Visual walkthrough — Solid state — crystalline vs amorphous; unit cell, Bravais lattices

This is the visual companion to the parent topic. Once you own this counting rule you unlock Packing efficiency and density of unit cell and Close packing in solids — HCP, CCP, void fraction.

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Step 1 — What is a unit cell, really?

WHAT. We start with the single word repeat. A crystal is a pattern that copies itself forever in three directions. The smallest chunk you can copy to rebuild everything is the unit cell. Picture a wall of identical bricks: you don't memorise the whole wall, you memorise one brick and the rule "keep laying it down."

WHY. If a pattern is infinite but repeating, storing the whole thing is impossible and wasteful. Storing one block + a copy rule is complete. So every quantity we care about (atoms, mass, volume) we compute per one block.

PICTURE. Below, the same square tile is copied left–right and up–down. The bold tile is our unit cell; everything else is a copy. Notice the corners: they are touched by four tiles at once in 2-D.

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Step 2 — The core idea: a corner atom is a shared pizza

WHAT. Take one atom sitting exactly at a corner of a cube. Ask: does that atom belong only to this cube? No. Eight cubes meet at that corner (four above, four below), and they all touch the same atom.

WHY. We must never double-count. If cube A counts the whole corner atom and cube B counts the same whole atom, then across the crystal we'd count each atom 8 times. To make the total honest, each cube may claim only its fair slice of the atom.

PICTURE. The atom is sliced like a pizza into 8 equal wedges; each of the 8 surrounding cubes takes exactly one wedge. So each cube's fair share is $\frac{1}{8}$ of a corner atom.

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Step 3 — Edge, face, body: the full sharing table

WHAT. We now repeat the fair-slice argument for an atom sitting on an edge, on a face, and in the body centre. The only thing that changes is how many cubes touch that spot.

WHY. Different positions are shared by different numbers of cubes, so each earns a different fraction. Counting how many cubes touch the atom is the entire trick — nothing else.

PICTURE. For each position we shade the neighbouring cubes that meet there:

• Edge — 4 cubes run along one edge → the atom is quartered → $\frac{1}{4}$.
• Face — exactly 2 cubes share a flat face → the atom is halved → $\frac{1}{2}$.
• Body centre — the atom sits deep inside one cube, no neighbour reaches it → whole → $1$.

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Step 4 — Simple (primitive) cubic: why $Z=1$

WHAT. Put atoms only on the 8 corners of the cube and nowhere else. Count the total.

WHY. This is the barest possible cubic cell — the "control case." Getting $Z=1$ here confirms our slice rule matches reality: eight eighth-slices reassemble into exactly one whole atom.

PICTURE. Eight corner atoms, each glowing at $\frac{1}{8}$ strength; collect all 8 slices and they merge into one complete atom floating in the centre of the reasoning.

$Z_{\text{SC}}=\underset{\text{corners}}{\underbrace{8}}\times \underset{\text{share each}}{\underbrace{\frac{1}{8}}}=1$

• $8$ ← the eight corners of a cube.
• $\frac{1}{8}$ ← each corner's fair slice (Step 2).
• Product $=1$ ← eight slices rebuild one atom.

Recall Check yourself

Why isn't $Z=8$ for simple cubic? ::: Because each corner atom is shared by 8 cells, so each contributes only $\frac{1}{8}$; $8\times \frac{1}{8}=1$.

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Step 5 — Body-centred cubic (BCC): why $Z=2$

WHAT. Keep the 8 corner atoms, then drop one extra atom in the exact centre of the cube.

WHY. The centre atom is our first unshared atom — it lives entirely inside this one cube, so no neighbour can claim a slice of it. It counts as a full $1$. Add it to the corners' total from Step 4.

PICTURE. The 8 corners contribute their combined $1$ (faded), and a bold, fully-solid atom sits dead centre contributing another whole $1$.

