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