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NE-CAT guides · fluorescence from the sample

X-ray fluorescence at NE-CAT

A protein crystal is nearly all carbon, nitrogen, oxygen and hydrogen, and none of those shows up in the fluorescence detector. What does show up is the handful of heavier atoms: the selenium grown into the protein to solve its structure, a metal the protein holds, or a heavy atom soaked in. Each gives off X-rays at energies of its own once the beam is above its absorption edge. This guide explains where that fluorescence comes from, which atoms give it, what it is used for at NE-CAT, and where the detector has to sit to see it clearly.

Each section has a plain explanation and a figure you can adjust. The equations are folded away under the figures. Open any of them, or all of them with the button at the top, if you want the detail.

Looking down on a protein crystal in the beam. The gray dots stand for the carbon, nitrogen and oxygen that make up nearly all of it, and the green dots are selenium atoms. Most X-rays in the beam pass straight through. Now and then one is absorbed by a selenium atom, which flashes and sends out an X-ray of its own, in a direction that has nothing to do with the beam's. The few that happen to head for the fluorescence detector are counted. They pile up at selenium's own two energies, below the energy of the beam. The drawing has far too few atoms, and far too many of them are selenium.

The Settings menu at the top chooses 24-ID-C or 24-ID-E, before or after the APS Upgrade. The full table of values, with their sources and the assumptions behind them, is in the guide How NE-CAT works.

An atom absorbs one X-ray and gives back another

The electrons in an atom sit in shells. The two in the innermost shell, the K shell, are held the most tightly. An X-ray can knock one of them out of the atom, but only if it carries more energy than it takes to free that electron. That threshold is the atom's K absorption edge, and every element has its own.

Losing the electron leaves a hole in the K shell. In about a millionth of a billionth of a second an electron from a shell further out drops in to fill it. The energy it gives up is the difference between the two shells, and that difference is fixed for each element.

The atom gets rid of that energy in one of two ways. It can send it out as an X-ray: this is fluorescence. Or it can hand it to another of its electrons, which leaves the atom instead. That one is called an Auger electron, and the detector never sees it. Light atoms mostly take the second route, and heavy atoms mostly the first.

The fluorescence X-ray always has less energy than the X-ray that started it, and its energy does not depend on the beam's. An electron dropping in from the L shell gives the line called Kα. One from the M shell gives Kβ, a little higher in energy and several times weaker. Heavy atoms such as platinum and mercury do the same thing one shell further out. Within the energy range of the beamline it is their L shell that is emptied, and the lines are called Lα and Lβ.

The atom keeps no memory of where the first X-ray came from, so fluorescence goes out equally in all directions.

Element

Beam energy 12.66 keV

K edge

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Kα line

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Kβ line

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X-rays per 100 atoms hit

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Left: one atom, with its three inner shells of electrons. An X-ray from the beam knocks out a K electron, an electron from the L shell drops into the hole, and the atom gets rid of the energy either as an X-ray or by throwing out another electron. The row of squares keeps score. Right: the same events as energy levels. The number under each shell is the energy that holds its electrons in the atom. The blue arrow is the beam X-ray, which has to reach the dashed line to free the electron. The green arrows are the two fluorescence lines. Choose a lighter element and fewer atoms answer with an X-ray. Take the beam energy below the edge and nothing happens at all. The shells are drawn to look clear, not to scale, and the outgoing X-ray is sent left or right only to keep it in the frame.

The math, if you want it

Equation 1.1The edge and the lines

E_{\text{beam}}>E_K\qquad\qquad E_{K\alpha}=E_K-E_L\qquad\qquad E_{K\beta}=E_K-E_M

$E_{\text{beam}}$: Energy of the X-ray that arrives.
$E_K,\ E_L,\ E_M$: How tightly an electron is held in the K, L and M shells: the energy it takes to remove it from the atom.
$E_{K\alpha},\ E_{K\beta}$: Energies of the two fluorescence lines.

What it saysThe first condition is the absorption edge. Below it the X-ray cannot free a K electron, and no fluorescence follows. The other two are bookkeeping: the X-ray given off carries the difference between the shell the electron leaves and the shell it lands in. Both lines lie below the edge, which is why fluorescence can always be told apart from the beam that caused it.

In numbersFor selenium the K edge is at 12.658 keV. The L and M shells are held by about 1.45 and 0.16 keV, so Kα is at 11.21 keV and Kβ at 12.50 keV. Each shell is really a few levels some tens of electron-volts apart, which is why tables list Kα₁ and Kα₂ separately.

Equation 1.2How often the atom answers with an X-ray

\omega_K\approx\frac{Z^{4}}{Z^{4}+a}\qquad\qquad a\approx 8.5\times10^{5}

$\omega_K$: The fluorescence yield: of the atoms left with a hole in the K shell, the fraction that give off an X-ray. The rest give off an Auger electron.
$Z$: The atomic number.
$a$: A number chosen to fit measured yields. This value fits the elements from sulfur to bromine to within about 10%.

