In my course we showed the principle of X-ray diffraction in crystals.

We have a X-ray beam going on a powder of crystals on we look in function of $2 \theta $ the intensity scattered, where $ \theta $ is the incidence angle (so $ 2 \theta $ is the angle between the scattered beam and the incident beam).

we have a figure like this for example : http://www.eserc.stonybrook.edu/ProjectJava/Bragg/SiC1.gif

We know from Bragg law that $ p \lambda = 2d \sin \theta $ where $p$ is a relative integer.

The thing that I don't understand is that we assume that the first intensity peak always correspond to $p=1$ in the Bragg law.

But why could'nt it be a case where $p=6$ for example ?

Indeed we could have a crystal where nearest parallel planes have a huge spacing between them ?

[edit] : to better understand my confusion, take this picture : https://upload.wikimedia.org/wikipedia/commons/thumb/7/74/BraggPlaneDiffraction.svg/1920px-BraggPlaneDiffraction.svg.png

Now imagine that $d=10 \lambda$, and that there are not any plane between the two planes represented in the picture (they are the closest ones). And imagine that I am looking for a peak of intensity corresponding to the reflection of thoose planes.

The bragg law will not be possible with $p=1$ ($p$ will have a value of at least $p=10$).


1 Answer 1


Let us assume we can see every peak. Then, it is clear that as we increase $\theta$ that the first peak encountered will be $p=1$ since $\sin \theta$ increases continuously from $0$. This can be seen from $p \propto \sin \theta$.

Now, in the real-world situation it may not be true that you can see every peak, namely the first. How then can you tell? This is where you have to fit the data of multiple peaks against $\sin\theta$, and see what the best guess for $p$ is for each peak. Since the relationship is not linear, the spacing between peaks becomes more dense for higher $p$, so you can actually differentiate between different $p$ in this case.

  • $\begingroup$ I don't understand/agree your first paragraph. I agree that if I increase $ \theta $, for a given set of parallel plane I will at a given angle have the $p+1$ diffraction. ** But ** my first peak could be at $ p=3 $. We just have to imagine a crystal where the closest parallel planes are at $ d=3*\lambda $ for example. The march difference for beams can't be equal to $ \lambda $ in this example, it must be at least $ 3 \lambda $. $\endgroup$
    – StarBucK
    Nov 11, 2016 at 11:14
  • $\begingroup$ You should take a look at a derivation of Bragg's law again, such as in en.wikipedia.org/wiki/Bragg%27s_law. $2d\sin\theta$ corresponds to the path difference, yes, but I think your image of what the path difference should be is different from what it actually is. Remember that you need to match points on each ray connected by a perpendicular line between them. $\endgroup$
    – Aaron
    Nov 11, 2016 at 16:46
  • $\begingroup$ I don't understand what you mean by matching points on each ray with a perpendicular line between them ? To better explain my confusion, take this picture : upload.wikimedia.org/wikipedia/commons/thumb/7/74/… and imagine that d=10 lambda. I will never have the case p=1 for diffraction with thoose planes because the march difference will necesseraly be greater than lambda. What is wrong with it ? $\endgroup$
    – StarBucK
    Nov 11, 2016 at 16:53
  • $\begingroup$ In that image you sent, $d\sin\theta$ is the difference that is marked out between the arrows. Now, imagine $\theta$ to be very small, so that the two solid lines marking out the distance are nearly vertical. In that case, this distance can be very small, much smaller that $d$. $\endgroup$
    – Aaron
    Nov 11, 2016 at 17:19
  • $\begingroup$ yeaah ok I get it ! My mistake was that I only thought about the path done by the refracted beam and I forgotten to substract it to the incoming beam. In fact another way to answer is to say $sin(\theta)=\frac{d}{\lambda}$ always admit a solution if $\frac{d}{\lambda}<1$, and this solution will correspond to a lower angle than all the other solutions $sin(\theta)=n*\frac{d}{\lambda}$. Thanks ! $\endgroup$
    – StarBucK
    Nov 11, 2016 at 17:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.