The ratio of a number of atoms present in a simple cubic, body-centered cubic and face-centered cubic structure are, respectively:

 

  • Option 1)

    8 : 1 : 6

  • Option 2)

    1 : 2 : 4

  • Option 3)

    4 : 2 : 1

  • Option 4)

    4 : 2 : 3

 

Answers (1)

 

No. of atoms(z) for simple cubic unit cell -

Lattice points: at corners

For simple cubic (SC), z=1

- wherein

 

 

 

No. of atoms(z) for face centered unit cell -

Lattice points: at corners and face centers of unit cell.

For face centered cubic (FCC), z=4.

- wherein

 

 

 

z for body centered unit cell -

Lattice points: at corners and body center of unit cell.

For body centered cubic (BCC), z=2

- wherein

 

 

 

In SC- lactic point present only at corner =\frac{1}{8}\times 8=1

In BCC batting point present at corner + Body centered = \frac{1}{8}\times 8+1=2

In FCC lattice point present at corner + face centre = \frac{1}{8}\times 8+\frac{1}{2}\times 6=4

1 : 2 : 4


Option 1)

8 : 1 : 6

Option 2)

1 : 2 : 4

Option 3)

4 : 2 : 1

Option 4)

4 : 2 : 3

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