What it saysThe two ways of filling the hole compete. Giving off an X-ray becomes rapidly more likely as the nucleus gets heavier, and the Auger route does not, so the yield climbs steeply with atomic number. This is the first reason the light atoms of a protein are invisible to the fluorescence detector.

In numbersTabulated yields: carbon 0.3%, oxygen 0.7%, sulfur 8%, calcium 17%, iron 35%, zinc 49%, selenium 60%, bromine 63%. The rule gives 7%, 16%, 35%, 49%, 61% and 64% for the last six.

Equation 1.3Fluorescence goes out in every direction

\frac{dN_{\text{fluor}}}{d\Omega}=\frac{N_{\text{fluor}}}{4\pi}

$N_{\text{fluor}}$: The number of fluorescence X-rays the atoms give off.
$d\Omega$: A small patch of directions. The whole sphere is 4π.

What it saysThe atom gives off its X-ray with no memory of the direction or the polarization of the one it absorbed, so the fluorescence is spread evenly over the whole sphere. At a given distance a detector collects the same share of it wherever it sits. This matters later in the guide, where the scatter turns out to depend strongly on direction.

Which atoms in a protein crystal fluoresce

A protein crystal is mostly carbon, nitrogen, oxygen and hydrogen, in the protein and in the water that fills about half of it. These atoms do absorb the beam now and then, but they almost never answer with an X-ray, and the few X-rays they do give off have so little energy that the sample and the air stop them. The fluorescence detector does not see them at all.

Sulfur, in methionine and cysteine, and phosphorus, in DNA and RNA, do a little better. Their lines are near 2 keV. Most of that is lost on the way out of the sample and through the air, and what arrives is at the very bottom of the detector's range. Atom for atom, sulfur sends the detector less than a thousandth of what selenium does. A 50 kDa protein has fifteen or twenty sulfur atoms, so all of its sulfur together amounts to about a hundredth of one selenium atom.

What the detector does see is the handful of heavier atoms.

Selenium. Selenomethionine is methionine with its sulfur replaced by selenium. A protein grown with it carries one selenium atom for roughly every 5 to 10 kDa. It is by far the most common label for phasing, and so the most common reason to measure fluorescence.

Metals the protein holds. Iron in hemes and iron–sulfur clusters, zinc, copper, manganese, nickel, cobalt and molybdenum. There may be only one such atom in a large protein.

Heavy atoms added on purpose. Mercury, platinum, gold, lead, osmium, the lanthanides, bromide and others, soaked into the crystal for phasing.

Atoms that came along with the solution. Zinc, calcium, bromide, arsenic from cacodylate buffer, or nickel left over from purification.

Fluorescence says that an element is in the beam. It does not say where the atoms are, or whether they sit in the same place in every copy of the protein. A metal in the liquid around the crystal fluoresces just as well as one bound in an active site.

Beam energy 12.66 keV

Air on the way to the sensor 15 mm

Sample crossed on the way out 50 µm

Sulfur reaching the sensor, against selenium

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Iron reaching the sensor, against selenium

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Edge above the beam

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Given off by the atomReaching the sensorBlue: the protein and its solution. Green: heavier atoms.

How much fluorescence one atom of each element sends toward the detector, compared with one selenium atom just above its edge. The scale is logarithmic: each line is ten times less than the one above, and a cross means less than a millionth. Carbon, nitrogen and oxygen give off almost nothing, and none of it arrives. Sulfur loses most of the little it gives. Take the beam energy above 13.5 keV and bromine switches on. Every other bar shrinks a little, because an atom absorbs less the further the beam is above its edge. Calculated from tabulated cross-sections (equation 6.1). The detector's window and its efficiency are left out.

The math, if you want it

Equation 2.1What the sample and the air take away

T=e^{-\mu_{\text{sample}}\,t}\;e^{-\mu_{\text{air}}\,d}

$T$: The fraction of the fluorescence X-rays that reach the sensor.
$\mu$: How strongly the sample, or the air, absorbs X-rays of that energy. It rises steeply as the energy falls, roughly as 1/E³.
$t$: How much sample and liquid the X-ray crosses on its way out.
$d$: The length of air between the sample and the sensor.

What it saysLow-energy X-rays do not travel far. This is the second reason the light atoms go unseen: what little fluorescence they give is stopped before it arrives. The detector's own window and its electronics set a further limit at the low end, which is left out here.

In numbersThrough 50 µm of sample and 15 mm of air: carbon and oxygen, nothing. Sulfur Kα at 2.31 keV, 7%. Calcium Kα at 3.69 keV, 50%. Iron Kα at 6.40 keV, 87%. Selenium Kα at 11.21 keV, 97%.

