Original English by Jim Hefferon — 34 validated sections. The original mathematics and supplied answers below are preserved. This is a partial-book reading edition, not the complete book or an Everyday-English rewrite.

Source, reuse and conversion details

Source revision df2262e089a02651c127f1dd12649c4622ee1383; CC BY-SA 2.5 option, with original component credits retained. This is not an Everyday-English rewrite. The complete active source topic is included. Previously given crystal data and diagram bindings travel with the applicable modular exercises. Supplied answers remain separate.

Four notes about the original supplied answers

These AI-assisted internal checks are separate from the unchanged original text and diagrams. They may reveal parts of solutions. Physical lattice constants have not been independently verified; this is not exhaustive certification or human review.

  1. Source note 1: Exercise 1’s supplied answer uses 3.34 × 10⁻¹⁰ cm, but the topic’s own 3.34 Å and Å-to-metre definition give 3.34 × 10⁻⁸ cm. It also divides one edge length even though the question asks for regions in a square face; using the printed spacing, the face count is (0.1/(3.34 × 10⁻⁸))², approximately 8.96 × 10¹². This checks the internal conversion and area count, not the physical lattice constant.
  2. Source note 2: Exercise 2(a)’s displayed system gives c₁ = 47447/17466 ≈ 2.72 and c₂ = 314/123 ≈ 2.55, not c₁ ≈ 2.74. Part (b) then reverses the rounded coefficients to 2.55β₁ + 2.72β₂; both original diagram definitions use that reversed pair. The prose and diagrams are retained unchanged rather than silently repaired.
  3. Source note 3: Exercise 3 lists (0.25,0.25,0.25) as an atom a quarter of the way down from the top. That point has height 0.25, not 0.75. The original diagram instead explicitly places this atom at (0.25,0.25,0.75), consistent with the prose description. The original supplied answer remains unchanged.
  4. Source note 4: Exercise 4(a) prints 195.08/6.02 × 10²³ = 3.239 × 10⁻²². Its denominator needs grouping, and the question gives 6.022 rather than 6.02. Using the stated data gives 195.08/(6.022 × 10²³) ≈ 3.239455 × 10⁻²² grams per atom. The other printed values are retained; no independent physical-constant verification is claimed.

Source-preserving rebuild, navigation, source packaging and current deterministic checks: OpenAI Codex — GPT-6 Astra, Ultra effort. Jim Hefferon remains the author of the mathematics. Earlier intermediate-conversion runtime identity is not established by its retained receipts and is not reassigned to this rebuild. No human review or exhaustive proof certification is claimed.

Crystals

Everyone has noticed that table salt comes in little cubes.

Original photograph of table-salt crystals, showing their approximately cubical shapes.

This orderly outside arises from an orderly inside— the way the atoms lie is also cubical, these cubes stack in neat rows and columns, and the salt faces tend to be just an outer layer of cubes. One cube of atoms is shown below. Salt is sodium chloride and the small spheres shown are sodium while the big ones are chloride. To simplify the view, it only shows the sodiums and chlorides on the front, top, and right.

Original cubic salt unit: small sodium spheres and larger chloride spheres on the front, top and right faces. The remaining faces are omitted to keep the view clear.

The specks of salt that we see above have many repetitions of this fundamental unit. A solid, such as table salt, with a regular internal structure is a crystal.

We can restrict our attention to the front face. There we have a square repeated many times giving a lattice of atoms.

A rectangular portion of the salt lattice repeats the square unit in three columns and two rows; corner and centre atoms are larger than the edge-midpoint atoms.

The distance along the sides of each square cell is about 3.34  Ångstroms (an Ångstrom is 10 − 10  meters). When we want to refer to atoms in the lattice that number is unwieldy, and so we take the square’s side length as a unit. That is, we naturally adopt this basis.

⟨ ( 3.34 0 ) , ( 0 3.34 ) ⟩

Now we can describe, say, the atom in the upper right of the lattice picture above as 3 β → 1 + 2 β → 2 , instead of 10.02  Ångstroms over and 6.68  up.

Another crystal from everyday experience is pencil lead. It is graphite, formed from carbon atoms arranged in this shape.

A single graphene plane: carbon atoms form a repeating lattice of connected hexagonal rings.

This is a single plane of graphite, called graphene. A piece of graphite consists of many of these planes, layered. The chemical bonds between the planes are much weaker than the bonds inside the planes, which explains why pencils write—the graphite can be sheared so that the planes slide off and are left on the paper.

We can get a convenient unit of length by decomposing the hexagonal ring into three regions that are rotations of this unit cell.

The rhombus-shaped unit cell used to divide a graphite hexagonal ring into three rotated regions. A hexagonal ring partitioned into three rotated rhombus-shaped cells, shaded differently to distinguish the three regions.

The vectors that form the sides of that unit cell make a convenient basis. The distance along the bottom and slant is 1.42  Ångstroms, so this

⟨ ( 1.42 0 ) , ( 0.71 1.23 ) ⟩

is a good basis.

Another familiar crystal formed from carbon is diamond. Like table salt it is built from cubes but the structure inside each cube is more complicated. In addition to carbons at each corner,

Diamond cube with carbon atoms at all eight corners; cube edges provide the three coordinate directions.

there are carbons in the middle of each face.

The same diamond cube with carbon atoms at the six face centres. The previously shown corner atoms are reduced to dots.

(To show the new face carbons clearly, the corner carbons are reduced to dots.) There are also four more carbons inside the cube, two that are a quarter of the way up from the bottom and two that are a quarter of the way down from the top.

Diamond cube showing the four interior atoms: two at height one-quarter and two at height three-quarters. Previously shown corner and face-centre atoms are dots.

