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Copper FCC: packing density and atomic concentration

Short answer

FCC has 4 atoms per cell, atoms touch on face diagonal: a = 2√2 R. For Cu with R=0.128nm, packing density = π/(3√2) ≈ 0.74 (74%). Atomic concentration = 4/a³ = 8.43×10²⁸ m⁻³ (8.43×10²² cm⁻³).

Copper has FCC (Face-centered cubic) structure. Find the packing density and atomic concentration for copper if radius of copper atom is 0.128nm.

Copper has an FCC structure: 4 atoms per unit cell (8×18+6×128\times\frac{1}{8} + 6\times\frac{1}{2}), with atoms touching along the face diagonal, so 2 a=4R\sqrt{2}\,a = 4R.

Given: R=0.128R = 0.128 nm.

Lattice constant

a=22 R=22×0.128=0.3620 nm=3.620×10−10 ma3=(3.620×10−10)3=4.745×10−29 m3\begin{aligned} a &= 2\sqrt{2}\,R = 2\sqrt{2}\times0.128 = 0.3620\ \text{nm} = 3.620\times10^{-10}\ \text{m} \\ a^3 &= (3.620\times10^{-10})^3 = 4.745\times10^{-29}\ \text{m}^3 \end{aligned}

Packing density

APF=4×43πR3a3=4×43π×(0.128×10−9)34.745×10−29=3.513×10−294.745×10−29=0.74\begin{aligned} \text{APF} &= \frac{4\times\frac{4}{3}\pi R^3}{a^3} = \frac{4\times\frac{4}{3}\pi\times(0.128\times10^{-9})^3}{4.745\times10^{-29}} \\ &= \frac{3.513\times10^{-29}}{4.745\times10^{-29}} = 0.74 \end{aligned}

(In general, for FCC, APF=π/(32)=0.74\text{APF} = \pi/(3\sqrt{2}) = 0.74.)

Atomic concentration

nat=4a3=44.745×10−29=8.43×1028 atoms m−3n_{at} = \frac{4}{a^3} = \frac{4}{4.745\times10^{-29}} = 8.43\times10^{28}\ \text{atoms m}^{-3}

Answer: packing density =0.74= 0.74 (74 %); atomic concentration =8.43×1028= 8.43\times10^{28} m⁻³ =8.43×1022= 8.43\times10^{22} cm⁻³.

✓ Answer checked by Claude Opus 5.5 (high effort)

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