Edges and lines within reach at 24-ID

Both the edge and the lines move up steadily with atomic number, so each element has its own place on the energy scale. For an element to fluoresce, the beam energy has to be above its edge. For an edge scan, the beamline also has to be able to step its energy across that edge.

24-ID-C can be set anywhere in 6–20 keV. That takes in the K edges of the metals from manganese to zinc, and of arsenic, selenium and bromine. It also takes in the L edges of the lanthanides and of the heavy atoms used for soaks, such as platinum, gold and mercury.

Elements whose edges lie below that range still fluoresce when they are in the beam. Calcium, iodine and xenon are common examples. Their lines can be seen, but their edges cannot be scanned here.

The line is always well below the edge: 1.45 keV below for selenium. That gap is what lets the detector separate the fluorescence from the beam's own scattered X-rays, which come later in this guide.

Element

Beam energy 12.66 keV

Edge

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Lines to look for

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Usually there because of

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On this beamline

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Every flag is an absorption edge. K edges are above the line, and the L edges of heavier atoms are below it. Green flags are below the beam energy, so those elements fluoresce. White flags are above it and stay silent. The yellow band is what the beamline can be set to. Choose an element from the menu, or click its flag, to see its lines on the lower scale: they always sit below its edge. The elements shown are the ones met most often in crystallography, not the whole periodic table. For the heavy atoms only the lowest of the three L edges is shown.

Set for 24-ID-C after the upgrade: the beam energy can be set anywhere in 6–20 keV.

The math, if you want it

Equation 3.1Why the lines march up with atomic number

E_{K\alpha}\approx 10.2\ \text{eV}\times(Z-1)^{2}

$Z$: The atomic number.
$10.2\ \text{eV}$: Three quarters of 13.6 eV, the energy that holds the electron in a hydrogen atom. The three quarters comes from the jump between the second shell and the first.
$Z-1$: The charge the falling electron feels: the nucleus, less the one electron still in the K shell.

What it saysMoseley's law. Each step up the periodic table moves the Kα line up by a predictable amount, so the lines of neighboring elements are laid out in order and never coincide. The same holds for the edges, and for the L lines of heavy atoms, with different numbers.

In numbersIron, Z = 26: 6.38 keV (measured 6.40). Zinc, Z = 30: 8.58 keV (8.63). Selenium, Z = 34: 11.11 keV (11.21).

The edge scan: finding the energies for phasing

The most common use of fluorescence at a crystallography beamline is the edge scan. The beam energy is stepped across the absorption edge of one element, about an electron-volt at a time. At each step the detector counts the X-rays in a narrow window around that element's line. Below the edge there is nothing to count. As the beam crosses the edge, the fluorescence switches on.

The fluorescence at each energy is in proportion to how strongly the element absorbs there. That absorption is one of the two numbers that describe how an atom's scattering changes near its edge. It is called f″. The other, f′, is how much the atom's ordinary scattering is reduced, and it can be worked out from the whole f″ curve. A program such as CHOOCH does this in a moment once the scan ends.

The scan is needed because the edge is not in exactly the same place, or the same shape, in every sample. The chemistry around the atom shifts it by a few electron-volts and often puts a sharp peak, called the white line, at the top of the rise. A table can say where the edge of the bare element is. Only a scan says where it is in this crystal.

Two energies come out of the scan. The peak is where f″ is largest. It gives the largest differences between Friedel pairs, and it is the energy for a SAD experiment. The inflection, partway up the rise, is where f′ is most negative. A MAD experiment adds data there, and at a remote energy well above the edge, to get the differences between energies as well.

Edge scans are run on 24-ID-C, where the energy can be changed. 24-ID-E stays at 12.662 keV, just above the selenium edge, so selenium data can be collected there without one.

Beam energy 12.6605 keV

Height of the peak at the top of the edge typical

f″ at this energy

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f′ at this energy

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Peak: f″ largest

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Inflection: f′ lowest

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Top: a selenium edge scan, the fluorescence counted at each one-electron-volt step. Bottom: the two curves worked out from it. f″, in red, has the same shape as the scan. f′, in blue, dips sharply where f″ rises. The buttons move the marker to the energies used for data collection. The second slider changes the height of the peak at the top of the edge, which depends on the chemistry around the selenium. The best energies move with it. This is a model with the sizes typical of a selenomethionine protein, not a measurement.

The math, if you want it

Equation 4.1From absorption to f″

f''(E)=\frac{E\,\sigma_a(E)}{2\,hc\,r_e}

$f''$: The absorbing part of the atom's scattering factor, in electrons.
$\sigma_a$: How strongly one atom absorbs X-rays of energy E. The fluorescence counted in an edge scan is in proportion to the part of this that comes from the K shell.
$hc$: 12.398 keV·Å.
$r_e$: The classical electron radius, 2.818×10⁻⁵ Å.

