Material Science Exam 1 Practice Test

Which expression correctly represents the APF for FCC, using r = (√2 a)/4?

(4/3)π((√2 a/4)^3)/a^3 2(4/3)π((√2 a/4)^3)/a^3 3(4/3)π((√2 a/4)^3)/a^3 4(4/3)π((√2 a/4)^3)/a^3

Correct Answer:

4(4/3)π((√2 a/4)^3)/a^3

Explanation:

APF is the fraction of the unit cell volume actually filled by atoms. For a face-centered cubic lattice, there are 4 atoms occupying each unit cell, so the total volume of the atoms is 4 times the volume of a single atom: 4 × (4/3)πr^3. The unit cell volume is a^3, so APF = [4 × (4/3)πr^3] / a^3. With FCC, the radius is related to the lattice parameter by a = 2√2 r, which gives r = (√2 a)/4. Substituting this into the APF expression yields APF = 4 × (4/3)π[(√2 a)/4]^3 / a^3. This is the expression that includes the factor 4 in front: 4(4/3)π((√2 a/4)^3)/a^3. As a check, the numeric APF for FCC is π/(3√2) ≈ 0.740.

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