English
Menu

Crystal structures and Miller indices

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.

  1. Aa = 2r
  2. Ba = 2√2r
  3. Ca = 4r/√3
Show answer and explanation

Answer: C. a = 4r/√3

In BCC, atoms touch along the body diagonal that joins a corner atom to the body-center atom. The body diagonal has length √3a and spans four atomic radii, so √3a = 4r, that is, a = 4r/√3 ≈ 2.31r. a = 2√2r is the FCC relation (contact along the face diagonal, √2a = 4r), and a = 2r is the SC relation (contact along the cube edge). If you overlook the fact that BCC atoms do not touch along the face diagonal, it is easy to apply the FCC relation by mistake.

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. 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?
  2. 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?
  3. 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?
  4. Which correctly pairs the close-packed planes and close-packed directions of an FCC metal?
  5. Which value is closest to the interplanar spacing of the (111) planes of aluminum (FCC, lattice parameter 0.405 nm)?