Question

In a unit cell atoms a b c and d are present at corners, face center, body centre and edge centre respectively

Answer 1

Lets first learn how much an atom contribute one unit cell when it is present in corner, face center, body centre and edge centre.

The contribution of atom present in corner is 1/8 because it shared among eight unit cell

so in a unit cell there are 8 corner, so total number of atom in unit cell would be 1/8*8= 1

similarly contribution of an atom present in face centre contributes 1/2 to the unit cell because it is shared between two unit cell

there are 6 faces so total number of atom in a unit cell is 1/2*6=3

There can be only one body centre atom.

Atom present in edge is shared among four unit cell. So, total contribution of atom to one unit cell is 1/4. there are total of 12 edges so

total number of atom in a unit cell is 1/4*12=3

According to question

a is present at corner which mean only one atom in unit a=1

b is present at face centre so number of b= 3

c is present at body centre c=1

and d is present in edge centre so number of d=3

so formula of unit cell would be $ab_{3} cd_{3}$

Answer 2

Question:

In a unit cell, atoms A, B, C and D are present at corners, face-centres, body-centre and edge-centre respectively in a cubic unit cell. The total number of atoms present per unit cell is ?

Answer:

A → at corners = 8

B → at face centres = 6

C → body centre = 1

D → edge centre = 12

Total atoms in a cube = 8 + 6 + 1 + 12 = 27

A =  8×1/8=1

B = 6 ×1/2=3

C = 1 × 1 = 1

D = 12×1/4=3

Total = 8 atoms per unit cell