# The Cause of X-ray crystal Diffraction

by fslab
Tags: bragg's law, crystallography, diffraction, scattering, xray
 P: 4 Hello, This is a multi-part question that stems from understanding how X-ray diffraction occurs in crystals (eg. protein crystals). 1. Diffraction occurs when Bragg's Law is satisfied, but I'm sure the waves aren't actually being reflected. The x-ray's are scattering. What type of scattering is it? (I think it is Rutherford scattering or Rayleigh scattering) 2. When this scattering occurs, is the electron absorbing and re-emitting the x-ray? If so, is this the type of absorption that occurs as electrons shift energy shells, or is the absorption and re-emission a result of electrons moving in the electron cloud (eg. oscillating)? 3. I assume that the electrons are re-emitting x-rays and those x-rays that are in phase result in diffraction. Electrons can emit x-rays in all directions, why is it that diffraction is strongest in the direction of the x-ray source/ x-ray generator. That is, why is the strongest diffraction around the beam stop on the detector? Why don't you get low resolution, high intensity diffraction anywhere else (eg. 90 degrees to the x-ray source)?
 P: 3 Think I can help with some of your questions, I'm sure I will be corrected where I'm wrong. 1. In regular XRD-measurements, the detector is usually positioned so that only elastically scattered photons are detected (Bragg-Brentano geometry). The elastic scattering process is called Thomson scattering. 2. When electrons from the beam collide with the solid, they are slowed down and this loss in energy results in radiation called Bremsstrahlung radiation. In addition some of the electrons with high energy will knock out inner shell electrons in atoms, which will cause an ionization process in which an X-ray photon with a characteristic energy of the material is emitted. 3. The x-rays that are in phase will interfere with each other and cause an interference pattern (this is basically bragg's law). I think the question you are asking can be answered in terms of an interference experiment, where you send light through a grid and see where the interference maxima occur.
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 Quote by fslab Hello, This is a multi-part question that stems from understanding how X-ray diffraction occurs in crystals (eg. protein crystals). 1. Diffraction occurs when Bragg's Law is satisfied, but I'm sure the waves aren't actually being reflected. The x-ray's are scattering. What type of scattering is it? (I think it is Rutherford scattering or Rayleigh scattering)
It is elastic scattering, or something very close to elastic. The scattered photon doesn't lose significant amounts of energy. If the photon did change energy, the scattered wave would change wavelength. The scattered wave would probably change phase. The scattered wave from an atom couldn't interfere with scattered waves from other atoms.
 Quote by fslab 2. When this scattering occurs, is the electron absorbing and re-emitting the x-ray? If so, is this the type of absorption that occurs as electrons shift energy shells, or is the absorption and re-emission a result of electrons moving in the electron cloud (eg. oscillating)?
No. The electron can't be absorbing and remitting the x-ray energy. If it did, the result "scattered" wave from one atom can't interfere with the scattered wave from other atoms.

There are x-rays that are created by the absorption and emission x-ray energy. This is called x-ray fluorescence. X-ray fluorescence occurs at a longer wavelength then the incident x-rays. Because the phase is randomized, there is no diffraction pattern associated with x-ray fluorescence. X-ray fluorescence is important in chemical analysis, but not in crystallographic analysis.

One caveat. Quantum mechanical analysis, which you probably won't need for a while, includes both real and virtual photons.

I have been talking about real photons. There is a concept that scattering occurs by absorption and re-emission of virtual photons. With virtual photons, energy is created and destroyed on a time scale where the Heisenberg uncertainty principle. However, the energy that is created and destroyed can't be measured.

One can say that the electrons in the atoms of the crystal, when the incident xray wave hits them, exchanges virtual photons. This can confuse discussions with experts between field. Virtual photon exchange is strictly an internal process within the atom, and doesn't directly effect the diffraction pattern.

The real photons contributing to the diffraction pattern do not change energy. Real photons that change energy can't contribute to the diffraction pattern.

Virtual photons are unimportant for crystal analysis. Quantum mechanics is generally unnecessary for determining the structure of a crystal.