What it saysAbsorption and f″ are the same thing in different units. So a fluorescence scan, which follows the absorption, gives the shape of f″ across the edge. The software sets the scale by matching the scan to the known values well below and well above the edge.

In numbersSelenium at 12.8 keV absorbs with σ = 2.0×10⁻²⁰ cm², which gives f″ = 3.7 electrons. Just below the edge it is 0.5. At the white line of a selenomethionine protein it is often 6 or more.

Equation 4.2From f″ to f′

f'(E)=\frac{2}{\pi}\,\mathcal{P}\!\!\int_{0}^{\infty}\frac{E'\,f''(E')}{E'^{2}-E^{2}}\,dE'

$f'$: The change in the atom's ordinary scattering near the edge, in electrons. It is negative there.
$\mathcal{P}\!\!\int$: An integral over all energies, taken so that the point E′ = E, where the fraction blows up, cancels from either side.

What it saysThe Kramers–Kronig relation. The f′ at one energy depends on f″ at every energy, with the nearest energies counting most. A sudden rise in f″ produces a sharp dip in f′ at the same place. This is how the f′ curve in the figure is obtained from the scan.

In numbersFor the edge in the figure, f′ falls from about −3 far below the edge to about −10 at the inflection, and is back to −5 by 50 eV above. These are the model's numbers, of the size found for selenomethionine.

Equation 4.3What the two numbers are used for

f=f_0+f'+i\,f''\qquad\qquad \Delta_{\text{Friedel}}\propto 2f''\qquad\qquad \Delta_{\text{energies}}\propto\left|f'_{1}-f'_{2}\right|

$f_0$: The atom's ordinary scattering factor: its number of electrons at small angles, falling at larger ones.
$\Delta_{\text{Friedel}}$: The difference between a reflection and its Friedel mate at one energy. Also called the anomalous or Bijvoet difference.
$\Delta_{\text{energies}}$: The difference in one reflection between two energies. Also called the dispersive difference.

What it saysThe two numbers are used differently. A SAD experiment relies on the Friedel differences alone, so it is run at the peak, where f″ is largest. A MAD experiment adds the differences between energies. The largest of these is between the inflection, where f′ is most negative, and a remote energy where it has recovered.

In numbersFor the edge in the figure: 2f″ is about 12 electrons at the peak and 7 well above the edge. Between the inflection and a point 50 eV above it, f′ changes by about 5 electrons. Selenium has 34 electrons in all, so these are large changes.

Finding out which metal is there

The second use needs no scan. Set the beam well above the edges of interest and record the whole spectrum: counts against the energy of the X-rays arriving. Each element in the beam puts a pair of peaks at its own energies, and a table of line energies names it.

This is the way to check that the selenium went in, to find out which metal a protein picked up when it was made, to tell zinc from nickel or copper in a binding site, or to confirm that a soak worked before spending beam time on a data set.

Some care is needed in reading a spectrum. The Kβ line of one element lies close to the Kα line of the next one up: iron Kβ and cobalt Kα are only 0.13 keV apart, which is about as close as the detector can separate. An element whose edge is above the beam energy stays silent, so the beam has to be set above every edge that matters. And the detector records whatever the beam and its scatter touch. Steel and brass parts near the sample can add iron, chromium, nickel, copper and zinc lines of their own.

Beam energy 12.66 keV

Strongest line of each

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Chosen but silent

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Closest lines of two elements

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The spectrum the fluorescence detector records with the beam at one energy: counts against the energy of each X-ray. The buttons put elements into the sample and take them out. Each one adds a pair of peaks. The two peaks at the right are the beam's own scatter. Try iron and cobalt together. Then switch on bromine and take the beam energy past 13.5 keV. Peak sizes follow from tabulated cross-sections, for one atom of each chosen element in every 150 kDa of sample, with the sample treated as water for the scatter. The weaker lines of the heavy atoms are left out.

Set for 24-ID-C after the upgrade: P = 0.982 at the sample. Detector at the ideal spot, 15 mm away through air, with a 9° sensor cone.

What else reaches the detector: scatter

Fluorescence is not the only thing that comes off the sample. Every electron in it also scatters the beam, and there are far more electrons than atoms of the element being looked for.

Elastic scatter. An X-ray bounces off the electrons in the sample and keeps all its energy. It shows up at the energy of the beam.

Compton scatter. An X-ray bounces off an electron and knocks it away, losing a little energy as it does. It shows up just below the energy of the beam.

The fluorescence detector is a small SDD beside the sample, which measures the energy of every X-ray it receives. It can tell scatter from fluorescence by energy. What it cannot do is count without limit. When too many X-rays arrive, the beam has to be turned down, and from then on every scatter count takes the place of a fluorescence count.

One selenium atom just above its edge gives off as much Kα as the scatter from a large amount of protein and water.

