— Interplanar spacing implies the magnitude of the distance between two adjacent and parallel planes of atoms.

The interplanar spacing is a function of the, Miller indices (h,k and /) as well as the lattice parameter(s).

Interplanar spacing with a given set of Miller indices is a useful bit of information.

Planes with large interplanar spacing have low indices and a high density of lattice points, whereas the reverse is true for planes of close spacing.

— If we consider the simple cube shown in Fig. 35.26, it is obvious that, in this case, the most widely spaced atomic planes are those spaced at intervals equal to the lattice parameters, i.e., the (100) planes. The spacing is

d100 = a                                                                                   …(i)

The dodecahedral planes (110) bisect the face diagonal for a spacing of

D110 =( ½)√(a2+a2)= a/√2                                      …(ii)

The octahedral planes (111) trisect the body diagonal for a spacing of

d111=(1/3)√(a2+a2+a2)=a/√3                             …(iii)

— General equations for interplanar spacing for various unit cells can be written in terms of the Miller indices; namely:

Cubic :          1/d2hkl = (h2+k2+l2) / a2            …(iv)

Tetragonal:      1/d2hkl = (h2+k2)/a2 +( l2 )/ c2      …(v)

Orthorhombic:   1/d2hkl = (h2)/a2 + k2/b2 + (l2) / c2 …(vi)

Hexagonal:      1/d2hkl = (4/3)(h2+hk+k2) / a2 +l2/c2…(viii)

N.B.: a is lattice parameter and c/a is axial ratio in HCP crystal struc­ture (refer Fig. 35.11). From (iv) above, for a cubic crystal, the interplanar spacing dhkl = a√/(h2+k2+l2) and so on for tetragonal, orthorhombic and hexagonal structures.

Derivation of interplanar spacing of cubic crystal

The interplanar spacing or distance ‘dhkl‘ between adjacent planes of Miller indices (hkl) is defined as the spacing between the first such plane and a parallel plane passing through the origin.

Refer Fig. 35.27 ais the lattice constant. Let h, k, and l are the Miller indices for a plane ABC. An equation is to be derived for interplanar distance‘d’ of plane ABC from a plane parallel to it

and passing through ‘o’ (origin) in terms ofh,k,land a. OD is the normal from O on plane ABC. OD – d = interplanar distance. The intercepts of the plane ABC on the three axis i.c.,X, Y and Z axis are OA, OB and OC respectively.

Now, OA=a/h, OB=a/k,OC=a/l  …(i)

Also,    cos α=OD/OA= d/(a/h) =dh/a     …(ii)

              cos β= OD/OA= d/(a/k)= dk/a     …(iii)

              cos γ=OD/OC = d/(a/l) = dl/a      …(iv)

Since, cos2 α+ cos2β + cos2γ = 1. therefore

(dh/a)2 +(dk/a)2 + (dl/a)2 = 1     

            d2h2/a2 + d2k2/a2   + d2l2/a2 = 1

d2(h2k2 + l2) =a2

d2=a2 /(h2k2 + l2)

d= a2 /√(h2k2 + l2)