Question

The lattice constant of a simple cubic primitive cell is 5.28 Å. Determine the distancebetween the nearest parallel ( a ) (100), ( b ) (110), and ( c ) (111) planes.

Answer 1

Answer:

a)5.28 Å , b)3.73 Å , c)3.048 Å

Explanation:

the atoms are situated only at the corners of cube.Each and every atom in simple cubic primitive at the corner is shared with 8 adjacent unit cells.

Therefore, a particular unit cell consist only 1/8th part of an atom.

The lattice constant of a simple cubic primitive cell is 5.28 Å

We know formula of distance,

d = $\frac{a}{\sqrt{h^{2}+k^{2}+l^{2}}}$

a)(100)

a=5.28 Å

Distance = $\frac{5.28 Å}{\sqrt{1^{2}+0^{2}+0^{2}}}$=5.28 Å

b)(110)

Distance = $\frac{5.28}{\sqrt{1^{2}+1^{2}+0^{2}}}$ = 3.73 Å

c)(111)

Distance= $\frac{5.28}{\sqrt{1^{2}+1^{2}+1^{2}}}$= 3.048 Å