Absorption Coefficient Calculator

https://xraytools.com/absorption

How strongly a compound absorbs X-rays, as the mass coefficient μ/ρ and the linear coefficient μ. It is μ that decides whether a measurement needs an absorption correction, and what crystal size to aim for.

You supply: A chemical formula — brackets and charges included — and a density, or tick the box and let it estimate the density from the formula.
Reading it: μ/ρ is the mass absorption coefficient, tabulated under that name in the crystallographic literature and as the mass attenuation coefficient by NIST — one quantity with two names in common use. It depends on the substance and on the photon energy, steeply, and it jumps at every absorption edge; what dividing by ρ removes is the density, so a gas and a solid of one substance share a μ/ρ and not a μ. It is μ that is the one to judge. For Cu radiation a common limit is 10 mm⁻¹, but absorption goes with the product μt — aim for μr of order 1.

Worked examples: NaCl · a nickel complex · C₆H₄Br₂ just above the bromine K edge · lead sulfide, 0.2 mm thick, where the crystal decides

Earlier on the path: Moseley Plot and Absorption Edges Next on the path: Friedif Calculator On Choosing the radiation, step 3 of 4

Notation here: μ, μ/ρ · f₀, f′, f″ — what each one means here

What each input changes

Sample thickness: The other half of the product. The coefficient above is a property of the material and the wavelength, and nothing about absorption follows from it until a path length is named — what a transmission states is μt. Doubling it squares the transmission.

Teaching with this page

Objective: After this page a learner can decide whether a sample absorbs too much for a given radiation, and name the two things that can be changed about it.
Start from: this worked example
Ask first: Your compound gives μ = 12 mm⁻¹ for Cu radiation, past the usual working limit. Does a smaller crystal help?
Watch for: “No — μ is a property of the compound and will not change”
Then: Friedif Calculator

Check yourself: Your compound gives μ = 12 mm⁻¹ for Cu radiation, past the usual working limit. Does a smaller crystal help?

Yes — absorption depends on μt, and t is yours No — μ is a property of the compound and will not change

The tempting answer is true and does not answer the question. What the beam experiences is the product μt: the working rule is μr of order 1 with r the crystal radius, so at 12 mm⁻¹ that is a radius of about 0.08 mm — a crystal roughly 0.17 mm across. There are exactly two levers, and this page shows both in one table: change the crystal, or change the radiation.

Input

Results

chemical formula:	C₃₂H₂₆N₂O₂Ni₁
molar mass:	529.252(19) g/mol
density:	1.237 g/cm³
mass absorption coefficient (Mo):	μ/ρ = 5.7052 cm²/g at 17.445 keV
mass absorption coefficient (Cu):	μ/ρ = 9.3723 cm²/g at 8.041 keV
linear absorption coefficient (Mo):	μ = 7.057 cm^-1 = 0.706 mm^-1
linear absorption coefficient (Cu):	μ = 11.594 cm^-1 = 1.159 mm^-1
through 0.1 mm (Mo):	μt = 0.071, transmission 0.9319
through 0.1 mm (Cu):	μt = 0.116, transmission 0.8905
thickness for μt = 1 (Mo):	1.417 mm where the beam is cut to 1/e, 0.3679
thickness for μt = 1 (Cu):	0.8625 mm where the beam is cut to 1/e, 0.3679

How much gets through

Transmission is a statement about the product μt and never about either factor alone. The curves below are the same exponential for every radiation; what differs is how much sample it takes to travel along it.

The dashed line is the 0.1 mm you entered. Point at a marked place on a curve to read it. Touch has no hover; the table below carries the same seven places.

How thick the sample may be, rung by rung of μt. CSV
μt	transmission	Mo / mm	Cu / mm
0.10	0.9048	0.1417	0.0863
0.25	0.7788	0.3542	0.2156
0.50	0.6065	0.7085	0.4313
1.00	0.3679	1.4170	0.8625
2.00	0.1353	2.8339	1.7251
3.00	0.0498	4.2509	2.5876
5.00	0.0067	7.0848	4.3127

Correcting for it

A is the fraction of the beam that survives the path in to a scattering point and out again, averaged over the crystal; A* = 1/A is what a measured intensity has to be multiplied by. It depends on the shape, on μR, and on the scattering angle — and it is that last dependence that makes absorption a correction rather than a scale factor. A quantity that does not vary with 2θ joins k, which is exactly what the corrections page says about a flat plate in Bragg–Brentano.