Looking down on the sample again, this time with the scatter drawn in. The beam comes in from the left with its electric field swinging across it. Green dots are fluorescence, which leaves in every direction. Red and blue dots are the two kinds of scatter. Hardly any of them go sideways along the line of the electric field, and that is where the fluorescence detector sits. The dots are drawn in the proportions the calculation gives for each direction. The numbers of each color are not to scale with one another.

SDD angle from the beam, horizontal 90°

Beam energy 12.662 keV

Selenium in the sample 1 Se per 20 kDa

Compton peak

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Scatter vs the ideal spot

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Fluorescence share

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What the fluorescence detector (SDD) records from a selenium-containing crystal: counts against photon energy. The green band is the window around Se Kα that an edge scan counts. Move the detector away from 90° and the two scatter peaks grow while the fluorescence stays the same. The Compton peak also slides in energy. Drop the beam energy below the Se K edge at 12.658 keV and the fluorescence switches off. The third slider sets how much sample there is in the beam for each selenium atom: protein, solvent and the liquid around the crystal together. The peak sizes follow from tabulated cross-sections, with the sample treated as water. Absorption in the sample and in air is left out.

Set for 24-ID-C after the upgrade: P = 0.982 at the sample.

The math, if you want it

Equation 6.1How much fluorescence one atom gives

\frac{d\sigma_{K\alpha}}{d\Omega}=\frac{\sigma_K\;\omega_K\;p_{K\alpha}}{4\pi}

$\sigma_K$: How strongly one atom absorbs X-rays by ejecting an electron from its innermost shell. It is zero below the K edge.
$\omega_K$: The fluorescence yield: the fraction of those events that end with an X-ray being given off.
$p_{K\alpha}$: The fraction of those X-rays that are Kα and not Kβ.
$4\pi$: The fluorescence is spread evenly over all directions (equation 1.3).

What it saysThe strength of the fluorescence from one atom, in the same units as the scatter from one electron. An atom just above its edge is a far stronger source than any single scattering electron. A sample has a great many more electrons than atoms of the element being looked for, and that is the only reason scatter can compete.

In numbersFor selenium at 12.7 keV: σK = 1.78×10⁻²⁰ cm², ωK = 0.60 and pKα = 0.86, which give 7.3×10⁻²² cm² per steradian. That is about 9,200 times what one free electron scatters in its brightest direction.

Equation 6.2The Compton energy shift

E'=\frac{E}{1+\dfrac{E}{mc^{2}}\,(1-\cos 2\theta)}

$E$: Energy of the incoming photon.
$E'$: Energy of the photon after it scatters from an electron that recoils.
$mc^{2}$: The electron's rest energy, 511 keV.
$2\theta$: The scattering angle between the beam and the direction to the detector.

What it saysWhen a photon knocks an electron away, it hands over some energy. The loss grows with the scattering angle: none straight ahead, most straight back. At these energies the loss is a few hundred electron-volts, enough for the detector to see Compton scatter as a separate peak just below the elastic one.

In numbersFor a 12.66 keV beam: 12.57 keV at 45°, 12.35 keV at 90°, 12.06 keV at 180°. Se Kα is at 11.22 keV, well below all of these. Se Kβ at 12.50 keV is not.

A polarized beam leaves one direction dark

The beam's electric field is horizontal and across the beam. It shakes every electron in the sample along that line. A shaken electron gives off light in the shape of a doughnut, with none at all along the line of the shaking. So there is one direction that receives almost no scatter: horizontal, at 90° to the beam. A fluorescence detector placed there still receives its full share of fluorescence. The guide Why the beam is polarized explains why the beam is polarized this way.

Real atoms add a dependence on angle of their own. In elastic scatter the electrons of an atom scatter in step with each other at small angles and fall out of step at large ones. So elastic scatter is very strong toward the beamstop and weak at 90° and behind. Compton scatter is the other way around. An electron can only be knocked free by a hard enough push, and a small-angle scatter is a gentle one, so Compton scatter is weak near the forward direction and strong at 90° and behind. At the ideal spot most of the little scatter that remains is Compton.

Together these mean that moving the detector toward the beamstop is worse than moving it the same amount toward the source.

SDD angle from the beam, horizontal 90°

SDD elevation 0°

Beam polarization P 0.980

Sensor half-angle α 9°

Scatter into the SDD vs the ideal spot

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Fluorescence into the SDD

unchanged

Top view, looking straight down. The sample is the small green crystal in its loop at the center, where the beam crosses the yellow lobes. The beam enters from the left and carries on to the diffraction detector on the right. The orange box is the fluorescence detector (SDD), which the sliders move around the sample. The yellow lobes are the single-electron scatter pattern in the horizontal plane, at the chosen elevation. At the ideal spot the SDD sits in the gap between them. The readout averages over the sensor's acceptance cone. It covers polarization only. The form-factor and Compton effects are in the next figure.

