Why X-ray scattering in crystalline solids is elastic?

[amjad-sh]

Hello,
I am wondering why in all X-ray diffraction experiments used to probe or know the crystal structure of the solid they assume that the scattering process is elastic, e.g, if an X-ray with wave vector $k\vec{n}$ is incident on a sample, it will diffract with a wave vector $k\vec{n}'$ of same magnitude but different direction.

Bragg's formula is based on the assumption that the scattering of X-ray is elastic, and this formula, I guess, is used to unpack the structure of all crystalline solids. Therefore, I can deduce from this that X-ray scattering in all crystalline solids is elastic, because otherwise they wouldn't use Bragg's formula to analyze their structure.

Is the reason behind this that the potential $V(\vec{r})$ that describes the interaction between the X-rays and any crystalline solid is very weak? and if so, why is it so weak? In some references they use Fermi's golden rule to derive the Laue condition which is equivalent to Bragg's formula, and Fermi's golden rule can't be reliable unless the potential $V(\vec{r})$ is assumed very small, because it is a first order approximation.

Thanks.

 

[anuttarasammyak]

When inelastic we need system to catch energy loss. X-ray diffraction is collision with photon and electron. I assume such a energy loss system is not so much active.
We may assume that Bragg’s formula is applied for not softened photon which still has maximum specter power after diffraction.

 

[amjad-sh]

When inelastic we need system to catch energy loss. X-ray diffraction is collision with photon and electron. I assume such a energy loss system is not so much active.
We may assume that Bragg’s formula is applied for not softened photon which still has maximum specter power after diffraction.

You mean that all types of photon-electron collisions are assumed elastic? because the energy lost from the photon would be negligible w.r.t to the initial energy of the photon? or it applies only for high-energy photons such as X-rays?

 

[hutchphd]

I believe you are missing an important point. It is only the elastically scattered x-rays which will be concentrated according to the Laue criteria. These show up as bright spots whose position implies the lattice type. There is some inelastic scattering (I have no idea how much) but it will be over a range of energies (an hence wavelengths) and be much more diffuse...it won't show up. So as long as there is some elastic scattering the technique works

 

[amjad-sh]

I believe you are missing an important point. It is only the elastically scattered x-rays which will be concentrated according to the Laue criteria. These show up as bright spots whose position implies the lattice type. There is some inelastic scattering (I have no idea how much) but it will be over a range of energies (an hence wavelengths) and be much more diffuse...it won't show up. So as long as there is some elastic scattering the technique works

According to Laue condition, unless $\vec{k}-\vec{k}^{'}=\vec{K}$, where $\vec{K}$ is a reciprocal lattice vector, no constructive interference can happen and hence no high intensity spots can be detected at the direction of $\vec{k}^{'}$.
However, Laue condition doesn't imply that $|\vec{k}|$ needs to be equal to $|\vec{k}^{'}|$. For Laue condition to be equivalent to Bragg's law, the assumption that $|\vec{k}|=|\vec{k}^{'}|$ needs to be imposed.

I believe that the scattered X-rays in bulk solids are not elastic in general, but they are regarded as elastic. why? because according to what I read in "The Oxford solid-state basics" written by professor Steven Simon, just a small amount of energy is lost from the photon scattered from a crystal. But because the amount is very tiny the scattering process is regarded as elastic.

I think that observing high intensity spots on the screen has nothing to do with the scattering process being elastic or not ( because as I said above it is in fact not elastic but regarded approximately as elastic),it however depends on satisfying the Laue condition, which is $\vec{k}-\vec{k}^{'}=\vec{K}$, and for this condition to be satisfied we need to choose the right magnitude for the wave vector $\vec{k}$ and the right incident angle $\theta$.

In Steven Simon's book, he explains why energy lost from scattered photon is tiny by saying that there is a maximum energy $\hbar \omega_{max}$ that a phonon can absorb, the maximum change in crystal momentum $\vec{k}$ that can occur, $\dfrac{\hbar \omega_{max}}{c}$ is tiny.

However I didn't understand his explanation very well.

 

[hutchphd]

This is a confusing but important point. Any change in momentum by $\vec K$ (a reciprocal lattice vector) involves resonant scattering by the entire sample. The entire crystal absorbs the recoil energy and momentum and it is effectively an elastic process. So photon energy is effectively conserved although the actual momentum can change by $\vec K$ (this is sometimes called "conservation of crystal momentum" as stated just involves coherent elastic scattering by the entire lattice which has a very very large mass).
The fact that the lattice is not completely rigid does show up in the scattering but mostly to make the diffraction very slightly fuzzy.