Interplanar cystal spacing of cubic crystal families is defined as

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

This source says that the interplanar spacing of the $(111)$ plane in FCC is $\frac{a}{\sqrt3}$, which is in agreement with the formula above.

However for BCC, interplanar spacing of $(111)$ is said to be $\frac{a}{2\sqrt3}$, which doesn't agree with the formula.

My question is: Does this mean that the formula for interplanar spacing of cubic structures is limited to special cases of crystal plane, for example $(111)$?


2 Answers 2


The formula $d_{hlk} = \frac{a}{\sqrt{h^2+k^2+l^2}}$ is a special case of the general formula:

$$d_{hkl} = \frac{2\pi}{|\vec{G}|}$$

for a reciprocal lattice vector $\vec{G}$ to the case of a cubic lattice. For FCC and BCC lattices you can express the reciprocal lattice vectors (usually written in terms of the Miller indices $(h,k,l)$) in terms of the reciprocal lattice vectors for a SC system BUT this is not quite correct!

In fact, the Miller index $(1,0,0)$ for a BCC lattice does not describe a family of lattice planes, because if you look at a BCC lattice, there are extra atoms at the centre of each SC cube. This means that the the correct lattice vector is in fact $(2,0,0)$.

This is summarised by the selection rules for these lattices which describe the Miller indices that are not valid $\vec{G}$ vectors. (These are important in x-ray diffraction since x-rays are scattered by exactly some $\vec{G}$ so these indices not being valid reciprocal vectors gives rise to systematic absences in the x-ray diffraction pattern. These are:

BCC: $h+k+l$ must be even

FCC: $h,k,l$ must have the same parity.

If you look at your table, you see your formula in the OP disagrees precisely when these conditions are not met.


The table shows the interplanar distances for cubic systems. That means simple cubic. It works also for (1,1,1) in FCC because the extra atoms in the center of each face doesn't change that distance. The planes become only more compact.


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