(As before, carbons shown earlier are reduced here to dots.) The distance along any edge of the cube is 2.18  Ångstroms. Thus, a natural basis for describing the locations of the carbons and the bonds between them, is this.

⟨ ( 2.18 0 0 ) , ( 0 2.18 0 ) , ( 0 0 2.18 ) ⟩

The examples here show that the structures of crystals is complicated enough to need some organized system to give the locations of the atoms and how they are chemically bound. One tool for that organization is a convenient basis. This application of bases is simple but it shows a science context where the idea arises naturally.

Exercises

  1. Exercise 1 Worked answer

    How many fundamental regions are there in one face of a speck of salt? (With a ruler, we can estimate that face is a square that is 0.1  cm on a side.)

    Back to Exercise 1

    Answer. Each fundamental unit is 3.34 × 10 − 10  cm, so there are about 0.1 / ( 3.34 × 10 − 10 ) such units. That gives 2.99 × 10 8 , so there are something like 300 , 000 , 000 (three hundred million) regions.

  2. Exercise 2 Worked answer

    In the graphite picture, imagine that we are interested in a point 5.67  Ångstroms over and 3.14  Ångstroms up from the origin.

    1. Express that point in terms of the basis given for graphite.

    2. How many hexagonal shapes away is this point from the origin?

    3. Express that point in terms of a second basis, where the first basis vector is the same, but the second is perpendicular to the first (going up the plane) and of the same length.

    Back to Exercise 2

    Answer.

    1. We solve

      c 1 ( 1.42 0 ) + c 2 ( 0.71 1.23 ) = ( 5.67 3.14 ) ⟹ 1.42 c 1 + 0.71 c 2 = 5.67 1.23 c 2 = 3.14

      to get c 1 ≈ 2.74 and c 2 ≈ 2.55 .

    2. Here is the point located in the lattice. In the picture on the left, superimposed on the unit cell are the two basis vectors β → 1 and β → 2 , and a box showing the offset of 2.55 β → 1 + 2.72 β → 2 . The picture on the right shows where that appears inside of the crystal lattice, taking as the origin the lower left corner of the hexagon in the lower left.

      Original supplied-answer diagram for Exercise 2: basis directions and a parallelogram for the printed 2.55 beta-one plus 2.72 beta-two offset. A separate source note explains the coefficient mismatch. The same original supplied-answer offset shown within the graphene lattice. Its definition uses the printed reversed coefficient pair, preserved unchanged and disclosed separately.

      So this point is two columns of hexagons over and one hexagon up.

    3. This second basis

      ⟨ ( 1.42 0 ) , ( 0 1.42 ) ⟩

      makes the computation easier

      c 1 ( 1.42 0 ) + c 2 ( 0 1.42 ) = ( 5.67 3.14 ) ⟹ 1.42 c 1 = 5.67 1.42 c 2 = 3.14

      (we get c 2 ≈ 2.21 and c 1 ≈ 3.99 ), but it doesn’t seem to have to do much with the physical structure that we are studying.

  3. Exercise 3 Worked answer

    Give the locations of the atoms in the diamond cube both in terms of the basis, and in Ångstroms.

    Back to Exercise 3

    Answer. In terms of the basis the locations of the corner atoms are ( 0 , 0 , 0 ) , ( 1 , 0 , 0 ) , …, ( 1 , 1 , 1 ) . The locations of the face atoms are ( 0.5 , 0.5 , 1 ) , ( 1 , 0.5 , 0.5 ) , ( 0.5 , 1 , 0.5 ) , ( 0 , 0.5 , 0.5 ) , ( 0.5 , 0 , 0.5 ) , and ( 0.5 , 0.5 , 0 ) . The locations of the atoms a quarter of the way down from the top are ( 0.75 , 0.75 , 0.75 ) and ( 0.25 , 0.25 , 0.25 ) . The atoms a quarter of the way up from the bottom are at ( 0.75 , 0.25 , 0.25 ) and ( 0.25 , 0.75 , 0.25 ) . Converting to Ångstroms is easy.

  4. Exercise 4 Worked answer

    This illustrates how we could compute the dimensions of a unit cell from the shape in which a substance crystallizes ([Ebbing], p. 462).

    1. Recall that there are 6.022 × 10 23 atoms in a mole (this is Avogadro’s number). From that, and the fact that platinum has a mass of 195.08 grams per mole, calculate the mass of each atom.

    2. Platinum crystallizes in a face-centered cubic lattice with atoms at each lattice point, that is, it looks like the middle picture given above for the diamond crystal. Find the number of platinum’s per unit cell (hint: sum the fractions of platinum’s that are inside of a single cell).

    3. From that, find the mass of a unit cell.

    4. Platinum crystal has a density of 21.45 grams per cubic centimeter. From this, and the mass of a unit cell, calculate the volume of a unit cell.

    5. Find the length of each edge.

    6. Describe a natural three-dimensional basis.

    Back to Exercise 4

    Answer.

    1. 195.08 / 6.02 × 10 23 = 3.239 × 10 − 22

    2. Each platinum atom in the middle of each face is split between two cubes, so that is 6 / 2 = 3 atoms so far. Each atom at a corner is split among eight cubes, so that makes an additional 8 / 8 = 1  atom, so the total is 4 .

    3. 4 ⋅ 3.239 × 10 − 22 = 1.296 × 10 − 21

    4. 1.296 × 10 − 21 / 21.45 = 6.042 × 10 − 23 cubic centimeters

    5. 3.924 × 10 − 8 centimeters.

    6. ⟨ ( 3.924 × 10 − 8 0 0 ) , ( 0 3.924 × 10 − 8 0 ) , ( 0 0 3.924 × 10 − 8 ) ⟩

References cited in this section

Ebbing

Darrell D. Ebbing, General Chemistry, fourth edition, Houghton Mifflin, 1993.