PHYSICS 315
Problem set 5 2004. Solution

4.2-1 Sodium metal is monovalent and body centered cubic with $a=4.23\times 10^{-10}$ Å. This gives with

$$\frac{N}{V}=\frac{2}{a^3}=\frac{k_F^3}{3\pi^2}$$

$$k_F=9.21\times 10^9 {\rm m^{-l}}$$

a: Substuting for $\hbar$ and the electron mass we get for the Fermi energy

$$\epsilon_F=5.18\;10^{-19} {\rm J}=3.23\;{\rm eV}$$

b: The Fermi velocity is

$$\frac{\hbar k_F}{m}=1.066\times 10^6 {\rm m s^{-1}}$$

c: With $k_B=1.381\times 10^{-23}$

$$T_F=3.75\times 10^4 {\rm K}$$

4.2-3
The kinetic energy of the gas is

$$2\sum_{k<k_F}\frac{\hbar^2k^2}{2m}=\frac{2V}{(2\pi)^3} 4\pi\i... ...k^4dk}{2m} = \frac{h^2k_F^5V}{5\pi^22m}= \frac{3}{5}N\epsilon_F$$

which is the desired result.
4.2-4
The Fermi temperature of Al is given in the text as $13.6\times 10^4$K, while the Debye temperature is 428 K. There are 3 conduction electrons per Al ion. The (Sommerfeld) electronic specific heat per Al ion is thus

$$\frac{3\pi^2k_BT}{2T_F}=1.1\times 10^{-4}k_B,\; 1.1\times 10^... ...1\times 10^{-2}k_B\;{\rm respectively \; for \; T=1,10,100 \;K}$$

The two lowest temperatures are much lower than the Debye temperature so we can use the formula

$$C_V=\frac{12\pi^4}{5}Nk_B\left(\frac{T}{\theta_D}\right)$$

giving for the specific heat per ion $3\times 10^{-6}k_B$ and $3\times 10^{-3}k_B$ for the two lowest temperatures. The highest temperature is about $0.23\theta_D$ and from fig 4.5 I read off a value of the specific heat of about $1.5k_B$ per ion. So, the conclusion is that at 1 K, the electronic specific heat dominates, at 100 K the lattice specific heat dominates, while at 10 K the two contributions are comparable