See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

No. of Si per unit cell = 8 Let radius of Si atom = R 2R = 3a 8R a 4 3   % of vacant space = 100 – 3 3 4 8 R 3 a   = 3 8 4 3 3 100 3 8     = 66%