$Z_{\text{BCC}}=\underset{\text{corners}=1}{\underbrace{8\times \frac{1}{8}}}+\underset{\text{body}=1}{\underbrace{1\times 1}}=2$

• First term $=1$ ← corners, exactly as in simple cubic.
• Second term $=1$ ← the body-centre atom is not shared, so its full share is $1$.
• Sum $=2$.

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Step 6 — Face-centred cubic (FCC): why $Z=4$

WHAT. Keep the 8 corners, and now place one atom in the middle of each of the 6 faces of the cube (no body atom).

WHY. Each face is the flat wall between exactly two cubes, so a face atom is halved between them — share $\frac{1}{2}$ (Step 3). Six of these plus the corners gives the answer.

PICTURE. Corners contribute their combined $1$; each of the 6 face atoms is sliced down the middle by the shared wall, glowing at $\frac{1}{2}$; the 6 half-atoms reassemble into 3 whole atoms.

$Z_{\text{FCC}}=\underset{\text{corners}=1}{\underbrace{8\times \frac{1}{8}}}+\underset{\text{faces}=3}{\underbrace{6\times \frac{1}{2}}}=4$

• $8\times \frac{1}{8}=1$ ← corners.
• $6\times \frac{1}{2}=3$ ← six faces, each atom split between 2 cells.
• Sum $=4$.

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Step 7 — Edge case: an edge-centred cell (and the zero case)

WHAT. Two "what if" situations that trip people up. (a) Suppose atoms also sat at the 12 edges of the cube — how much do they add? (b) What if a position is claimed by zero extra atoms — does the rule still hold?

WHY. Case (a) exercises the last fraction ($\frac{1}{4}$) so no reader meets it cold later (it appears in real structures like NaCl). Case (b) checks the rule degenerates gracefully: a position with no atom contributes $0$, so an empty cell has $Z=0$ — the arithmetic never breaks.

PICTURE. Left: a cube's 12 edges highlighted, each edge atom shared by 4 surrounding cubes ($\frac{1}{4}$ each); $12\times \frac{1}{4}=3$. Right: the same cube with edge sites empty — every term multiplies by $0$, total stays $0$.

$\text{edge total}=\underset{\text{edges}}{\underbrace{12}}\times \underset{\text{share}}{\underbrace{\frac{1}{4}}}=3,\text{empty cell: }Z=(\text{anything})\times 0=0$

• $12$ ← a cube has twelve edges.
• $\frac{1}{4}$ ← four cubes run alongside each edge (Step 3).
• Right equation: no atoms means every count is $0$, so $Z=0$ — the formula covers even the degenerate empty box.

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The one-picture summary

WHAT. One figure that stacks the three cubic cells side by side with their slice-arithmetic underneath, so the whole derivation reads left to right.

Recall Feynman retelling — explain the whole walkthrough to a 12-year-old

Imagine you're sharing pizzas with the neighbours whose kitchens all touch yours at the walls, edges, and corners of your house. A pizza sitting at the corner where 8 houses meet? You only get $\frac{1}{8}$ of it. On an edge shared by 4 houses? $\frac{1}{4}$. On a flat wall shared with 1 neighbour (2 houses total)? Half, $\frac{1}{2}$. A pizza sitting in the middle of your own room? It's all yours — a whole $1$. Now count the pizzas for your "house" (the cube). Corners only → the 8 corner-slices rebuild into 1 whole pizza (that's simple cubic). Add a whole pizza in the middle → 2 (body-centred). Or instead put a pizza on each of the 6 walls, each halved → 3, plus the 1 from corners → 4 (face-centred). That's the entire story: 1, 2, 4. You never counted a slice that wasn't really yours.

Recall Quick reveal checks

Share of a corner atom, and why? ::: $\frac{1}{8}$, shared by 8 cells. Share of an edge atom, and why? ::: $\frac{1}{4}$, shared by 4 cells. $Z$ for BCC and the two contributions? ::: $2$; corners $8\times \frac{1}{8}=1$ plus body $1\times 1=1$. $Z$ for FCC and the two contributions? ::: $4$; corners $=1$ plus faces $6\times \frac{1}{2}=3$.