Question 5 of 7: Microstructural Characterization

Question 5: Microstructural Characterization (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.

5(a) — XRD Peak Positions for FCC Nickel Powder

Given. Nickel, FCC, $R=0.125$ nm; Cu-Kα radiation, $\lambda=0.1542$ nm.

Find. The lattice parameter, the first three allowed $(hkl)$, their $d$-spacings, and the corresponding $2\theta$.

Approach. Get $a$ from the FCC touching condition along the face diagonal; apply the FCC structure-factor selection rule (reflections occur only when $h,k,l$ are all even or all odd); rank the allowed planes by $d$-spacing (largest $d$ = smallest $2\theta$ = first peak); apply Bragg's law.

1. Lattice parameter. FCC atoms touch along the face diagonal, $a\sqrt2=4R$: $$a=2\sqrt2R=2\sqrt2(0.125)=\boxed{0.3536\ \text{nm}}$$
2. Allowed planes (reflection rule). FCC reflects only unmixed-parity $(hkl)$ (all even or all odd). In order of decreasing $d$ (increasing $h^2+k^2+l^2$), the first three allowed planes are $(111)$, $(200)$, $(220)$ — $(100)$, $(110)$, $(210)$ etc. are systematically absent (mixed parity, destructive interference).
3. Interplanar spacings, $d=a/\sqrt{h^2+k^2+l^2}$. $$d_{111}=\frac{0.3536}{\sqrt3}=0.2041\ \text{nm}\qquad d_{200}=\frac{0.3536}{2}=0.1768\ \text{nm}\qquad d_{220}=\frac{0.3536}{\sqrt8}=0.1250\ \text{nm}$$
4. Bragg angles, $n=1$: $\sin\theta=\lambda/2d$. $$2\theta_{111}=2\arcsin\!\frac{0.1542}{2(0.2041)}=\boxed{44.4^{\circ}}\qquad 2\theta_{200}=2\arcsin\!\frac{0.1542}{2(0.1768)}=\boxed{51.7^{\circ}}$$ $$2\theta_{220}=2\arcsin\!\frac{0.1542}{2(0.1250)}=\boxed{76.2^{\circ}}$$

5(b) — Single Crystal with (100) Parallel to the Surface

A conventional powder pattern samples every crystallite orientation simultaneously, so all allowed $(hkl)$ families appear. A single crystal cut with $(100)$ parallel to the surface presents only planes parallel to that surface to the incident beam in a standard $\theta$–$2\theta$ scan (Bragg reflection requires the diffracting planes to lie parallel to the sample surface). Only the $(100)$-family orders $(h00)$ satisfy this and the FCC selection rule — i.e. only $(200)$, $(400)$, … would be observed. The $(111)$ and $(220)$ peaks, which come from planes not parallel to $(100)$, would be absent from this single-crystal scan.

5(c) — Distinguishing FCC from BCC by XRD

The two structures obey different selection rules: FCC allows only unmixed-parity $(hkl)$ (all even or all odd), giving $h^2+k^2+l^2$ in the ratio $3:4:8:11:12:16\ldots$; BCC allows only $h+k+l=$ even, giving the ratio $2:4:6:8:10:12\ldots$. Measuring the peak angles, computing $\sin^2\theta$ for each (which is proportional to $h^2+k^2+l^2$), and forming successive ratios $\sin^2\theta_n/\sin^2\theta_1$ reproduces one of these two characteristic integer sequences — matching the sequence identifies the structure without needing to know $a$ or $R$ in advance.