English
Menu

Crystal structures and Miller indices

Treating atoms as hard spheres of equal size, what is the atomic packing factor of the FCC structure (the fraction of the unit-cell volume occupied by atoms)?

  1. AAbout 0.68
  2. BAbout 0.74
  3. CAbout 0.52
Show answer and explanation

Answer: B. About 0.74

In FCC, atoms touch along the face diagonal, so √2a = 4r, that is, a = 2√2r. With 4 atoms per unit cell, the packing factor is 4 × (4/3)πr³ / (2√2r)³ = π/(3√2) ≈ 0.74. About 0.68 is the BCC value (atoms touch along the body diagonal, a = 4r/√3, 2 atoms), and about 0.52 is the SC value (a = 2r, 1 atom, π/6). HCP with the ideal c/a also reaches 0.74 and is the other close-packed structure alongside FCC. Note that a high packing factor does not mean that each individual interstitial site is large.

Report an error

Practice this topic Back to topic

Related calculators

XRD Calculator (Bragg's Law and Extinction Rules)Calculate cubic interplanar spacings, 2θ angles and the peaks allowed by extinction rules

More Crystal structures and Miller indices questions

  1. Which correctly relates the lattice parameter a and the atomic radius r in the BCC structure? Assume hard-sphere atoms with nearest neighbors in contact.
  2. Copper has the FCC structure with a lattice parameter of 0.3615 nm and a molar mass of 63.55 g/mol. Taking Avogadro's number as 6.022×10²³ mol⁻¹, which value is closest to the calculated density of copper?
  3. If the lattice parameter of α-iron (BCC) is 0.2866 nm, which value is closest to the atomic radius in the hard-sphere model?
  4. In a cubic crystal, a plane intercepts the x axis at 1a and the y axis at 2a, and is parallel to the z axis. What are its Miller indices?
  5. Which correctly pairs the close-packed planes and close-packed directions of an FCC metal?