Question 3 of 7: Question III: Crystal Structure I

Question III: Crystal Structure I (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

III.1(a, c) — [111] Direction and (132) Plane in a Simple Cubic Cell

[111] direction. Starting at the origin, the vector travels one full lattice translation along each of $+x$, $+y$, $+z$ — i.e. straight along the cube's body diagonal to the opposite corner $(1,1,1)$, drawn as the red arrow above.

(132) plane. Miller indices $(hkl)=(132)$ mean the plane intercepts the axes at $1/h=1$, $1/k=1/3$, $1/l=1/2$ (in units of the cell edge $a$) — the shaded blue triangle above, connecting $(1,0,0)$, $(0,\tfrac13,0)$ and $(0,0,\tfrac12)$.

III.1(b, d) — $[1\bar{2}10]$ Direction and $(\bar{2}\bar{2}10)$ Plane in a Hexagonal Cell

$[1\bar{2}10]$ direction. For a Miller–Bravais direction $[uvtw]$ the real-space vector reduces to $U\,a_1+V\,a_2+W\,c$ with $U=u-t$, $V=v-t$, $W=w$. Here $u=1,\ v=-2,\ t=1,\ w=0$ (consistent, since $u+v+t=1-2+1=0$ as required), giving $U=0$, $V=-3$, $W=0$: the vector is purely along $-a_2$, with $\gcd(0,3,0)=3$, so the written index is exactly 3 primitive $-a_2$ lattice steps, not a single primitive step (drawn above as the direction line, with the reduction noted).

$(\bar{2}\bar{2}10)$ plane. A valid Miller–Bravais plane $(hkil)$ requires $i=-(h+k)$. Here $h=-2$, $k=-2$ give $i=-(-2-2)=4$, but the printed third index is $1$ — the same class of printing error as in Question I.2 (the Cl atomic number). Taking the self-consistent correction $i=4$, the plane is $(\bar2\bar24\,0)$, which reduces (dividing by 2) to $(\bar1\bar120)$: since $l=0$, this is a prism-type plane, parallel to the $c$-axis, of the $\{10\bar10\}$ family (the standard side faces of the hexagonal prism).

III.2 — FCC vs. HCP Stacking Sequence

Both FCC and HCP are close-packed structures built from identical close-packed atomic planes (each atom touching 6 neighbours within its own plane, packing factor 0.74), and differ only in how successive planes are stacked. HCP repeats every second plane directly above the first: ABABAB…, so the third layer sits exactly above the first. FCC instead repeats only every third plane: ABCABC…, with the third layer occupying a third distinct set of interstitial positions before the pattern returns to A. Both stackings achieve the same maximum packing density, but the different stacking periodicity is what gives FCC its cubic (rather than hexagonal) symmetry and its larger number of independent close-packed planes (4 × {111} in FCC vs. a single (0001) basal plane in HCP) — directly relevant to Question VI.2(b)'s FCC slip-system count.

III.3 — Theoretical Density of GaAs

Given. Zinc blende structure (4 formula units GaAs per unit cell, same atom count as diamond cubic); $a=0.565$ nm; $M_{Ga}=69.72$ g/mol, $M_{As}=74.91$ g/mol.

Find. Theoretical density $\rho$.

Approach. Apply the appendix relation $\rho=\dfrac{n\cdot A_{wt}}{V_c\cdot N_A}$ with $n=4$ and $A_{wt}=M_{Ga}+M_{As}$ (mass of one GaAs formula unit).

1. Unit cell volume. $$V_c=a^3=(5.65\times10^{-8}\,\text{cm})^3=1.8036\times10^{-22}\ \text{cm}^3$$
2. Formula-unit mass and density. $$A_{wt}=69.72+74.91=144.63\ \text{g/mol}, \qquad n=4$$ $$\rho=\frac{4(144.63)}{(1.8036\times10^{-22})(6.023\times10^{23})}=\frac{578.52}{108.63}=\boxed{5.33\ \text{g/cm}^3}$$