The numbers below are for a sphere of radius 0.05 mm — half the 0.1 mm you entered, because that field is a path through the sample and this one is a radius.

A* for a sphere of this crystal’s size, and what skipping it costs. CSV
radiation	μR	A* at 2θ = 0	2θ at 0.6 Å⁻¹	A* there	apparent ΔB
Mo	0.035	1.054	50.5°	1.054	0.000 Å²
Cu	0.058	1.091	135.4°	1.089	-0.001 Å²

The spread down a row is the correction; the size of the numbers is not. On the Mo row A* falls by 0.007 % between 2θ = 0 and 50.5°. A crystal that absorbed the same amount at every angle would need no correction at all — the scale factor would swallow it.
Skip it and the structure refines with displacement parameters too small. A rises with angle, because at 2θ = 0 the path is the whole chord whatever the depth and at backscatter it is twice the depth — so the low-angle reflections are attenuated hardest, which is what a negative temperature factor looks like. Fitted against s² over the range above, this crystal on Mo shifts B by 0 Å², and the fit is a straight line to R² = 0.9994 — which is why it hides so well. Far enough down that road is the non-positive-definite ellipsoid the displacement parameters page reports on.
The same crystal, a different wavelength, a different bias. Mo gives ΔB = 0 and Cu gives -0.001 Å². Two things move at once and they do not cancel: μ differs, and so does the angle needed to reach the same resolution — 50.5° against 135.4° for 0.6 Å⁻¹. B is a slope against s², so a bias spread over less s is a steeper one.
Where the sphere numbers come from. At 2θ = 0 the path in plus the path out is the whole chord however deep the scattering happened, and the integral collapses to a closed form — A = (3/x³)[2 − e^−x(x² + 2x + 2)] with x = 2μR, which is 1 − (3/2)μR for a weakly absorbing crystal. Everywhere else it is a quadrature over the sphere, and that one exact case is what proves it: on this run they agree to 0.024 parts per million at μR = 0.035. International Tables volume C tabulates the same quantity.

Which correction, and what each one knows

Empirical, multi-scan — knows that symmetry-equivalent reflections measured in different orientations disagree. Fits a smooth function of the beam directions — in practice an expansion in spherical harmonics — to make equivalents agree. It needs real redundancy, and it absorbs every other orientation-dependent error along with absorption, which is both why it works on badly shaped crystals and why it cannot be checked against a measurement of the shape. Reach for it: the default on an area detector, and sound while μR is small enough that the correction is a perturbation.
Empirical, ψ-scan — knows how one reflection’s intensity varies as the crystal is rotated about its own scattering vector. A few reflections near χ = 90° are measured through a full azimuthal turn, and the resulting curve is applied to everything else. It predates area detectors, where redundancy is free, and is now mostly of historical interest. Reach for it: serial detectors, and low redundancy.
Analytical, from indexed faces — knows the crystal’s actual shape, as a set of indexed planes and their distances from a centre. The path in and the path out are integrated over the measured polyhedron for every reflection. It is the only correction whose input is a measurement of the crystal rather than of the data, so it is the only one that can be wrong for a reason you can see down a microscope. Reach for it: a well-formed crystal at μR past about 1, where the shape stops being a detail.
A sphere or a cylinder — knows the shape exactly, because it was ground into one. A* depends on μR and the scattering angle alone, which is what the table above computes. Grinding a crystal is destructive and takes skill, and for a strongly absorbing sample it is still the cleanest answer there is. Reach for it: heavy-atom work at large μR, and powders in a capillary.

A CIF records which was used in _exptl_absorpt_correction_type and what it did in _exptl_absorpt_correction_T_min and _T_max — the smallest and largest transmission the correction applied. A T_min / T_max ratio far from 1 on a crystal described as equant is worth a second look, and so is a multi-scan on a sample whose μR is anywhere near the numbers above.

When the sample fluoresces

An absorption edge is a binding energy, so a photon above it can empty that shell and one below it cannot. Everything on this page so far has treated the absorbed beam as gone; some of it comes back out, as the sample’s own characteristic radiation, at energies the diffractometer was not built to reject.

