Physics 315
Physics of Materials
Solution to midterm exam February 2004

1: (a) If the two types of ions in the structure were of the same size, the packing fraction would be the same as for a simple cubic structure with the side of the unit cube equal to a/2. Hence,

$$f=\frac{4\pi}{3}(\frac{a}{4})^3(\frac{2}{a})^3=0.524$$

(b) If the largest ions touch their radii would be

$$r_1=\frac{a}{2\sqrt{2}}=.3536$$

If the largest ions touch the smaller ones the radius of the largest ion would be

$$r_2=\frac{a}{3}=0.3333$$

The last result is the smallest and thus the correct answer. We have with 4 ions of each kind per unit cube

$$f=\frac{4\pi}{3}\frac{1}{3^3}4(1+\frac{1}{2^3})=0.698$$

2:
(a) A possible choice of direct lattice vectors is

$$\vec{a}=\frac{a}{2}(\hat{y}+\hat{z});\;\;\vec{b}=\frac{a}{2}(\hat{z}+\hat{x});\;\; \vec{c}=\frac{a}{2}(\hat{x}+\hat{y})$$

A set of primitive reciprocal lattice vectors would then be

$$\vec{A}=\frac{2\pi \vec{b}\times\vec{c}}{a\cdot(\vec{b}\times\vec{c})}=\frac{2\pi}{a}(-\hat{x}+\hat{y}+\hat{z})$$

Similarly

$$\vec{B}=\frac{2\pi}{a}(\hat{x}-\hat{y}+\hat{z})$$

$$\vec{C}=\frac{2\pi}{a}(\hat{x}+\hat{y}-\hat{z})$$

b: The $(\bar{1}11)$ and $(111)$ planes are equivalent so the distance between the $(111)$ planes is

$$d=\frac{2\pi}{\vert A\vert}=\frac{a}{\sqrt{3}}=2.08\;{\rm\AA}$$

c: The bulk density is

$$\rho=\frac{4}{a^3}$$

The number of atoms/unit area in the (111) planes of copper is thus

$$\rho d=\frac{4}{\sqrt{3}a^2}=0.177\;{\rm\AA^{-2}}$$

3: a: Since the pressure is uniform and the material isotropic we must have for the strains

$$e_1=e_2=e_3\equiv e;\;\;e_4=e_5=e_6=0$$

Hooke's law then becomes

$$\sigma_1=P=C_{11}e_1+C_{12}((e_2+e_3)=(3\lambda+2\mu)e$$

The fractional volume change is then

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

Substituting numerical values for $P,\;\lambda,\;\mu$ gives

$$\delta=7.9\times 10^{-3}$$

(b) We have with $T$ the tension, $f$ the force and $A$ the cross-sectional area

$$\sigma_1=T=\frac{f}{A}=(\lambda+2\mu)e_1+2\lambda e_2$$

There is no stress in the perpendicular directions

$$\sigma_2=0=\lambda e_1+(2\lambda+2\mu)e_2$$

The solutions to the two equations are

$$e_1=\frac{(\lambda+\mu)T}{\mu(3\lambda+2\mu)};\;\;e_2=\frac{-\lambda T} {2\mu(\lambda+2\mu)}$$

Substituting $f=100 N$, $A=10^{-6}$ m$^2$ and the provided values for $\lambda$ and $\mu$

$$e_1=0.0092,\;\;e_2=-0.0033$$

The fractional change in length is thus 0.0268 while the fractional change in volume is $e_1+2e_2=0.002.$