Set for 24-ID-C after the upgrade: P = 0.982 at the sample, taking an assumed 0.98 from the source through the mono of step 6 in the guide Why the beam is polarized.

The math, if you want it

Equation 7.1Thomson scattering by one electron

\frac{d\sigma}{d\Omega}=r_e^{2}\,\bigl|\hat{\boldsymbol{\varepsilon}}\cdot\hat{\boldsymbol{\varepsilon}}'\bigr|^{2}\;\;\longrightarrow\;\;r_e^{2}\Bigl[1-(\hat{\mathbf{n}}\cdot\hat{\boldsymbol{\varepsilon}})^{2}\Bigr]=r_e^{2}\sin^{2}\psi

$d\sigma/d\Omega$: The differential cross-section: how strongly one electron scatters into a given direction.
$r_e$: The classical electron radius, $e^2/4\pi\varepsilon_0mc^2$ = 2.818×10⁻¹⁵ m.
$\hat{\boldsymbol{\varepsilon}},\ \hat{\boldsymbol{\varepsilon}}'$: Polarization directions of the incoming and the scattered X-ray.
$\hat{\mathbf{n}}$: Direction from the sample to the detector.
$\psi$: Angle between that direction and the incoming polarization.

What it saysThe beam's field shakes the electron along ε, and equation 2.2 in the guide Why the beam is polarized gives what it radiates. The arrow marks the step of adding up both possible polarizations of the scattered X-ray, since the detector does not care which it receives. The result is the same sin²ψ doughnut, now with its hole along the beam's polarization.

Equation 7.2Where the detector is

\sin^{2}\psi=1-\cos^{2}\!\phi_{\text{el}}\,\sin^{2}\!\beta_h\qquad\begin{cases}\phi_{\text{el}}=0: & \sin^{2}\psi=\cos^{2}\!\beta_h\\[2pt] \beta_h=90^{\circ}: & \sin^{2}\psi=\sin^{2}\!\phi_{\text{el}}\end{cases}

$\beta_h$: The detector's horizontal angle from the beam direction. In the horizontal plane this is the scattering angle 2θ.
$\phi_{\text{el}}$: The detector's elevation above the horizontal plane.

What it saysThis is equation 7.1 with the detector's position written out. Swinging the detector in the horizontal plane gives cos² of the angle from the beam, which is zero at 90°. Lifting it at 90° gives sin² of the elevation. Near the ideal spot both behave as the square of the offset, so the first few degrees cost little and the cost then climbs quickly.

In numbersA 10° offset in either direction gives sin²10° = 0.030. A 45° swing gives 0.5. Straight up gives 1.

Equation 7.3A beam that is not fully polarized

\frac{d\sigma}{d\Omega}=r_e^{2}\left[\,P\,\sin^{2}\psi\;+\;(1-P)\,\frac{1+\cos^{2}2\theta}{2}\,\right]

$P$: Degree of linear polarization from equation 1.3 in the guide Why the beam is polarized.
$2\theta$: The scattering angle between the beam and the detector direction.

What it saysThe beam behaves as a fraction P of perfectly polarized light plus a fraction 1 − P of unpolarized light. The unpolarized part scatters with the familiar (1 + cos²2θ)/2 factor, which is ½ at 90° and never zero.

SDD position 90°

Beam energy 12.662 keV

Beam polarization P 0.980

Sensor half-angle α 9°

Total scatter vs the ideal spot

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Elastic share here

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Compton peak here

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Scatter reaching the fluorescence detector as it moves, compared with the ideal spot. The vertical scale is logarithmic: each grid line is ten times the one below. Swing horizontally: the detector stays in the horizontal plane and its angle from the beam changes. 90° is the ideal spot, smaller angles are toward the beamstop and larger ones toward the source. Lift vertically: the detector stays at 90° to the beam and rises out of the horizontal plane. The sample is treated as water, which is most of what a protein crystal and its surrounding liquid are made of.

Set for 24-ID-C after the upgrade: 12.662 keV, P = 0.982 at the sample.

The math, if you want it

Equation 7.4Elastic scatter from real atoms

\left.\frac{d\sigma}{d\Omega}\right|_{\text{elastic}}=r_e^{2}\;\mathcal{P}\;\sum_{\text{atoms}}f(x)^{2}\qquad\qquad x=\frac{\sin\theta}{\lambda}

$\mathcal{P}$: The polarization factor: the square bracket of equation 7.3.
$f(x)$: The atomic form factor: how many of an atom's electrons scatter in step with each other. It equals the atomic number at x = 0 and falls as x grows.
$x$: A measure of the scattering angle. θ is half the scattering angle and λ is the wavelength.

What it saysAt small angles all the electrons in an atom scatter together, so the intensity goes as the square of their number. At large angles their waves fall out of step and mostly cancel. So elastic scatter is strong toward the beamstop and weak at 90° and beyond. This sum treats atoms one at a time. It leaves out interference between neighboring molecules and the sharp Bragg spots from the crystal.