Mo, 17.445 keV — one element with a shell this beam can empty. CSV
element	share of μ_photo	deepest edge reached	E / edge	emits	1/μ at that line
Ni Z = 28	94.1 %	K, 8.3328 keV	2.09	Kα₁ 7.4839 keV predicted	0.7065 mm

Cu, 8.041 keV — one element with a shell this beam can empty. CSV
element	share of μ_photo	deepest edge reached	E / edge	emits	1/μ at that line
Ni Z = 28	56.4 %	L1, 1.0081 keV	7.98	—	—

The share column is exact and the ranking is not the whole story. Absorption is the mass-fraction-weighted sum this page already prints, and the derivation below shows that sum; the share here is the photoelectric part of it, which is the part that makes vacancies rather than scattering the photon away. What decides how many of those vacancies come back out as a photon instead of an Auger electron is the fluorescence yield, and this site holds none — it climbs steeply with atomic number, which is why sodium and chlorine cross their K edges under every laboratory radiation without being a fluorescence problem and iron under Cu Kα is.
E / edge near 1 is the bad case. A shell’s photoelectric cross-section is largest immediately above its own edge and falls away above it, so a beam that sits just over an edge puts most of its absorption into that shell. Under Mo the beam sits at 2.09 times Ni’s K edge, and Ni takes 94.1 % of the photoelectric absorption there.
No filter on the incident beam removes it. A characteristic line is the difference of two binding energies — Kα₁ is E(K) − E(L₃), because that is what the electron gave up falling — so it is always below the edge that produced it, and therefore below the beam that produced it. The sample is correspondingly transparent to its own fluorescence, which is what the last column says: it is the depth from which such a photon still escapes. Rejecting it is the detector’s job, not the optics’.
Choosing a longer wavelength works only against a K edge. Ni’s K edge is at 8.3328 keV, and of the six anodes this site carries, Kα from Cu (8.041 keV), Co (6.925 keV), Fe (6.400 keV), Cr (5.412 keV) fall below it. Dropping below an L edge of a heavy element does not stop it fluorescing; it only moves the vacancy to a shallower shell, whose softer line is reabsorbed inside the crystal.

An element with no edge inside the shipped table cannot be answered for here, and is left out of the rows above rather than reported as safe: the tabulation starts at 1 keV, so carbon, nitrogen and oxygen — whose K edges are below it — are absent from the table and not from the sample. Their fluorescence is soft enough to be reabsorbed within micrometres of where it was made.

Filtering out Kβ

A tube emits two K lines and a diffraction experiment wants one. A filter is a foil of the element whose K absorption edge falls between them: Kβ is above the edge and strongly absorbed, Kα is below it and much less so. What the foil buys is therefore not absorption but the difference between two absorptions, and that is the whole calculation — both lines are attenuated, and only the ratio improves:

I_β / I_α after = I_β / I_α before × exp[−(μ/ρ_β − μ/ρ_α) ρt]

So the answer is a mass thickness ρt, which is why filters are tabulated in mg cm⁻² and not in millimetres. A reader who picks the foil for its absorption of Kβ alone, ignoring μ/ρ_α, gets a thickness too small by whatever the Kα coefficient is.

Each radiation, its Kβ filter, and what the filter costs. CSV
radiation	filter	edge above Kα	μ/ρ at Kα	μ/ρ at Kβ	Kβ/Kα wanted	ρt	t	Kα left
Mo Kα	Zr Z = 40	553 eV	15.6	75.9	1 : 100	49.6 mg cm⁻²	0.0761 mm	46 %
Mo Kα	Zr Z = 40	553 eV	15.6	75.9	1 : 500	76.3 mg cm⁻²	0.117 mm	31 %
Cu Kα	Ni Z = 28	291 eV	46.8	282.1	1 : 100	12.7 mg cm⁻²	0.0143 mm	55 %
Cu Kα	Ni Z = 28	291 eV	46.8	282.1	1 : 500	19.6 mg cm⁻²	0.022 mm	40 %

The starting ratio is nominal — Kβ / Kα = 0.2 as emitted, which varies with the anode and with the tube voltage. The answer depends on its logarithm, so it survives that: being out by a factor of two moves ρt by 11.5 mg cm⁻² on the Zr row above. The millimetre column is the only place a density enters (6.52 g/cm³ for Zr), and it carries that column's rounding; the mass thickness beside it needs no density at all.

The edge above Kα column is what decides whether a filter is a good one, and it is the reason the Kα left column differs from row to row. A filter that passes its Kα comfortably has an edge a few hundred electronvolts above the line; where that margin is a few tens of eV the metal is already absorbing the radiation it is supposed to transmit, and the arithmetic above will still print a thickness for it.

