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For a cubic structure only one quantity is involved, the cell edge or the lattice parameter .
Equation (1) can be expressed in the form
From the observed values of , a table of values of should be prepared. For a cubic structure these should be in simple numerical proportion. The highest common factor should be /4a2, and hence a can be derived. (There is an exception to this rule which will be discussed later.)
Not all values of h2 + k2 + l2, which we shall call N, are possible. For example, there are no three numbers whose squares add up to 7, or, in general, to p2(8q - 1) where p and q are integers. Numbers such as 7, 15, 23, 28, 60 are said to be forbidden. The absences of numbers such as these are useful in checking the correctness of indexing for a large unit cell.
For small values of N, the values of h, k and l are easily deduced. Thus N = 1 corresponds to 100, 2 to 110 and 3 to 111. For some values of N, more than one set of indices exists: 9 is both 300 and 221. For values of N (up to 100) see Lipson and Steeple (1970) Table 5.
The intensities of the lines depend upon the arrangement of atoms in the unit cell, but they also depend upon the number of possible ways of combining the indices - the multiplicity factor . For simple structures, this factor is dominant. Thus 100 includes 010, 001, 00, 00, 00; the multiplicity factor is 6, corresponding to the six faces of a cube. For N = 14 (321) however there are 48 arrangements, and line 14 will usually be much stronger than line 1, even if the decrease in intensity with is allowed for.
If the lattice is not primitive, some values of N are not possible. For the lattice F (face-centred), the indices must be all odd or all even: thus the first few lines are N = 3(111), 4(200), 8(220), 11(311), 12(222), 16(400)These are shown diagrammatically in Fig. 4. For the body-centred lattice, I (Innenzentiert), N must be even, and so the possible lines are 2(110), 4(200), 6(211), 8(220), 10(310)
This raises the difficulty mentioned earlier: for a body-centred structure the common factor is 2/4a2 and not /4a2. We can make use of the forbidden numbers to deal with this problem: if a line appears to have N = 7, we know that it is not so; the line is probably 14 and the structure body-centred.
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