I'm studying X-rays crystallography. The basic assumption is that if I let some X-rays scatter on a crystal the atoms act as sources of new waves. Considering two waves that hit two "neighbouring" planes of the crystal I can have a constructive interference then if the phase difference between the two waves is equal to an integer multiple of the wavelenght:

$$ n\lambda=2d\sin{\theta} $$

where d is the distance between the two planes of the crystal. My question is: how can I practically determine the value of n? If I need to get to know the value of d I can't just put a random n in the equation as the value of d will change and give me an uncorrect result. Thank you in advance for the help.


Since $\lambda$ is fixed by the incoming light you send, each $n$ gives you a different angle $\theta_n$ such that $n \lambda = 2d~\sin\theta_n$. All of these secondary waves will be present because the incoming wave is scattered in all directions but only the directions $\theta_n$ verifying the above equation will interfere constructively.

Therefore each direction $\theta_n$ will correspond to a Bragg peak on your area detector. So the game is then to index those peaks, i.e. to find which value of $n$ correspond to each peak. But nobody does indexing by hand anymore: this will be performed by the software provided by the manufacturer of your diffractometer.

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    $\begingroup$ I think you might want to elaborate on OP's question as to how to actually determine an $n$ for a specific peak or sth along those lines ... $\endgroup$ – Sanya Jul 8 '17 at 12:32
  • $\begingroup$ Or am I totally off here? Are you doing some kind of "school" project for which you are suppose to do everything "by hand"? $\endgroup$ – user154997 Jul 8 '17 at 14:48
  • $\begingroup$ @LucJ.Bourhis Thank you for your answer. I still don't get why each n should give me a different angle $\theta$, I'm sure I'm missing something. $\endgroup$ – Federico Mastellone Jul 8 '17 at 16:06
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    $\begingroup$ @LucJ.Bourhis to me, your edited answer explains a lot more :) $\endgroup$ – Sanya Jul 9 '17 at 12:37
  • $\begingroup$ You can thank @Alexis Prel for that: he authored the changes! $\endgroup$ – user154997 Jul 9 '17 at 22:27

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