PHYSICS 315
Solution to problem set 2:

2.2-4:
(a)
The reciprocal lattice vectors can be obtained by requiring that

$${\bf\vec{A}\cdot\vec{a}}={\bf\vec{B}\cdot\vec{b}}=2\pi$$

$${\bf\vec{A}\cdot\vec{b}}={\bf\vec{B}\cdot\vec{a}}=0$$

We find

$${\bf\vec{A}}=\frac{2\pi}{a}\left({\bf\hat{x}}-\frac{\bf\hat y}{\sqrt{3}}\right)$$

$${\bf\vec{B}}=\frac{4\pi}{a\sqrt{3}}{\bf\vec{y}}$$

The reciprocal lattice is then

$${\bf\vec{G}}=h{\bf\vec{A}}+k{\bf\vec{B}}$$

b:The Wigner Seitz cell looks like fig 2.15 a of class notes. The corners of the cell are at

$$[0,\pm\frac{a}{\sqrt{3}}];\;\;[\pm\frac{a}{2},\pm\frac{a}{2\sqrt{3}}]$$

c: The Brillouin zone looks like the Wigner Seitz cell except the reciprocal lattice is rotated by 30$^o$ with respect to the direct lattice. The Brillouin zone is also rotated by 30$^o$ with respect to the Wigner Seitz cell. The coordinates of the corners are at

$$\frac{2\pi}{3a}(\pm 1,\pm\sqrt{3});\;\;(\pm\frac{4\pi}{3a},0)$$

2.2-6:

$$2d\sin\theta =n\lambda$$

Differentiating yields

$$\Delta d\sin{\theta}+\Delta\theta d\cos\theta =0$$

$\Delta d=150\alpha$ where $\alpha$ is the linear expansion coefficient. We need to convert $\Delta\theta$ to radians. We find

$$\alpha=\frac{6.4\pi}{60\times 180\times 150}=1.24\times 10^{-5}$$

3.2-2: KCl is face centered cubic with $a=6.29\;{\rm\AA}=6.29\times 10^{-10}$ m. The distance between nearest neighbor positive and negative ions is $d=a/2$. The Coulomb energy in eV is then

$$-\frac{e}{2\pi\epsilon_0a}=-4.57\;{\rm eV}$$

We need to find the parameters for the Lennard-Jones potential of Argon. Using

$$r_{nn}=1.09\sigma;\;\;\sigma=\frac{3.71}{1.09}=3.4\;{\rm\AA}$$

From the formula for the internal energy

$$U=-8.6N\epsilon;\;\;\epsilon=\frac{0.089}{8.6}=0.0104\;{\rm eV}$$

This gives

$$u_{LJ}=4\epsilon\left((\frac{\sigma}{d})^{12}-(\frac{\sigma}{d})^6\right)=0.27\; {\rm eV}$$