 Quote by fslab 3. I assume that the electrons are re-emitting x-rays and those x-rays that are in phase result in diffraction. Electrons can emit x-rays in all directions, why is it that diffraction is strongest in the direction of the x-ray source/ x-ray generator. That is, why is the strongest diffraction around the beam stop on the detector? Why don't you get low resolution, high intensity diffraction anywhere else (eg. 90 degrees to the x-ray source)?
It is a little unclear what you mean by the diffraction is strongest in the direction of the x-ray source and generator. A diffraction pattern can be seen over a broad range of angles. I think that a high resolution diffraction pattern, if not high intensity, can even be seen at 90 degrees. I conjecture that you are talking about the extinction component.

The extinction component is the "shadow" of the atoms. The photons that are scattered can't proceed in the direction of the source. For the limit of measurements done on a classical scale, the extinction profile is the shadow of the atoms. Ray optics is usually used to analyze the shadow of large objects. However, fine angular measurements would show that the shadow contains a lot of structure that can't be determined by ray optics.

The extinction component is a diffraction phenomenon. The conventional shadow of large objects is merely the first dark band of the diffraction pattern. The extinction component is caused by the interference between the scattered light and the incident radiation.

There are three components of energy flow in a scattering experiment. They are the incident flux, the scattered flux, and the extinction flux. The incident flux comes from the source. Diffraction patterns are associated with the aperture. The scattered flux comes only from the atoms of the object. Interference between waves scattered from different atoms dominates the diffraction pattern of the scattered flux. The extinction flux is the shadow of the atoms. The diffraction pattern of the scattered flux is dominated by interference between the waves scattered from the atoms and the incident wave.

Most of what you learn about in classes on diffraction probably concerns the scattered radiation. Most diffraction experiments are done at extreme scattering angles. I think that a lot of experiments are done at 90 degrees. However, at very small scattering angles the extinction flux is important. Since it is hard to analyze, most diffraction experiments include a baffle to block off the extinction flux.

The baffle at small angles has two purposes. First, it blocks off the shadow of the atoms. The shadow of the atoms has some information on the source, which crystallographers don't care about. Second, the baffle blocks off the incident radiation. The incident radiation contains information about the source and contains no information about the crystal. Therefore, the baffle can be considered a "filter" that screens out information about the source of xray radiation.

Beer's Law comes from the extinction flux. The incident flux minus the extinction flux is the transmitted beam. It decreases due to both absorption of radiation and scattering. It may be useful to consider the experimental differences between using Beer's Law and using the law of diffraction.

An extinction coefficient is determined by the extinction flux. Beer's Law is the exponential decay of energy in a beam with time. The exponential drop in intensity of xray radiation from one end of the crystal can be described on two scales of angular resolution. One one hand, the extinction coefficient is determined by the interference between incident radiation and scattered radiation. When measuring the extinction coefficient, one doesn't use a baffle.

The scattered radiation may peak at small scattering angles. However, small scattering angles is also where the shadow lies. Actual experiments involve a compromise. Angles should be small enough to get intense scattering but large enough to avoid the shadow.

Note all the incident photons are scattered by the crystal. If the sample is thin and the xray photons very high in energy, most photons pass through the crystal without scattering. The atoms are small so there is plenty of space between the atoms. The value of the cross section is small, and the spacing between atoms high. Most photons just pass through. Hence, the total amount of light passing through the sample at small angles will mostly be due to incident flux. However, there will be structure to the diffraction of the crystal added by the shadow of the atoms. If all you want is the structure of the crystal, the transmitted beam is just unwanted background.

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## The Cause of X-ray crystal Diffraction

 Quote by Darwin123 It is elastic scattering, or something very close to elastic. There are x-rays that are created by the absorption and emission x-ray energy. This is called x-ray fluorescence.
How does elastic scatter occur? I assumed elastic scatter is a result of energy being absorbed by an electron, entering a higher energy shell, then falling, in a single step, back to its original shell. In this manner, I assumed the x-ray released would have the same energy as the one absorbed.

 Quote by Darwin123 It is a little unclear what you mean by the diffraction is strongest in the direction of the x-ray source and generator.
Sorry if this was unclear. What I meant is, in normal xray diffraction experiments, the x-ray source, crystal (which I want to determine the structure of) and diffraction detector are positioned in a linear path. The x-ray generator fires x-rays at/through the crystal and the diffraction is read perpendicular to the x-ray beam.

My problem comes from a weak understanding of the scatter that occurs at the crystal. I presume two atoms that scatter x-rays, will do so in a sphere. The points of constructive interference will then cause diffraction, which can occur in all directions. Why is it that the optimal direction to measure diffraction is in line with the xray beam?

