PHYSICS 315
Solution to problem set 3:

3.2-3:
(a)
Let us define

$$b=\frac{\alpha e^2}{4\pi\epsilon_0r}$$

$$u=\left\vert\frac{U}{N}\right\vert$$

We have

$$\frac{C}{r^m}=b-u$$

The cohesive energy has a minimum at the nearest neighbor distance

$$0=\frac{\partial U}{\partial r}=\frac{b}{r}-\frac{mC}{r^{m+1}}$$

giving

$$m=\frac{b}{b-u}$$

$$C=(b-u)r^m$$

Substituting

$$\alpha=1.7476;\;\;a=5.63\times 10^{-10} \;{\rm m};\;\; e=1.6... ...psilon_0=8.8542\times 10^{-12} {\rm Fm^{-1}};\;\;r=\frac{a}{2}$$

gives $m=8.84$, $C=0.585\times 10^{-103} SI units$.
b:
The volume of the primitive unit cell is $v=a^3/4=2r^3$. The pressure is

$$P=\frac{\partial u}{\partial v}=\frac{\partial r}{\partial v}\frac{\partial u}{\partial r}$$

with

$$\frac{\partial r}{\partial v}=\frac{1}{6r^2}$$

we have

$$B=-v\frac{\partial P}{\partial v}=\frac{v}{6r^2}\frac{\partial }{\partial r} (\frac{1}{6r^2}\frac{\partial u}{\partial r})$$

Using the fact that

$$\frac{\partial u}{\partial r}=0$$

at the equilibrium distance

$$B=\frac{1}{18r}(-\frac{2b}{r^2}+m(m+1)\frac{C}{r^{m+2}}$$

Finally, with

$$b=\frac{mC}{r^m}$$

we get the desired result

$$B=\frac{b(m-1)}{18r^3}$$

c: Substituting numbers I get $B=2.8\times 10^{10}$ Pa.
3.3-1
Since the material is isotropic we have $e_1=e_2=e_3\equiv e$ when it is subject to a uniform pressure. Hooke's law then becomes

$$P=\sum_{i=1}^6C_{ij}e_j=(3\lambda+2\mu)e$$

Hence

$$B=\frac{P}{\delta}=\frac{P}{3e}=lambda+\frac{2mu}{3}$$

3.3-2
We now have

$$P=(C_{11}+2C_{12})e$$

$$B=\frac{P}{3e}=\frac{C_{11}+2C_{12}}{3}$$