Cubic systems — SCC, BCC, FCC; packing fraction calculations: Visual walkthrough

We will need three ideas you may already have met: Pythagoras Theorem (to measure the diagonals of a cube), Avogadro Number and Molar Mass (only at the very end, for density), and the counting rule from the parent topic. Everything else we draw from scratch.

Step 1 — What are we even measuring?

WHAT we are doing: defining the single ratio the whole page computes.

WHY a ratio and not just "how many atoms"? Because a big box holds more atoms than a small box — that tells us nothing about how tightly they sit. A fraction cancels the box size, so it measures pure tightness. That is the only honest way to compare SCC, BCC and FCC on equal terms.

PICTURE — the same box, one nearly empty, one nearly full. The red-shaded region is atom; white is wasted air.

Figure — Cubic systems — SCC, BCC, FCC; packing fraction calculations

Step 2 — The box, and the one length that controls everything

WHAT: we named the box's edge $a$ and got its volume $a^{3}$ .

WHY start with $a$ ? Because $a$ will be our ruler. Every distance inside the cube — face diagonals, body diagonals, atom radii — will be written in terms of $a$ . If we can write both the atom volume and the box volume in the same ruler, the ruler cancels and $ϕ$ becomes a pure number.

PICTURE — the cube with all three edges labelled $a$ , and the volume formula sitting inside.

Step 3 — Counting the atoms the box truly owns ( $Z$ )

For our three cases:

$Z_{SCC} = 8 \times \frac{1}{8} = 1, Z_{BCC} = 8 \times \frac{1}{8} + 1 = 2, Z_{FCC} = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$

$8 \times \frac{1}{8}$ — the eight corners, each giving one-eighth → one whole corner-atom.
$+ 1$ (BCC) — the lone atom sitting at the very centre, wholly owned.
$+ 6 \times \frac{1}{2}$ (FCC) — six faces, each with a half-atom → three more atoms.

WHAT: we found $Z = 1, 2, 4$ .

WHY fractions and not whole counts? Because if every cube claimed a corner atom in full, we would count each corner atom $8$ times across the crystal — the total atom count would be eight times too big and $ϕ$ would exceed $1$ , which is nonsense.

PICTURE — one shared corner sphere sliced by the eight meeting cubes; the red wedge is the $\frac{1}{8}$ this cube keeps.

Step 4 — Volume of the atoms: turning $Z$ into filled space

$r$ — the atom's radius, the distance from its centre to its surface. This is the only new length we introduce.
$\frac{4}{3} π r^{3}$ — the standard volume of one sphere.
$Z$ — how many whole atoms are inside, from Step 3.

WHAT: we wrote the numerator of $ϕ$ .

WHY does $r$ appear and not the diameter? Because the sphere-volume formula is defined with the radius; and, as we'll see next, the touching condition is naturally written with $r$ too. Keeping one variable ( $r$ ) for atoms and one ( $a$ ) for the box sets up the crucial link in Step 5.

Right now $ϕ = \frac{Z \cdot \frac{4}{3} π r ^{3}}{a ^{3}}$ has two unknowns, $r$ and $a$ . We cannot yet get a number. We need one equation connecting them — that is the entire job of Steps 5–7.

PICTURE — a single sphere with its radius $r$ drawn in red, and the volume formula beside it.

Step 5 — The key move: which atoms actually touch?

The touch direction is different for each structure, and that difference is what makes $52.4%$ , $68%$ , $74%$ different numbers. We take them one at a time.

SCC — touch along the edge. Only corners are occupied. Two neighbouring corner atoms lie a full edge apart and just touch, so the edge spans two radii: $a = 2 r \Rightarrow r = \frac{a}{2}$ Here $2 r$ = radius of the left atom + radius of the right atom = one full diameter across the edge.

PICTURE — the bottom edge of the cube with two red half-spheres kissing at its midpoint.

Step 6 — BCC touches along the body diagonal (and how we measure it)

In BCC the corner atoms do not reach each other — there is a gap. The atom parked at the centre bridges the gap, touching two opposite corners. So the contact runs along the body diagonal, the longest straight line through the cube (corner → opposite corner, passing through the centre).

To measure that diagonal we use Pythagoras Theorem twice.

$a^{2} + a^{2}$ — the two equal edges of a face; their squares add (Pythagoras) to give the diagonal squared.
$(a 2)^{2} = 2 a^{2}$ — the face diagonal, squared, used as one leg of the second triangle.
$a 3$ — the full corner-to-corner distance.

Along this diagonal sit three atom-parts in a row: half of one corner atom, a whole centre atom, half of the far corner atom = $r + 2 r + r = 4 r$ . Therefore $a 3 = 4 r \Rightarrow r = \frac{a 3}{4}$

WHY $4 r$ and not $3 r$ ? Line up the centres: corner → centre is $2 r$ (they touch), centre → far corner is another $2 r$ . Total $4 r$ . The word "diagonal $= 4 r$ " hides that it is really two touching pairs nose-to-tail.

