Question 1 of 7: Miller Indices of Planes A & B; FCC Rhodium Density

Question 1: Miller Indices of Planes A & B; FCC Rhodium Density (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.

Given. (a) Two adjacent unit cells; plane A is the triangle through the front-bottom-left corner, the midpoint of the far vertical edge, and the shared front-bottom corner; plane B is the vertical plane at the mid-depth mark of the second cell. (b) FCC rhodium, $a_0=3.796$ Å; $M_{Rh}=102.9$ g/mol (page-1 table); $N_A=6.02\times10^{23}$ mol$^{-1}$.

Find. (a) Miller indices of A and B; construct the [012] line from their intersection. (b) Density of rhodium.

Approach

Assign local axes with the front-bottom-left corner of the first cell at $(x,y,z)=(1,0,0)$ (the figure's $+x$ arrow points toward this corner, so $x$ increases toward the viewer) and the hidden back corner at the origin. Each plane's intercepts on the three axes follow directly from the marked "1/2" fractions, and the Miller indices are the reciprocals of those intercepts, cleared of fractions. The two planes' line of intersection is found by solving their equations simultaneously; part (b) is a standard FCC unit-cell mass/volume density calculation.

1. Plane A — identify the three vertices and fit a plane. Plane A's triangle runs from $P_0=(1,0,0)$ (the front-bottom-left corner) to $V_2=(0,0,\tfrac12)$ (the "1/2" mark halfway up the far vertical edge) to $V_3=(1,1,0)$ (the shared front-bottom corner with the next cell). Using edge vectors $\mathbf e_1=V_2-P_0=(-1,0,\tfrac12)$ and $\mathbf e_2=V_3-P_0=(0,1,0)$, the normal is $$\mathbf n=\mathbf e_1\times\mathbf e_2=\left(-\tfrac12,0,-1\right)\ \parallel\ (1,0,2),$$ giving the plane $x+2z=1$ (checked against $P_0$). Setting $y=z=0$ gives $x$-intercept $1$; setting $x=y=0$ gives $z$-intercept $\tfrac12$; there is no $y$-term, so the $y$-intercept is at infinity (the plane is parallel to $y$). Intercepts $(1,\infty,\tfrac12)$ ⇒ reciprocals $(1,0,2)$: $$\text{Plane A}=\boxed{(102)}.$$
2. Plane B — a face-parallel plane. Plane B is vertical and parallel to both $y$ and $z$, cutting the depth axis at the "1/2" marks on the top and bottom faces of the second cell, i.e. at $x=\tfrac12$. Intercepts $(\tfrac12,\infty,\infty)$ ⇒ reciprocals $(2,0,0)$: $$\text{Plane B}=\boxed{(200)}.$$ (Not simplified to (100): a plane parallel to a cube face but offset to the half-cell mark is a genuinely different, more closely spaced plane than the true face plane, and Miller indices distinguish the two.)
3. Intersection of A and B, and the [012] construction. Substituting $x=\tfrac12$ into $x+2z=1$ gives $z=\tfrac14$ (independent of $y$), so the two planes meet along the line $x=\tfrac12,\ z=\tfrac14$. Within the bounded triangle A, this line enters at $s=\tfrac12$ along edge $P_0V_2$, i.e. at the point $$Q=\left(\tfrac12,0,\tfrac14\right),$$ the midpoint of edge $P_0V_2$ — marked with the red dot in Fig. Q1. The requested $[012]$ direction is $(\Delta x,\Delta y,\Delta z)=(0,1,2)$: from $Q$, move purely in $+y$ and $+z$ with twice as much rise in $z$ as run in $y$. Drawn to the top face ($z=1$, a rise of $\tfrac34$) this reaches $y=\tfrac12\times\tfrac34/1=0.375$, i.e. the line drawn in red in Fig. Q1 from $(\tfrac12,0,\tfrac14)$ to $(\tfrac12,0.375,1)$ — the same $x=\tfrac12$ depth throughout, since the $[012]$ direction has no $x$-component.
4. (b) FCC rhodium density. An FCC cell contains $n=4$ atoms (8 corners $\times\tfrac18$ + 6 face centres $\times\tfrac12$). With $a_0=3.796\times10^{-8}$ cm, $$\rho=\frac{nM_{Rh}}{N_Aa_0^3}=\frac{4\times102.9}{6.02\times10^{23}\times(3.796\times10^{-8})^3} =\boxed{12.50\ \text{g/cm}^3}.$$ (The accepted density of rhodium is 12.41 g/cm$^3$ — the close agreement confirms the FCC assumption and the arithmetic.)