Question 4 of 7: Fermi Energy, Fermi Temperature and Occupation Probability of Helium-3
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — Phys-A6: Solid State Physics — May 2014 (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 Helium-3 (20 marks)
Given. $He^3$ density $\rho=0.081\ \text{g/cm}^3$ near $T=0$; each atom has mass $\approx3\,\text{amu}$ (2 protons + 1 neutron); paper's Fermi-energy formula, eq. (13): $\epsilon_F=\dfrac{\hbar^2}{2m}\!\left(\dfrac{3\pi^2N}{V}\right)^{2/3}$; Fermi-Dirac distribution, eq. (12): $f(\epsilon)=\dfrac1{\exp[(\epsilon-\mu)/k_BT]+1}$.
Find. (a) $\epsilon_F$; (b) $T_F=\epsilon_F/k_B$; (c) $f(\epsilon=2.5\epsilon_F)$ at $\mu=1.5\epsilon_F$, $T=45$ K.
Fig. 4 — the 3-D Fermi surface: a sphere of radius $k_F$ in $\mathbf k$-space, filled up to energy $\epsilon_F$ at $T=0$ (reproduces Figure P4).
Approach. Treat the $He^3$ liquid as a free (non-interacting) Fermi gas of atoms: compute the number density $N/V$ from the given mass density and atomic mass, substitute into eq. (13) for $\epsilon_F$, divide by $k_B$ for $T_F$, then evaluate eq. (12) directly for part (c).
Part (a) — number density and Fermi energy. One $He^3$ atom has mass $m=3\times1.66053\times10^{-24}\ \text{g}=4.9816\times10^{-24}\ \text{g}=4.9816\times10^{-27}\ \text{kg}$. The number density is
$$\frac{N}{V}=\frac{\rho}{m}=\frac{0.081\ \text{g/cm}^3}{4.9816\times10^{-24}\ \text{g}}=1.626\times10^{22}\ \text{cm}^{-3}=1.626\times10^{28}\ \text{m}^{-3}$$
Substituting into eq. (13) with $\hbar=1.05459\times10^{-34}\ \text{J}\cdot\text{s}$ and $m=4.9816\times10^{-27}\ \text{kg}$:
$$\epsilon_F=\frac{(1.05459\times10^{-34})^2}{2(4.9816\times10^{-27})}\Big(3\pi^2\cdot1.626\times10^{28}\Big)^{2/3}$$
$$\boxed{\epsilon_F=6.86\times10^{-23}\ \text{J}=6.86\times10^{-16}\ \text{erg}\approx7\times10^{-16}\ \text{erg}}$$
matching the target value (the small residual is rounding in the source's "about $7\times10^{-16}$").
Part (b) — Fermi temperature. $T_F=\epsilon_F/k_B$ with $k_B=1.38062\times10^{-23}\ \text{J/K}$:
$$T_F=\frac{6.86\times10^{-23}\ \text{J}}{1.38062\times10^{-23}\ \text{J/K}}$$
$$\boxed{T_F\approx4.97\ \text{K}}$$
— this lands close to the accepted experimental Fermi temperature of liquid $He^3$ ($\sim5\ \text{K}$), which is exactly the check one expects since $He^3$'s low-temperature normal-liquid behaviour is well described by Landau Fermi-liquid theory built on this free-fermion estimate.
Part (c) — occupation probability at $\epsilon=2.5\epsilon_F$. With $\mu=1.5\epsilon_F$, $\epsilon-\mu=1.0\epsilon_F$, so
$$\frac{\epsilon-\mu}{k_BT}=\frac{\epsilon_F}{k_B(45\ \text{K})}=\frac{T_F}{45\ \text{K}}=\frac{4.97}{45}=0.1104$$
Substituting into eq. (12):
$$f(\epsilon)=\frac1{e^{0.1104}+1}=\frac1{1.1167+1}$$
$$\boxed{f(2.5\epsilon_F)\approx0.472\ \ (47.2\%)}$$
only slightly below $1/2$. At $T=0$ a level above $\mu$ would be empty ($f=0$), but here $k_BT=(45/4.97)\,\epsilon_F\approx9\epsilon_F$ is about nine times the gap $\epsilon-\mu=\epsilon_F$, so the Fermi-Dirac step is smeared almost flat over this energy range.
