Problem 9 - Entrance Test

For a face-centered cubic (FCC) structure, the relationship between the edge length (a) and the atomic radius (r) is: 4r = a√2 Therefore, a = 4r/√2 = 2√2 r Given the atomic radius of copper (r) = 128 pm. First, calculate the edge length 'a': a = 2√2 × 128 pm = 2 × 1.414 × 128 pm = 362.00 pm (approximately) The length of the body diagonal of a cubic unit cell is given by the formula: Body diagonal = a√3 Now, substitute the value of 'a': Body diagonal = (2√2 r) × √3 = 2√6 r Body diagonal = 2 × 2.449 × 128 pm = 627.00 pm (approximately) Let's re-calculate using the calculated 'a': a = 2√2 × 128 = 2 × 1.41421 × 128 = 362.037 pm. Body diagonal = a√3 = 362.037 pm × 1.73205 = 626.96 pm. This value is very close to 625 pm. Let's check the options again. The calculation yields approximately 627 pm. Option D is 625 pm. Let's be precise: a = 2 * sqrt(2) * 128 = 362.038 pm. Body diagonal = a * sqrt(3) = 362.038 * sqrt(3) = 362.038 * 1.73205 = 626.96 pm. Final answer is 626.96 pm, which is closest to 625 pm.