Why the reflected angle is the same as incident in x-ray diffraction?

  • #1

Summary:

When deriving Bragg's law for diffraction of x-rays off crystal lattices, it is assumed that the x-rays are reflected with the same angle as the incident one. Why? Aren't x-rays scattered in all directions?

Main Question or Discussion Point

I'm reading about x-ray diffraction in the context of crystal structure determination. Usually this discussion begins with Bragg's law, $$2d\sin\theta=n\lambda,$$ where ##\theta## is the angle of incoming and "reflected" x-rays. This is the bit that bothers me. I understand that the diffraction of x-rays off the electrons is described by Thomson scattering where the electrons are accelerated by the electric field portion of the incoming x-rays. These accelerating electrons then emit EM radiation of their own with the same frequency as the incoming wave. My math is a but rusty, and I don't understand why this re-emitted x-ray should have the same angle and the incoming wave, as if the plane of atoms were like a solid wall and the electrons just bounced off them. I've read in some places that the electrons emit x-rays in all directions.

So basically, I was wondering if someone has a nice conceptual explanation as to why Bragg's theory works? Why does the x-ray bounces off the crystal plane like a ball off a wall?
 

Answers and Replies

  • #2
BvU
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  • #3
Lord Jestocost
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One has to distinguish between "reflection" and "diffraction". In his book “Elements of X-ray Diffraction, Second Edition”, B. D. Cullity writes the following:

"At first glance, the diffraction of x-rays by crystals and the reflection of visible light by mirrors appear very similar, since in both phenomena the angle of incidence is equal to the angle of reflection. It seems that we might regard the planes of atoms as little mirrors which "reflect" the x-rays. Diffraction and reflection, however, differ fundamentally in at least three aspects:
(1) The diffracted beam from a crystal is built up of rays scattered by all the atoms of the crystal which lie in the path of the incident beam. The reflection of visible light takes place in a thin surface layer only.
(2) The diffraction of monochromatic x-rays takes place only at those particular angles of incidence which satisfy the Bragg law. The reflection of visible light takes place at any angle of incidence.
(3) The reflection of visible light by a good mirror is almost 100 percent efficient. The intensity of a diffracted x-ray beam is extremely small compared to that of the incident beam.

Despite these differences, we often speak of "reflecting planes" and "reflected beams" when we really mean diffracting planes and diffracted beams. This is common usage and, from now on, we will frequently use these terms without quotation marks but with the tacit understanding that we really mean diffraction and not reflection.*

To sum up, diffraction is essentially a scattering phenomenon in which a large number of atoms cooperate. Since the atoms are arranged periodically on a lattice, the rays scattered by them have definite phase relations between them; these phase relations are such that destructive interference occurs in most directions of scattering, but in a few directions constructive interference takes place and diffracted beams are formed. The two essentials are a wave motion capable of interference (x-rays) and a set of periodically arranged scattering centers (the atoms of a crystal).

* For the sake of completeness, it should be mentioned that x-rays can be totally reflected by a solid surface, just like visible light by a mirror, but only at very small angles of incidence (below about one degree). This phenomenon is of little practical importance in x-ray metallography and need not concern us further."
 
  • #4
jambaugh
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I think that you start by identifying a specific incident direction and the direction of the diffracted x-rays corresponding to a peak. You then choose the frame where you draw your planes normal to the bisector of the incident and diffracted directions. These must be planes containing regular arrays of atoms by the nature of the constructive interference.
 
  • #5
Thank you to everyone for your replies. Sorry it took me a while to check back in. I understand the interference part very well, and that we get a diffraction pattern at very specific angles because of how the thousands of planes lead to destructive interference unless Bragg's condition is satisfied.

I just didn't understand why the reflected angle was the same as the incident angle, since the x-rays are not bouncing off a "mirror" and are emitted in all directions. So I was asking myself, why not have an incident ray at 30 degrees and a reflected one at 63 that lead to the path difference between both rays to be an exact multiple of the x-ray wavelength.

Your insight jambaugh has cleared this up for me. Thank you!!! This was bugging me for a while.
 

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