Another post rationalized this to a double slit experiment, where the strongest interference is in the direction of the incident beam. Does this hold true for atoms scattering x-rays despite the fact that the scattered waves propagate in all directions?
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 Quote by fslab How does elastic scatter occur? I assumed elastic scatter is a result of energy being absorbed by an electron, entering a higher energy shell, then falling, in a single step, back to its original shell. In this manner, I assumed the x-ray released would have the same energy as the one absorbed. Sorry if this was unclear. What I meant is, in normal xray diffraction experiments, the x-ray source, crystal (which I want to determine the structure of) and diffraction detector are positioned in a linear path. The x-ray generator fires x-rays at/through the crystal and the diffraction is read perpendicular to the x-ray beam.
Perpendicular to the x-ray beam is not the same as in-line with the x-ray beam. Perpendicular to the x-ray beam can be considered the very opposite of in-line with the x-ray beam.

My understanding was that very often diffraction is read perpendicular or at an angle to the x-ray beam. However, this isn't always the case.

Sometimes the diffraction pattern is read at small angles to the x-ray beam. However, the angles where the diffraction is read must be large enough to exclude the shadow of the crystal. You seem to think that this is the only way that x-ray diffraction patterns are read.

The scattering of x-rays is strongest in the forward direction. Therefore, most of the scattered radiation is going to be at small angles to the x-ray beam. Therefore, the brightest diffraction peaks are going to be at small angles to the x-ray beam.

One can read the diffraction patterns at right angles to the x-ray beam. However, one then has to deal with some polarization effects. No energy is scattered in the direction of the polarization beam. So the polarization of the x-rays scattered at right angles is always "up".

 Quote by fslab My problem comes from a weak understanding of the scatter that occurs at the crystal. I presume two atoms that scatter x-rays, will do so in a sphere. The points of constructive interference will then cause diffraction, which can occur in all directions. Why is it that the optimal direction to measure diffraction is in line with the xray beam? Another post rationalized this to a double slit experiment, where the strongest interference is in the direction of the incident beam. Does this hold true for atoms scattering x-rays despite the fact that the scattered waves propagate in all directions?
You are making a mistake in what "spherically scattered" means. "Spherically scattered is not the same as "isotropically scattered".

Both light and x-rays elastically scatter with preferred directions. They usually scatter by inducing a dipole in the particle. The dipole has a preferred direction with regards to intensity of scatter.

The wave fronts may be close spherical. However, a spherical wave front does not mean that the intensity is the same at all angle.
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 Quote by fslab My problem comes from a weak understanding of the scatter that occurs at the crystal. I presume two atoms that scatter x-rays, will do so in a sphere. The points of constructive interference will then cause diffraction, which can occur in all directions. Why is it that the optimal direction to measure diffraction is in line with the xray beam? Another post rationalized this to a double slit experiment, where the strongest interference is in the direction of the incident beam. Does this hold true for atoms scattering x-rays despite the fact that the scattered waves propagate in all directions?
You don't understand how waves scatter from a single particle. The wave vectors point in all directions. However, the wave vector is not the same as the energy flux. The energy flux does not the same in all directions.

Elastic scattering has a preferred direction. In the limit of immobile particles much smaller than the wavelength of the wave, elastic scattering is Rayleigh scattering. Larger particles may cause Mie scattering. There are all sorts of elastic scattering. However, I don’t think there is any elastic scattering where the energy flux is isotropic. For most types of elastic scattering, the intensity is weighted in the forward direction.

You are mixing up intensity and phase. For “spherical waves”, the phase varies only with distance from the source. For “isotropic scattering”, the amplitude of the wave varies with the distance from the source. Elastic scattering usually results in a spherical wave, meaning the wave fronts are spherical (phase). Elastic scattering is almost never isotropic because of conservation of momentum.

Here are a few links concerning different types of elastic scattering. Note that the intensity is always biased in certain directions. However, the phase is implicitly a function of distance. So the wave fronts are spherical.

http://en.wikipedia.org/wiki/Rayleigh_scattering
“The intensity I of light scattered by a single small particle from a beam of unpolarized light of wavelength λ and intensity I0 is given by:

where R is the distance to the particle, θ is the scattering angle, n is the refractive index of the particle, and d is the diameter of the particle. “

http://en.wikipedia.org/wiki/Mie_scattering
“Rayleigh scattering describes the elastic scattering of light by spheres which are much smaller than the wavelength of light. The intensity, I, of the scattered radiation is given by

where I0 is the light intensity before the interaction with the particle, R is the distance between the particle and the observer, θ is the scattering angle, n is the refractive index of the particle, and d is the diameter of the particle.”

