Question
Calculate the density of a face-centred cubic (FCC) metal with edge length a=4×10−8 cm and atomic mass M=60 g/mol. Take NA=6.022×1023.
Solution — Step by Step
Each corner atom is shared among 8 unit cells, contributing 1/8. Each face-centre atom is shared between 2 cells, contributing 1/2.
Z=8×81+6×21=1+3=4 atoms per unit cell
ρ=a3⋅NAZ⋅M
ρ=(4×10−8)3×6.022×10234×60
=64×10−24×6.022×1023240
=64×6.022×10−1240=38.54240≈6.23 g/cm3
Final answer: ρ≈6.23 g/cm3.
Why This Works
The density formula ρ=ZM/(a3NA) packages crystallography. ZM is the mass of all atoms in a unit cell (in grams, after dividing by Avogadro). a3 is the unit cell volume. Their ratio is the density.
Memorise Z for the three common cubics: simple cubic Z=1, BCC Z=2, FCC Z=4.
Alternative Method
Compute mass per unit cell first: 4×60/(6.022×1023)≈3.99×10−22 g. Volume: (4×10−8)3=6.4×10−23 cm3. Density: 3.99×10−22/6.4×10−23≈6.23 g/cm3. Same answer with explicit intermediate quantities.
For NEET, watch for unit traps: edge length might be in pm or Å, density expected in g/cm3. Convert everything to cm before computing a3, otherwise the powers of 10 spiral out of control.
Common Mistake
Students sometimes use Z=8 for FCC, mistakenly counting only corner atoms without dividing by 8. The face-centre atoms are crucial — they bring the count to 4. Always tally corner contribution and face contribution separately.