I don't know what it is to 6 decimal places, and I'm pretty sure it isn't known. The sixth decimal place for ~8 keV is 0.008 eV, which is about the energy of a thermally-induced vibration. Changes in chemical environment of the copper atom (and hence the "starting energy" of the electron that fills theK-shell core hole) will shift emission lines even more than this. So, unless your copper anode is a chemically pristine single crystal of copper that is cooled by liquid Helium, the width of the thermal Doppler shifts for Kalpha1 and Kalpha2 peaks are going to make them VERY broad with respect to the sixth decimal place.

As for references:

I generally use the "little orange book"
http://xdb.lbl.gov/Section1/Table_1-2.pdf

which reports CuKa1 = 8047.78 eV (1.54060 A) and CuKa2 = 8027.83 eV (1.54443 A)
this is largely taken from the original literature reference:
J. A. Bearden, “X-Ray Wavelengths,” /Rev. Mod. Phys. /*39*, 78 (1967)


Another reference is the widely-used "mucal" data compiled by Pathikrit Bandyopadhyay
http://csrri.iit.edu/cgi-bin/mucal-form?name=Cu&ener=

Which reports CuKa1 = 8046.99993 eV (1.54075 A)
This is taken from the classic book:
"Compilation of x-ray cross sections" W.H. McMaster et. al. (1969) OSTI ID: 4794153 which I think is largely taken from the monumental body of measurements made by J. H. Hubbell.

The "last word" on relative yields of fluorescence emission lines appears to be
/M. O. Krause, J. Phys. Chem. Ref. Data. 8, 307(1979)/


You may notice that these two popular sources do not report the same value for CuKa1. The second is given to 8 decimal places, but something tells me that the uncertainty is higher than that.

It is interesting to note that there is definitely inconsistency in the way x-ray wavelengths are defined. for example, to convert from photon energy to wavelength, you use lambda=12398.42/energy, or is it just 12398.0? Sometimes I have seen 12398.45. Gathering together the physical constants (http://physics.nist.gov/cuu/Constants/): speed of light: c = 299792458.00000000 m/s (exactly! what are the chances of that? :) Plank's constant: h = 4.13566733(10) x 10^-15 eV*s (known to 8 decimal places)
Angstroem: A = 1e-10 m (exact)
gives: c*h*1e10 = 12398.4187(3) eV*Angstroem
This ought to be the correct conversion factor, but rounded-off values are often taken as acceptable in the literature and in beamline software. For example: http://heasarc.gsfc.nasa.gov/cgi-bin/Tools/energyconv/energyConv.pl) uses a slightly different value. It would appear that there is some disagreement as to the unit charge of an electron in the sixth decimal place.


So, I guess if you really want to know what CuKa1 is to six decimal places, you have to measure it yourself. I can tell you how you might go about doing that. You can buy chunks of Si from NIST that are certified to have a unit cell spacing of 5.4311946 ± 0.0000092 Angstroem. A little better than 6 decimal places.
https://srmors.nist.gov/view_detail.cfm?srm=640c

Bounce your favorite rotating-anode beam off of one of these babies and then put your detector somewhere in the next building to see what the absolute angle between the incident beam and the diffracted ray is. If your detector pixels are 0.1 mm, then you will need to place the detector 100 meters away from the Si crystal get the sixth decimal place accurate. You will also need to know the direct beam position to the same accuracy. In fact, all three edges of the triangle between the sample, direct beam and the Si(111) diffraction spot must be known to six decimal places for this to work. This means you will need a 48-meter long stage (sin(2*theta)*100 m) to translate the detector. The peaks will be broad (look up "core hole lifetime" to read about x-ray emission line widths), but you should be able to fit them to Gaussians and get a little better than 1 pixel accuracy. If you can do that, you might only need a 10 meter path.


For the record, I personally define the wavelength of my beamline (ALS 8.3.1) so that the "inflection" of the absorption edge of a metal Cu foil is 8979.000 eV. This is taken from the "electron binding energy" in the "little orange book"
http://xdb.lbl.gov/Section1/Table_1-1a.htm
and agrees with the original literature:
J. A. Bearden and A. F. Burr, “Reevaluation of X-Ray Atomic Energy Levels,” /Rev. Mod. Phys./ *39*, 125 (1967)

Years ago, I measured the positions of 17 different absorption edges from 14 different metal foils and compared them to those of Bearden & Burr.
http://bl831.als.lbl.gov/~jamesh/mono_calib.png
The scatter in these points is a bit alarming. However, Bearden and Burr were using a much more monochromatic beam than can be achieved with Si(111). On most protein crystallography beamlines, the energy spread acts as a blurring fliter on the absorption edge, changing the position of maximum slope, which is what Bearden and Burr defined as the "edge".

Since the best-fit straight line to the "error" I measured in all the edge positions has a very shallow slope, I am assuming that much of this "error" is due to offsets induced by these fine-structure effects. So, I am further assuming (hoping) that the spacings on the Heidenhain rotary encoder in my monochromator are accurate. I am using Cu because it is convenient, radiation-hard and seems to have a minimal amount of fine structure in its absorption edge. Se might sound like a good idea, but the position of the Se edge moves around a lot (+/- 3 eV) depending on the chemical state, and the chemical state of many Se preparations (such as SeMet for example), changes with radiation damage.

This does mean that the energy/wavelength written into image headers at my beamline may be off by as much as 7 eV relative to another facility that uses a different metal edge as their reference standard. That said, I would like to add that even this difference ultimately corresponds to a 0.05% change in bond lengths. So, I'm not all that worried about it.

-James Holton
MAD Scientist

Jim Pflugrath wrote:

It has come to my attention that the wavelength of a Copper Kalpha may have changed over the years. At least this appears to be true if you look at the International Tables.

What is the currently approved wavelength in Angstrom of a Copper Kalpha X-ray produced by a X-ray generator with a copper anode to 6 decimal points? In your response, please tell us how you arrived at that wavelength and how you weighted any alpha1, alpha2, etc components.

Thanks! Jim

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