This reference uses the phrase “spherical wave.” However, it is clear that they are not talking about isotropic scattering. The energy has a preferred direction. Most of it is going forward.
http://farside.ph.utexas.edu/teachin...s/node103.html
“Consider a plane electromagnetic wave incident on a spherical obstacle. In general, the wave is scattered, to some extent, by the obstacle. Thus, far away from the sphere the electromagnetic fields can be expressed as the sum of a plane wave and a set of outgoing spherical waves.”
 P: 4 Thanks, greatly appreciate your help.
P: 3,269
 Quote by Darwin123 Elastic scattering is almost never isotropic because of conservation of momentum.
I don't think that's the reason, at least for single particle scattering. E.g. Rayleigh scattering of longitudinal sound waves from small particles should be isotropic. The variation with theta in the case of electromagnetic waves is due to the vector character of the electromagnetic field: e.g. in the dipole approximation (long wavelength limit) there is no scattering in the direction of the induced dipole moment which is at 90 deg with direction of incidence. For unpolarized radiation this makes up for the minimum at theta=90 deg in the formula cited from wikipedia in the long wavelength limit. This may be looked at as a simple example of an atomic form factor.
 Emeritus Sci Advisor PF Gold P: 11,154 It's possible the OP's confusion about the diffracted intensity arises because s/he primarily works with proteins. For something with a lattice parameter on the order of 100A, the reflections happen at very small angles (ie, very close to the path of the direct transmitted beam).
 Sci Advisor P: 3,269 Yes, Gokul, that is a possibility. So the largest scattering angles are due to reflection from Bragg planes with a small distance. These have large Miller indices and the density of scatterers per area is very low, so is the intensity of the scattered wave. This is because the number of scatterers per area of the Miller plane divided by the minimal distance of the planes is a constant - the volume density of the scatterers. This is especially accute in the case of protein crystallography, where for reasonable Miller indices only small scattering angles arise.
P: 4
 Quote by Gokul43201 It's possible the OP's confusion about the diffracted intensity arises because s/he primarily works with proteins.
I do indeed. I'm a protein crystallographer/biochemist with very little physics background. In trying to understand how diffraction occurs, I came across the questions above.

While I now have a vague understanding of scatter and prominent direction of diffraction, I'm now wondering how x-rays induce a dipole in the electron density/molecule to then re-radiate x-rays.
 Sci Advisor P: 3,269 For X-rays it is not only the dipole moment, but for Rayleigh scattering it is. The point is that an electromagnetic field generally couples to matter via the product of the dipole moment operator d=e r and the electric field vector E. If the wavelength of the electric field is much larger than the dimension of the molecules (as is the case for visible light), then the dependence of E on r can be neglected. However with X-rays, the wavelength is comparable to atomic dimensions, hence one has to take into account that E varies as E exp(ikr). Then, in combination with the dipole operator, also dependence on higher multipole moments r^2, r^3 etc becomes important. That is the origin of the structure factors which are just the corresponding Fourier transforms of $\rho(r)r$.
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 Quote by fslab I do indeed. I'm a protein crystallographer/biochemist with very little physics background. In trying to understand how diffraction occurs, I came across the questions above. While I now have a vague understanding of scatter and prominent direction of diffraction, I'm now wondering how x-rays induce a dipole in the electron density/molecule to then re-radiate x-rays.
If xray wavelength is generally larger than the diameter of the atom, an explanation using classical electromagnetic theory is sufficient. The xray wave consists of an oscillating electric field and an oscillating magnetic field at right angles to each other and at 90° in phase. At the intensities of a typical crystallographic experiment, the magnetic field is negligible. The atom has an electric polarizability which is basically phenomenological. The electric field pulls the electrons in one direction and pushes the nucleus in the other. The separation between the centers of charge of electron and nucleus causes the dipole.

If the xray wavelength is less than the diameter of the atom, then you may have to treat each electron separately. The electron oscillates. More detailed calculations may be necessary.