In numbersAt 12.66 keV and 90°, x = 0.72 Å⁻¹. Oxygen's form factor has fallen from 8 to about 1.6 there, so its elastic scatter is down about 25 times from the forward direction.

Equation 7.5Compton scatter from real atoms

\left.\frac{d\sigma}{d\Omega}\right|_{\text{Compton}}\approx r_e^{2}\;\mathcal{P}\;\Bigl(\frac{E'}{E}\Bigr)^{2}\sum_{\text{atoms}}S(x)

$S(x)$: The incoherent scattering function: how many of an atom's electrons can be knocked free at this angle. It is zero at x = 0 and rises to the atomic number at large x.
$E'/E$: The energy ratio from equation 6.2. It is within a few percent of 1 here.

What it saysCompton scatter obeys the same polarization factor as elastic scatter, so the ideal spot suppresses it too. Its dependence on angle is the reverse of elastic scatter. A bound electron cannot recoil from a gentle push, so there is little Compton scatter near the forward direction and a lot at 90° and behind. For light atoms at 90° it is the larger of the two.

In numbersFor water at 12.66 keV, seen from the ideal spot with a 9° sensor cone: about 74% of the scatter is Compton and 26% is elastic.

How close? The sensor sees a cone, not a line

The sensor is a small disc, not a point. Seen from the sample it covers a cone of directions. Only the middle of that cone looks straight along the dark direction. Its edges look a few degrees to one side, where there is some scatter.

Moving the detector closer widens the cone. The fluorescence collected goes up with the area of the cone: at half the distance, four times as much. The scatter goes up faster, because a wider cone reaches further into the bright lobes. There is also a limit to how clean the mix can get. The beam is not perfectly polarized, and the unpolarized part scatters in every direction. Once the cone is narrow enough for that part to dominate, backing away loses fluorescence and gains nothing.

Which end to favor depends on the sample. More fluorescence collected means less beam, and so less radiation damage, for the same counts. A cleaner mix matters when the fluorescence is weak. The design study for the new detector on 24-ID-C settled on about 15 mm from the sample to the sensor, behind an opening 2.4 mm in radius. That gives a cone of 9° half-angle, which the other figures in this guide use. These are design values, not measurements of the installed detector.

Sample to sensor distance 15.0 mm

Cone half-angle α

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Cone covers

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Fluorescence collected vs 15 mm

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Scatter per fluorescence vs 15 mm

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Left, top: the sample, the sensor and the cone between them, with the detector at the ideal spot so that it looks along the electric field. The opening in front of the sensor is 2.4 mm in radius. Left, bottom: the scatter in each direction as seen from the sample, over 22° either side of the dark direction. The solid circle is the patch the sensor covers. As the detector moves in, the circle grows into the brighter bands. Right: what the sensor collects at each distance, compared with 15 mm. The vertical scale is logarithmic. Both curves rise as the detector moves in, and the scatter curve rises faster. The last readout is the gap between the curves: how much scatter arrives with each fluorescence X-ray.

Set for 24-ID-C after the upgrade: 12.662 keV, P = 0.982 at the sample.

The math, if you want it

Equation 8.1The cone the sensor sees

\tan\alpha=\frac{r}{d}\qquad\qquad \Omega=2\pi\,(1-\cos\alpha)\approx\frac{\pi r^{2}}{d^{2}}

$r$: Radius of the opening in front of the sensor.
$d$: Distance from the sample to that opening.
$\alpha$: Half-angle of the cone of directions the sensor accepts.
$\Omega$: The solid angle of that cone. The whole sphere is 4π.

What it saysFluorescence goes out evenly, so the sensor collects the fraction Ω/4π of it. Halve the distance and the fraction is four times as large.

In numbersFor r = 2.4 mm and d = 15 mm: α = 9.1° and Ω = 0.079, which is 0.63% of all directions. At 8 mm: 16.7° and 2.1%.

Equation 8.2The floor at the ideal spot

\left.\frac{S}{S_{\max}}\right|_{\text{ideal}}\approx\;P\,\frac{\alpha^{2}}{2}\;+\;\frac{1-P}{2}

$S/S_{\max}$: Scatter into the detector, relative to the worst direction.
$\alpha$: Half-angle of the cone the sensor accepts, in radians.

What it saysEven at the ideal spot two things let scatter in. The sensor has a finite size, so most of it looks slightly off the axis, and sin²ψ averaged over a small disc is α²/2. And the unpolarized fraction scatters at half strength. A smaller sensor cone or a purer beam lowers one term each.

In numbersFor α = 9.1° = 0.159 rad and P = 0.98: 0.98 × 0.0126 + 0.010 = 0.022. The two terms are about equal. The pointing offset that doubles this floor is about 8.5°.

