JEE Main 2019 — Solid State Question with Solution

${\mathrm{Z}}_{\mathrm{SC}} = 8 \times \frac{1}{8}= 1$

[Atoms are only at the corners]

${\mathrm{Z}}_{\mathrm{BCC}} = 8 \times \frac{1}{8} +1= 1 + 1=2$ 

[Atoms are at the corners and at the body center]

${\mathrm{Z}}_{\mathrm{FCC}} = 8 \times \frac{1}{8} + 6 \times \frac{1}{2}= 1 + 3 = 4$  

[Atoms are at the corners and at the face centers]

Here, Z= effective number of atoms in the unit cell.

Therefore, ${\mathrm{Z}}_{\mathrm{SC}}: {\mathrm{Z}}_{\mathrm{BCC}}: {\mathrm{Z}}_{\mathrm{FCC}}=1 : 2 : 4$

$\mathrm{SC}:\mathrm{BCC}:\mathrm{FCC}=1:2:4$