PICTURE — the cube with its red body diagonal drawn, the two Pythagoras triangles marked, and the $r + 2 r + r$ bead-string along it.

Step 7 — FCC touches along the face diagonal

In FCC every face carries an atom at its centre. A corner atom and the face-centre atom of the same face just touch, so contact runs along the face diagonal (which we already measured in Step 6: $a 2$ ).

Along that face diagonal sit three atom-parts: half a corner atom, a whole face-centre atom, half the other corner atom = $r + 2 r + r = 4 r$ . So $a 2 = 4 r \Rightarrow r = \frac{a 2}{4} = \frac{a}{2 2}, a = 22 r$

$a 2$ — the diagonal of one square face (Pythagoras, Step 6).
$4 r$ — the same nose-to-tail chain of touching atoms, now on the face instead of through the body.

WHY does FCC end up densest? The touch chain of length $4 r$ fits into the shortest diagonal that still contains it — the face diagonal is shorter than the body diagonal. A shorter chain-holder for the same atoms means bigger atoms for the same box → less air. Ranking the diagonals: $SCC edge a < FCC face a 2 < BCC body a 3$ but SCC crams only $2 r$ into its short edge while FCC crams $4 r$ into a modestly longer face — FCC wins.

PICTURE — one face of the cube, red diagonal, corner-halves and the full central atom touching along it.

Step 8 — Plug in and collect the three numbers

Now every structure has both pieces: its $Z$ (Step 3) and its $r$ – $a$ link (Steps 5–7). Substitute into $ϕ = \frac{Z \cdot \frac{4}{3} π r ^{3}}{a ^{3}}$ .

Notice $r^{3}$ cancels top and bottom in every case — exactly as promised in Step 4. The answer is a pure number, independent of how big the atoms actually are. That is why all copper crystals, and all iron crystals, share their structure's packing fraction.

Empty space is just $1 - ϕ$ : SCC $47.6%$ , BCC $32%$ , FCC $26%$ .

Step 9 — Edge and degenerate cases (never leave a gap)

Intuition The corner cases a good derivation must survive

$r = 0$ (point atoms): $V_{atoms} = Z \cdot \frac{4}{3} π \cdot 0 = 0$ , so $ϕ = 0$ . An empty box — the formula behaves sanely.
"Can $ϕ$ exceed 1?" No. If two spheres overlapped, the touch equation $= 4 r$ would become $< 4 r$ , forcing a larger $r$ than geometry allows — the model forbids it. Real spheres cannot interpenetrate, so $ϕ \leq 0.74$ for these cubics.
Mixing up the touch direction (the classic BCC error): using $a = 2 r$ for BCC assumes corner atoms touch, but in BCC they don't — there's visible air between them (Step 6 picture). Only the centre atom bridges the gap. Wrong direction → wrong $a$ → wrong density.
SCC vs FCC same corners, different answer: both have corner atoms, yet FCC's face atoms change the touch line from the edge to the shorter-per-atom face diagonal. Same corners, different neighbours actually touching — that alone triples the packing efficiency of the extra atoms.

The one-picture summary

One diagram, three columns. Each column shows: the occupied sites, the red touch line, the $r$ – $a$ equation it produces, and the resulting $ϕ$ . Read it left to right and the whole derivation replays.

Recall Feynman retelling — the walkthrough in plain words

We want to know how much of a box is really filled by round atoms and how much is air. First we name the box's edge $a$ — that's our only ruler. Then we count how many whole atoms the box owns, sharing the ones on corners and faces with neighbours: that's $Z = 1, 2, 4$ for the three cubes. The atoms take up $Z$ times a sphere's volume, which uses the radius $r$ — but now we have two unknowns, $a$ and $r$ , and only one equation. The trick that saves us: atoms are biggest when they just kiss. Where they kiss tells us $r$ in terms of $a$ . In a simple cube they kiss along the edge, so the edge is two radii. In a body-centred cube the centre atom bridges opposite corners along the long body diagonal (measured by Pythagoras twice, $a 3$ ), holding four radii. In a face-centred cube they kiss along the shorter face diagonal ( $a 2$ ), also four radii — and because that chain fits in a shorter line, the atoms end up biggest and the box tightest, $74%$ . Plug each $r$ – $a$ link into (atoms ÷ box), watch the $r^{3}$ vanish, and out drop three pure numbers: 52, 68, 74 — more atoms, more floor.

Connections

Coordination Number in Crystals — the same touch geometry decides how many neighbours each atom kisses.
Voids — Tetrahedral and Octahedral — the leftover $26%$ / $32%$ / $47.6%$ air lives in these gaps.
Density of a Unit Cell — feeds $Z$ and $a$ into $ρ = Z M / (a^{3} N_{A})$ .
Pythagoras Theorem — the tool behind both diagonals.
Close Packing — HCP vs CCP — FCC's $74%$ is the closest-packing limit.
Avogadro Number and Molar Mass — needed to turn $Z$ into real mass.