Question

Find the d-spacing of (111) plane in FCC crystal of
lattice constant 4.938 A
al 2469 Å​

Answer 1

Explanation:

check attachment for solution

Answer 2

Answer:

The d-spacing of (111) plane in FCC crystal of

1)   lattice constant 4.938 Å =  2.509 Å

2)  lattice constant 2469 Å =   1425.4778 Å

Explanation:

The d- spacing or the distance between the consecutive lattice planes of a Face centered cubic lattice  is given by,

$d_{h k l}=\frac{a}{\sqrt{h^{2}+k^{2}+l^{2}}}$          ...(1)

Where,

h,k,l    -  Miller indices

a         - lattice constant

Given,

( hkl )  = (100)

h = 1

k = 1

l  = 1

1)

Lattice constant a = $2.814$ Å

Equation (1) becomes,

$d_{111}=\frac{4.938 \AA}{\sqrt{1^{2}+1^{2}+1^{2}}}$  

       $=\frac{4.938}{\sqrt{3} } \AA= 2.8509\ \AA$

2)

Lattice constant a = $2469$ Å

Equation (1) becomes,

$d_{111}=\frac{2469 \AA}{\sqrt{1^{2}+1^{2}+1^{2}}}$

       $=\frac{2469}{\sqrt{3} } \AA= 1425.4778\ \AA$