JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Solid State

In solid 2 the corner atoms have radius $r$ and the body-centre atom has radius $2r$. Along the body diagonal the atoms touch, so $\sqrt{3}a=r+2(2r)+r=6r$, giving $a=\frac{6r}{\sqrt{3}}=2\sqrt{3}r$. The cell holds eight corner atoms that add up to one whole atom of radius $r$, plus one central atom of radius $2r$, so the occupied volume is $\frac{4}{3}\pi r^3+\frac{4}{3}\pi (2r{)}^3=\frac{4}{3}\pi r^3(1+8)=12\pi r^3$. The cell volume is $a^3=(2\sqrt{3}r{)}^3=24\sqrt{3}{r}^3$. The packing efficiency is $\frac{12\pi r^3}{24\sqrt{3}{r}^3}\times 100=\frac{\pi }{2\sqrt{3}}\times 100\approx 90\%$.