How this is calculated

$μ = \frac{μ}{ρ} \times ρ$

$μ (Mo) = 5.70523 {cm}^{2} / g \times 1.237 g/ {cm}^{3} = 7.057 {cm}^{- 1}$

$\frac{I}{I_{0}} = \exp (- μ t)$ — the fraction of the beam that gets through a sample of thickness t.

For a sample 0.1 mm across (t = 0.01 cm): $\frac{I}{I_{0}} = \exp (- 7.057 {cm}^{- 1} \times 0.01 cm) = 93.2 %$ transmitted.

The other radiations differ only in the mass absorption coefficient, which is a property of the elements present and of the wavelength. The thickness is the one you entered, so the transmission is about your sample; the coefficient above it is about the material.

Notes and references

Generally, Mo radiation provides a higher resolution and more reflections compared to Cu radiation and should thus be considered to be the standard case. Cu radiation yields better reflection intensities, which is useful especially for weakly scattering crystals.
When the absolute configuration of a chiral compound is to be resolved (see also: Friedif Calculator) or a very large unit cell is expected, Cu radiation may be preferable.
The linear absorption coefficient of matter towards Cu radiation is generally higher, and a common working limit is 10 mm⁻¹. That bound carries a crystal size inside it: what absorption actually depends on is the product μt, which is what the exponential above takes, so the same μ is harmless on a small crystal and severe on a large one. The usable form couples the two — aim for μr of order 1, with r the crystal radius, which at 10 mm⁻¹ means r = 0.1 mm — a crystal about 0.2 mm across. Beyond that an absorption correction stops being a refinement and starts being the measurement.
f′ and f″ are Chantler’s relativistic tabulation (NIST FFAST, J. Phys. Chem. Ref. Data 24 (1995) 71 and 29 (2000) 597); the absorption coefficient is f″ through the optical theorem plus the tabulated scattering term. Compilations of these quantities differ from one another by a few per cent, and that is the accuracy to read them at. Tabulated from 1 keV to 150 keV — 0.0827 Å to 12.398 Å. Outside that range no value is given, because a cross-section continued past the last tabulated point is a smooth curve with no physics in it. Elements 1 ≤ Z ≤ 92, up to U. Until August 2026 this page read three tabulated columns — Ag, Mo and Cu Kα only, from H. P. Klug and L. E. Alexander, X-Ray Diffraction Procedures, Wiley, 2nd edn., 1974 — and its coefficients differ from those by a few per cent. At Cu Kα in particular the difference against NIST XCOM is one-sided: this dataset runs 4–6 % below XCOM for every element spot-checked (carbon, oxygen, silicon, lead), while at Mo Kα the two agree to about 2 %. That is a disagreement between two compilations at low energy, not a fault in either; quote which one a number came from.
atomic weights are the IUPAC standard atomic weights shipped with this site (1 ≤ Z ≤ 108, up to hassium), and the molar mass is quoted to the precision they support rather than to a fixed number of decimals. For fourteen elements — including H, C, N, O, S, Cl and Pb — IUPAC publishes an interval rather than a value, because the isotopic composition of normal materials genuinely varies; the midpoint is used here and the CHN calculator shows the interval itself. The Isotope Pattern Calculator answers a different question and sums the NIST representative isotopic composition instead, so its average mass is the mean of the pattern it draws and will differ from the molar mass here in the last figures. The 26 elements with no stable isotopes have no standard atomic weight at all; they carry the mass number of their longest-lived isotope, which is a nominal figure with no uncertainty rather than a precise one, and the results say so when one is used.
for the estimation of crystal densities, the very rough assumption of an atomistic volume of 18 Å³ per non-hydrogen atom is made

Where this comes from

XrayDB
Matthew Newville and contributors · on the reading list under “The tables this site computes from”
Where the coefficients come from. This page computes them from Chantler’s tabulated f″ through the optical theorem, plus the tabulated scattering term, for 1 ≤ Z ≤ 92 at any wavelength between 0.083 and 12.4 Å. Until August 2026 it read three tabulated columns instead — Ag, Mo and Cu Kα only, from Klug and Alexander — and its numbers differ from those by a few per cent.
Detailed tabulation of atomic form factors, photoelectric absorption and scattering cross section, and mass attenuation coefficients
C. T. Chantler, J. Phys. Chem. Ref. Data 2000, 29, 597–1056 · doi:10.1063/1.1321055
The calculation itself — a relativistic computation of the form factors and cross-sections of every element, which is what the numbers on this page are, one interpolation away.