Question 1 of 7: Miller Indices from a Unit-Cell Sketch; FCC Rhodium Lattice Constant and Density

Question 1: Miller Indices from a Unit-Cell Sketch; FCC Rhodium Lattice Constant and 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 cubic unit cell drawn with $+x$ toward the viewer (lower left), $+y$ to the right and $+z$ upward, carrying two shaded planes. Three fractional marks are printed on the sketch, all reading ½:

• Plane A is the shaded triangle whose corners are the near-bottom cell corner on the $x$-axis, the ½ mark on the $y$-axis, and the ½ mark on the $z$-axis — i.e. intercepts $a$, $b/2$, $c/2$.
• Plane B is the shaded parallelogram parallel to the $z$-axis. Its vertical left edge stands on the $y$-axis at the same ½ mark (it shares that point with Plane A), and its vertical right edge stands on the far bottom edge at the third ½ mark, i.e. at $x=a/2$, $y=b$. Plane B therefore contains the base line joining $(0,\,b/2,\,0)$ and $(a/2,\,b,\,0)$.

Find. (a) Miller indices $(hkl)$ for planes A and B, and a construction of the $[012]$ direction from the point where the two planes intersect. (b) Lattice constant confirmation and theoretical density of FCC rhodium given $a_0 = 3.796$ Å.

Approach

(a) The standard Miller-index procedure is: read the plane's intercepts on each axis (in units of the cell edge), take reciprocals, then clear to the smallest integer triplet. Solving the two planes' equations simultaneously inside the cell locates the tail point for the requested $[012]$ direction, which is then drawn with components $0$ (along $x$), $1$ (along $y$), $2$ (along $z$) from that tail — scaled down as needed so the segment stays inside the cell. (b) Rhodium's FCC cell holds $Z=4$ atoms; density follows from $\rho = ZM/(N_A a_0^3)$ using the page-1 atomic mass and Avogadro's number.

1. (a) Plane A — intercepts and indices. The triangle cuts the $x$-axis at the full edge and both of the other axes at the printed ½ marks: intercepts $(1,\ \tfrac12,\ \tfrac12)$ in units of $a,b,c$. Reciprocals: $$\frac{1}{1},\ \frac{1}{1/2},\ \frac{1}{1/2} \;=\; 1,\ 2,\ 2$$ — already the smallest integer triplet. $$\boxed{\text{Plane A} = (122)}$$
2. (a) Plane B — intercepts and indices. Plane B is parallel to $z$, so its $z$-intercept is at infinity and its third index is $0$. Its base line passes through $(0,\,\tfrac12,\,0)$ and $(\tfrac12,\,1,\,0)$, so in the base plane it is $$y - x = \tfrac12 .$$ Setting $z=0$ and $y=0$ gives the $x$-intercept $x=-\tfrac12$; setting $x=0$ gives the $y$-intercept $y=+\tfrac12$. Intercepts $(-\tfrac12,\ \tfrac12,\ \infty)$, reciprocals $-2,\ 2,\ 0$, and dividing through by the common factor $2$: $$\boxed{\text{Plane B} = (\bar 1 1 0)}$$ (equivalently $(\bar 2 2 0)$ before reduction, or $(1\bar 1 0)$ if the opposite normal — the same plane — is taken; the negative index simply records that the plane must be extended to $x=-a/2$ to reach the $x$-axis.)
3. (a) Locating the intersection point of A and B. In cell-edge units the two planes are $$\text{A:}\quad \frac{x}{1}+\frac{y}{1/2}+\frac{z}{1/2}=1 \ \Rightarrow\ x+2y+2z=1, \qquad \text{B:}\quad y-x=\tfrac12 .$$ Substituting $y=x+\tfrac12$ into A gives $x+2x+1+2z=1$, i.e. $3x+2z=0$. Since $x\ge0$ and $z\ge0$ inside the cell, the only solution is $x=z=0$, hence $y=\tfrac12$: $$\boxed{P = \left(0,\ \tfrac{b}{2},\ 0\right)}$$ The two planes therefore touch the cell at exactly one point — which is why the question says "the intersection point" — and the sketch shows it: the right-hand corner of triangle A and the foot of Plane B's left edge are the same ½ mark on the $y$-axis.
4. (a) Constructing the $[012]$ direction. A direction symbol $[uvw]$ specifies a vector with components $u\cdot a$, $v\cdot b$, $w\cdot c$ from a chosen tail point. Starting at the tail $P=(0,\,b/2,\,0)$ found above, move $0$ along $x$, $+1$ unit along $y$ and $+2$ units along $z$; halving the vector (which does not change the direction) keeps the whole segment inside the cell: $$Q = P + \tfrac12\,(0,\,b,\,2c) = \left(0,\ b,\ c\right).$$ So the $[012]$ line runs from the ½ mark on the $y$-axis to the cell corner $(0,b,c)$, lying entirely in the $x=0$ face — shown dashed in the figure below.
5. (b) Confirm the FCC assignment is self-consistent, then find $\rho$. The FCC cell holds $Z=4$ atoms (8 corners $\times\tfrac18$ + 6 face-centres $\times\tfrac12 = 1+3=4$). Using the page-1 table $M_{Rh}=102.90$ g/mol, $N_A=0.602\times10^{24}$ mol$^{-1}$, and converting $a_0 = 3.796\times10^{-8}$ cm so that $a_0^3 = 5.4699\times10^{-23}$ cm$^3$: $$\rho = \frac{ZM_{Rh}}{N_A a_0^3} = \frac{4(102.90)}{(6.02\times10^{23})(5.4699\times10^{-23}\,\text{cm}^3)}$$ $$\boxed{\rho \approx 12.50\ \text{g/cm}^3}$$ This is close to rhodium's accepted handbook density ($12.41$ g/cm$^3$), confirming the FCC assignment and the given lattice constant.

[Figure not reproduced: Fig. Q1(a) — the unit cell redrawn to the printed sketch: Plane A $(122)$ cutting $a,\ b/2,\ c/2$; Plane B $(\bar 1 1 0)$ parallel to $z$ through $(0,b/2,0)$ and $(a/2,b,0)$; their single common point $P=(0,b/2,0)$; and the $[012]$ direction drawn from $P$ to the cell corner $(0,b,c)$. See the official exam paper.]