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Density from lattice constant

Short answer

Using ρ = nA/(a³Nₐ): FCC Cu (n=4) atomic radius R = (√2/4)×0.3615 nm = 0.128 nm, density 8.93 g/cm³. BCC W (n=2) density 19.26 g/cm³.

(a) Copper has an FCC structure with lattice constant 0.3615 nm and atomic mass 63.54 g/mol. Calculate its atomic radius and theoretical density. (b) Tungsten is BCC with a = 0.3165 nm and atomic mass 183.85 g/mol. Calculate its theoretical density. Take Avogadro's number as 6.022×10236.022 \times 10^{23} atoms/mol.

Formula: theoretical density

ρ=nAVcNA\rho = \frac{n A}{V_c N_A}

where nn = atoms per unit cell, AA = atomic mass, Vc=a3V_c = a^3, NAN_A = Avogadro's number.

(a) Copper (FCC, n=4n = 4)

Atomic radius, from 4R=2 a4R = \sqrt{2}\,a:

R=2×0.36154=0.1278 nmR = \frac{\sqrt{2}\times 0.3615}{4} = 0.1278\ \text{nm}

Cell volume: a=0.3615×10−7a = 0.3615 \times 10^{-7} cm, so Vc=(0.3615×10−7)3=4.724×10−23 cm3V_c = (0.3615\times10^{-7})^3 = 4.724\times10^{-23}\ \text{cm}^3.

ρ=4×63.54(4.724×10−23)(6.022×1023)=254.1628.45=8.93 g/cm3\rho = \frac{4 \times 63.54}{(4.724\times10^{-23})(6.022\times10^{23})} = \frac{254.16}{28.45} = 8.93\ \text{g/cm}^3

(b) Tungsten (BCC, n=2n = 2)

a=0.3165×10−7a = 0.3165\times10^{-7} cm, so Vc=3.171×10−23 cm3V_c = 3.171\times10^{-23}\ \text{cm}^3.

ρ=2×183.85(3.171×10−23)(6.022×1023)=367.719.10=19.26 g/cm3\rho = \frac{2 \times 183.85}{(3.171\times10^{-23})(6.022\times10^{23})} = \frac{367.7}{19.10} = 19.26\ \text{g/cm}^3

The copper value agrees with the handbook value of 8.94 g/cm3^3 and tungsten with 19.25 g/cm3^3.

Answer: (a) R=0.128R = 0.128 nm, ρCu=8.93\rho_{Cu} = 8.93 g/cm³ (8930 kg/m³); (b) ρW=19.26\rho_{W} = 19.26 g/cm³ (19 260 kg/m³).

✓ Answer checked by Claude Opus 5.5 (high effort)

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