar^2}{2m}\left(\dfrac{3\pi^2N}{V}\right)^{2/3}$, with $\hbar=1.05459\times10^{-27}\ \text{erg}\cdot\text{s}$ (CGS, from the constants page) and $m=m_{\text{He3}}=4.982\times10^{-24}\ \text{g}$:
$$\epsilon_F=\frac{(1.05459\times10^{-27})^2}{2(4.982\times10^{-24})}\Bigl[3\pi^2(1.626\times10^{22})\Bigr]^{2/3}\ \text{erg}$$
$\boxed{\epsilon_F\approx6.86\times10^{-16}\ \text{erg}\approx7\times10^{-16}\ \text{erg}}$ — matches the target.
Part (b) — Fermi temperature. $T_F=\epsilon_F/k_B$, with $k_B=1.38062\times10^{-16}\ \text{erg/K}$ (CGS):
$$T_F=\frac{6.86\times10^{-16}\ \text{erg}}{1.38062\times10^{-16}\ \text{erg/K}}$$
$\boxed{T_F\approx4.97\ \text{K}}$ — this lands remarkably close to the accepted experimental Fermi temperature of liquid He-3 ($\approx5\ \text{K}$), which is a strong internal check on Part (a).
Part (c) — occupation probability at $\epsilon=20\epsilon_F$, $T=45\ \text{K}$. Using eq. (12) with $\mu\approx\epsilon_F$:
$$f(\epsilon)=\frac{1}{\exp\!\left[\dfrac{\epsilon-\mu}{k_BT}\right]+1}$$
$$\frac{\epsilon-\mu}{k_BT}=\frac{19\epsilon_F}{k_B(45\ \text{K})}=\frac{19(6.86\times10^{-16})}{(1.38062\times10^{-16})(45)}=2.10$$
$$f=\frac{1}{e^{2.10}+1}=\frac{1}{8.16+1}$$
$\boxed{f(20\epsilon_F)\approx0.109\ (10.9\%)}$ — because $T=45\ \text{K}\gg T_F=4.97\ \text{K}$, He³ at this temperature is only weakly degenerate, so a level well above $\epsilon_F$ still carries a non-negligible thermal occupation, unlike a metal's electron gas at room temperature where $T\ll T_F$ and such a level would be occupied with vanishing probability.