What scatter costs: a noisier edge scan

Scatter does not land in the window an edge scan counts, because its energy is too high. What it does is fill the detector. The beam is turned down until the detector is counting at a rate it can handle, and if most of what it counts is scatter, most of that rate is wasted. The fluorescence counts are then few, and the scan is noisy.

How much this matters depends on how strong the fluorescence is to begin with. A selenomethionine crystal gives so much that the detector can be badly placed and the scan still looks fine. With one metal atom in a large complex, or a small crystal in a large drop of liquid, the placement can decide whether the edge is seen at all.

If the beam is so weak that the detector never reaches its working rate, scatter costs almost nothing. The figure shows the usual case at an undulator beamline, where the beam is strong and has to be turned down.

SDD angle from the beam, horizontal 90°

Selenium in the sample 1 Se per 300 kDa

Counts the detector takes per point 3.0k

Fluorescence share of the counts

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Kα counts at the peak

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Noise on that point

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Beam allowed, vs the ideal spot

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A simulated scan across the selenium K edge. The pale green band is the true curve. The dots are what the detector counts in the Se Kα window, with the noise that goes with that many counts. The beam is turned down so that the detector takes the chosen number of counts, of all kinds, at the top of the edge. Swing the detector away from 90° and scatter takes a larger share, so fewer of the counts are fluorescence. Try 1 Se per 20 kDa, which is like a selenomethionine crystal, and then 1 per 1000 kDa. The edge shape is an illustration, not a measurement, and the same random numbers are used throughout so that only the size of the noise changes.

Set for 24-ID-C after the upgrade: P = 0.982 at the sample, 9° sensor cone.

The math, if you want it

Rules of thumb

What will fluoresce. Any element heavier than about calcium that is in the beam, once the beam energy is above its edge. The light atoms of the protein itself will not.

Which energy. For an edge scan, the element's edge has to lie within the range of the beamline. For a spectrum, set the beam above every edge that matters, and remember that each line sits well below its own edge.

Where the detector goes. In the horizontal plane at 90° to the beam. The first few degrees away from that cost little, and after that the cost climbs quickly. Staying within about 5° keeps the scatter under one and a half times its minimum, and within about 8° keeps it under twice. If the detector cannot go there, moving it toward the source costs the least, lifting it costs slightly more, and moving it toward the beamstop costs the most.

How close. Closer collects more fluorescence, and farther collects a cleaner mix. For a sensor opening of 2.4 mm radius, about 15 mm is a reasonable middle. Go closer for the lowest dose, and farther back when the fluorescence is weak.

What this guide leaves out. The sample is treated as water for scatter, with every atom scattering on its own. A real sample also sends sharp reflections and diffuse scatter toward the area detector, and the loop, the pin and any ice add scatter of their own. The air along the beam path scatters too, and the detector sees some of that path unless it is shielded. Fluorescence that has to cross a thick sample is absorbed more than the figures here allow for, and so is the beam on its way in. None of this moves the ideal spot. It does change the totals.

How many times more scatter the fluorescence detector receives than at the ideal spot, for the settings chosen at the top of the page. Calculated with the same water model as the figure above, at 12.662 keV with a 9° sensor cone. Fluorescence is the same in every row.

Where to go deeper

A. C. Thompson et al., X-ray Data Booklet, Lawrence Berkeley National Laboratory. Tables of absorption edges, emission lines and fluorescence yields.
T. Schoonjans et al., "The xraylib library for X-ray–matter interactions", Spectrochim. Acta B 66, 776 (2011). The source of the edge and line energies, yields and cross-sections used here.
M. O. Krause, "Atomic radiative and radiationless yields for K and L shells", J. Phys. Chem. Ref. Data 8, 307 (1979). Fluorescence and Auger yields.
G. Evans and R. F. Pettifer, "CHOOCH: a program for deriving anomalous-scattering factors from X-ray fluorescence spectra", J. Appl. Cryst. 34, 82 (2001). From an edge scan to f′ and f″.
W. A. Hendrickson, "Anomalous diffraction in crystallographic phase evaluation", Q. Rev. Biophys. 47, 49 (2014). How f′ and f″ are used in SAD and MAD.
J. Als-Nielsen and D. McMorrow, Elements of Modern X-ray Physics. Chapter 1 for Thomson and Compton scattering and the polarization factor, and the chapters on absorption and resonant scattering for edges.
J. H. Hubbell et al., "Atomic form factors, incoherent scattering functions, and photon scattering cross sections", J. Phys. Chem. Ref. Data 4, 471 (1975). The tables behind the water model.
R. Jenkins, R. W. Gould and D. Gedcke, Quantitative X-ray Spectrometry. Fluorescence detection and scatter background.

Last updated November 11, 2026. Release 1.2. What changed.
The NE-CAT beamlines are funded by the National Institute of General Medical Sciences from the National Institutes of Health (P30 GM124165).
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