Question

Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic, (b) fcc, (c) bcc, and (d) diamond lattice. As- suming that nearest atoms are touching each other, what is the lattice constant of each lattice?

Answer 1

(a) A =
`3.90 \AA`

(b)
`A = 4.50 \AA`

(c)
`A = 5.51 \AA`

(d)
`A = 9.02 \AA`

Solution:

As per the question:

Radius of atom, r = 1.95
`\AA = 1.95* 10^(- 10) m`

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A =
`2* 1.95 = 3.90 \AA`

(b) For body centered cubic lattice:

`A = (4)/(√(3))r`

`A = (4)/(√(3))* 1.95 = 4.50 \AA`

(c) For face centered cubic lattice:

`A = 2{√(2)}r`

`A = 2{√(2)}* 1.95 = 5.51 \AA`

(d) For diamond lattice:

`A = 2* (4)/(√(3))r`

`A = 2* (4)/(√(3))* 1.95 = 9.02 \AA`