Which lattice has an APF of approximately 0.68?

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Multiple Choice

Which lattice has an APF of approximately 0.68?

Explanation:
APF is the fraction of the crystal’s volume actually taken up by atoms. In a body-centered cubic lattice there are two atoms per unit cell: eight corner atoms contribute 1/8 each (adding up to 1) plus one atom in the center, giving two atoms per cell. The atomic volume is 2 × (4/3)πr^3. The cell edge is related to the atomic radius by a = 4r/√3, so the cell volume is a^3 = (4r/√3)^3. Putting it together, APF = [2 × (4/3)πr^3] / [ (4r/√3)^3 ] = (8/3)πr^3 / (64r^3 / (3√3)) = π√3 / 8 ≈ 0.680. For simple cubic the APF is lower, about π/6 ≈ 0.524. Close-packed structures like FCC and HCP have higher APFs, around 0.740, so 0.68 is best matched by the body-centered cubic arrangement.

APF is the fraction of the crystal’s volume actually taken up by atoms. In a body-centered cubic lattice there are two atoms per unit cell: eight corner atoms contribute 1/8 each (adding up to 1) plus one atom in the center, giving two atoms per cell. The atomic volume is 2 × (4/3)πr^3. The cell edge is related to the atomic radius by a = 4r/√3, so the cell volume is a^3 = (4r/√3)^3.

Putting it together, APF = [2 × (4/3)πr^3] / [ (4r/√3)^3 ] = (8/3)πr^3 / (64r^3 / (3√3)) = π√3 / 8 ≈ 0.680.

For simple cubic the APF is lower, about π/6 ≈ 0.524. Close-packed structures like FCC and HCP have higher APFs, around 0.740, so 0.68 is best matched by the body-centered cubic arrangement.

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