Solid State: Tricky Questions Solved (1)

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Question

Calculate the density of a face-centred cubic (FCC) metal with edge length a=4×108a = 4 \times 10^{-8} cm and atomic mass M=60M = 60 g/mol. Take NA=6.022×1023N_A = 6.022 \times 10^{23}.

Solution — Step by Step

Each corner atom is shared among 88 unit cells, contributing 1/81/8. Each face-centre atom is shared between 22 cells, contributing 1/21/2.

Z=8×18+6×12=1+3=4 atoms per unit cellZ = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4 \text{ atoms per unit cell}

ρ=ZMa3NA\rho = \frac{Z \cdot M}{a^3 \cdot N_A}

ρ=4×60(4×108)3×6.022×1023\rho = \frac{4 \times 60}{(4 \times 10^{-8})^3 \times 6.022 \times 10^{23}}

=24064×1024×6.022×1023= \frac{240}{64 \times 10^{-24} \times 6.022 \times 10^{23}}

=24064×6.022×101=24038.546.23 g/cm3= \frac{240}{64 \times 6.022 \times 10^{-1}} = \frac{240}{38.54} \approx 6.23 \text{ g/cm}^3

Final answer: ρ6.23\rho \approx 6.23 g/cm3^3.

Why This Works

The density formula ρ=ZM/(a3NA)\rho = ZM / (a^3 N_A) packages crystallography. ZMZ M is the mass of all atoms in a unit cell (in grams, after dividing by Avogadro). a3a^3 is the unit cell volume. Their ratio is the density.

Memorise ZZ for the three common cubics: simple cubic Z=1Z = 1, BCC Z=2Z = 2, FCC Z=4Z = 4.

Alternative Method

Compute mass per unit cell first: 4×60/(6.022×1023)3.99×10224 \times 60 / (6.022 \times 10^{23}) \approx 3.99 \times 10^{-22} g. Volume: (4×108)3=6.4×1023(4 \times 10^{-8})^3 = 6.4 \times 10^{-23} cm3^3. Density: 3.99×1022/6.4×10236.233.99 \times 10^{-22} / 6.4 \times 10^{-23} \approx 6.23 g/cm3^3. Same answer with explicit intermediate quantities.

For NEET, watch for unit traps: edge length might be in pm or Å, density expected in g/cm3^3. Convert everything to cm before computing a3a^3, otherwise the powers of 10 spiral out of control.

Common Mistake

Students sometimes use Z=8Z = 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.

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