If you are talking gamma rays, then maybe you may need to use quantum electrodynamics to describe the scattering.

Some classical analogies may be useful. Basically, an xray is like a light wave. One can adapt light wave optics to the xray. The atoms is like a spherical lump of dialectric. Or like a radio wave speeded up. The atom is just like a radio antennae. Quantum mechanics may changed the strength of the interaction somewhat, but the visual picture of these analogies is basically good.

The quantum mechanical details won't help in most crystallography applications. Although the quantum mechanics may be more accurate in some ways, the extra complexity is worth it. Classical electromagnetic theory may help a little. Classical electromagnetic theory can get too complicated, too. However, it sounds like you may be ready for it.
 P: 3 Dear all i am struggling on a question regarding powder XRD. Can any body please help me on this. Question: I have lambda(X-ray wavelength used) and 2theta(100% intense peak) value of a crystal compound; with that data can any one calculate the "d(distance between two planes)" value of that crystal using braggs law(i.e. nλ = 2 d sinΘ) theoritically. If so please explain me with an example. Note: What is the n value we need to apply in that formula.
 Sci Advisor P: 3,269 n is an integer number. For any value of n, you will get a corresponding Bragg peak.
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 Quote by phanivaka Dear all i am struggling on a question regarding powder XRD. Can any body please help me on this. Question: I have lambda(X-ray wavelength used) and 2theta(100% intense peak) value of a crystal compound; with that data can any one calculate the "d(distance between two planes)" value of that crystal using braggs law(i.e. nλ = 2 d sinΘ) theoritically. If so please explain me with an example. Note: What is the n value we need to apply in that formula.
Crystallographers usually solve the inverse problem. They analyze the diffraction pattern to find the structure of the crystal.

The crystallographer finds values of "n" which are consistent with both the Bragg law and the diffraction pattern being formed. From the set of n which satisfy the Bragg law, and explain the diffraction pattern, the crystallographer figures out the structure of the crystal.

There is a uniqueness problem here. Sometimes, there is more than one model consistent with a diffraction pattern. Sometimes one can discard some of these models using data other than diffraction patterns. However, sometimes even the data from other experiments are not sufficient to select a unique model. These and other fun complications have to be dealt with using higher mathematics.

The crystallographer has references with sets of n calculated for different crystal structures. If diffraction peaks are missing, this may mean there is a greater crystal structure overlying the crystal structure that are not in the references.

The integers "n", are usually the input data from a diffraction experiment. The user tries to find a series of distances, "d", consistent with the integers "n" that are measured. The set of "d" is what I would call a crystallographic model.

Diffraction patterns usually require a number of integers: n, m, ... The one dimensional case is where all the planes of reflection are parallel. Therefore, the crystallographer basically needs only one series of "n" to figure out the spacing between the planes. Two and three dimensional crystals require more than one series of "n".

There were two crystallographic problems that I worked on before I moved out of the field of crystallography. In both cases, the "n" was part of the experimental results. A hypothetical model was developed to explain the "n". I worked with a "simple" problem involving artificial multilayer structures. Although detected peaks were consistent with the desired "d", the intensity of the peaks were less that calculated and desired. By looking at the intensities as a function of "n", I was able to show that there were unexpected bumps on the substrate that the multilayer structure was deposited on. I also worked with a natural beryl crystal. The structure of beryl is known. However, in my data some of the peaks were missing. Therefore, there were "n" with no diffraction peaks. I determined there was a cesium atom in some of the gaps of crystal structure.

There were uniqueness problems associated with both problems. In the case of the beryl, it was a choice between four water molecules in for interstitial points or one cesium atoms at one interstitial point. It was probably water molecules not cesium atoms. I damaged the beryl crystal trying to drive out the water. That was the end of my crystallographic career. But enough about me...

The designer of gratings often solves the problem that you are asking about. He knows the spacing between grooves that he wanted. He carves a series of grooves on a matrix. Then, he does a diffraction experiment. He checks the spacing by seeing if the visible peaks are consistent with the spacing. He tries to find spacing, "d", consistent with all the "n" that he can detect. Generally, he tries to make the shape of each grove in such a way that all but one diffraction peak vanishes. Therefore, he looks for what values of "n" don't have a diffraction peak for the spacing that the designer chose.
 P: 3 Hi DrDu Thanks for your response.

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