Diffraction, wave length I read that when we want to study the structure of a crystal we have to use an electromagnetic wave which has a wave length of approximately the distance between two consecutive atoms in the crystal.
Why is that ? why can't we use a wave length that is 10 time bigger/shorter (or more) ? The calculations still hold , don't they ? (and the diffraction image has still the good dimensions )
Thanks. 
 A: According to Bragg's law:
$$ \sin\theta = \frac{n\lambda}{d} $$
In a typical X-ray diffractometer it's experimentally convenient to have $\theta$ around $\pi/2$ (we actually measure $2\theta$ in the diffractometer so $\theta = \pi/2$ means the line appears at 90° to the incoming beam). If $\theta$ is much bigger the diffraction line disappears off the end of our film, and if $\theta$ is too small all the lines are clustered together at small angle and we can't see the small detail.
Suppose we choose $\lambda = 10d$, then we get:
$$ \sin\theta = \frac{10nd}{d} = 10n \ge 10$$
because $n \ge 1$, and there is no solution i.e. we get no line. OK, lets suppose we choose $\lambda = d/10$, then we get:
$$ \sin\theta = \frac{nd}{10d} = \frac{n}{10} $$
In this case we can get values for $\sin\theta$ in a useful range by choosing lines with large values of $n$. The trouble is that the line intensity drops rapidly with increasing $n$ and by the time we get to $n = 5 - 10$ the intensity has dropped so much we need long exposure times.
So wavelengths of around $\lambda = d$ are the only ones that are experimentally useful.
Footnote:
$d$ isn't (necessarily) the distance between atoms. It's the distance between planes in the lattice and will typically between 1 and 0.1 times the unit cell dimensions. The distance between atoms can't be bigger than the unit cell but it can be a lot smaller if the structure is complex.
