Dear Erik,

quick and dirty, i.e. neglecting any effects of non homogeneous density or composition you can simply apply Lambert-Beer I(x)=I(0)exp(-mu x) where I(0) is the initial intensity, x your penetration depth, mu the mass attenuation coefficient and I(x) what comes out.

With your 50keV you will penetrate approx. 300 microns but this is not the number you are looking for. You will only observe the L and M fluorescence lines of Au most probably Lalpha @ 9.7 keV. If you want to calculate the depth just use the above equation. Let's say you want 1% of the fluorescence lines to come out of your precious nugget, then simply do I(x)/I(0)=exp(-mu x) and apply ln.

This gives ln(0.01)=-mu x. or x = -ln(0.01)/mu . The quantity mu you can find @ http://physics.nist.gov/PhysRefData/XrayMassCoef/ElemTab/z79.html

with approx. 130 cm²/g (which is mu divided by the density). This yields x= 18 um. Not very deep, isn't it ? Not to say that for most spectrometers you will need more than this 1% to get reasonable results. From my experience in pure gold you have to live with only 10 um.

Best regards,

Joerg

Thanks for taking the time to reply Jan. I have a question though. My XRF instrument has a 50 kV tube. Do you mean the instrument energy, or the K-alpha energy for gold? I saw that gold has a K-alpha1 of  68.79.Since you state that the energy is important, I want to make sure I am using the right one. Forgive my ignorance on this. I am not a materials scientist, and this is very new to me. Many Thanks, --Erik

Hm - too slow. With a 50 keV tube you can only excite the L and M lines as I said before. The energies you have to insert in the calculations are the energies of the fluorescence lines. This gives you the information depth as Jan stated. A 50 keV tube means that the highest excitation energy from Bremsstrahlung is approx. 50keV. Thus, the K-alpha line of Gold cannot be excited.

As Jan said,   X-rays of energy 50keV have a 1/e penetration depth into Au of 71 microns.

That is, mu for Au at 50 keV is 7.26 cm^2/gr.  For Other sources, see  the tables of Elam, Ravel, and Seiber:  http://dx.doi.org/10.1016/S0969-806X(01)00227-4, and available  from either https://github.com/XraySpectroscopy/XrayDB  or   https://github.com/xraypy/xraylarch   or https://github.com/bruceravel/demeter) or see https://github.com/tschoonj/xraylib/wiki.

With a density of 19.37 gr/cm^3,  that gives a mu = 7.26 * 19.37 cm^-1, and so a 1/mu depth of 0.0071 cm = 71 microns.

However, the escape depth for the Au Lalpha1 line (at 9.7 keV) would only be 4 microns, and for Au Lbeta1 line (at 11.4 keV) would only be 6 microns.  

Many thanks to you all for the help. It seems that the short answer is that my handheld XRF unit will give results for the surface composition of the gold nuggets, plus or minus a few microns. This is good to know. We have been testing some large nuggets in museums, as it is a non-destructive test. Now I know exactly what we are measuring with this technique, and why my results were different from whole-nugget assays.

Best Regards,

--Erik