I proposed on 27 October 2019, 13:04:39https://www.chemicalforums.com/index.php?topic=101272.msg356752#msg356752
to use atoms near an energy absorption edge, but how brilliant are they?
With the detection of flame retardant in mind, I consider the KαI emission by 15P at 2014eV=616pm with 0.2µm coherence time
- The K (1s->valence) absorption edge of 15P is at 2146eV or 1.066× above the emission line. Numerically, strong reflection would take much more than 0.2µm phosphorus thickness, but the decoherence prevents it.
- The L (2s->valence) edge of 39Y is at 2080eV or 1.033×. Here the 0.2µm are a few times too short, imperfect.
- The M (3d->valence) edge of 77Ir is at 2040eV or 1.013× and the path in a mirror is but more than 0.2µm. Considered here.
- The emission by 17Cl instead of 15P fits the absorption edge of unstable elements only, bad luck.
- Freely chosen emitting and resonating elements would fit much better than 1.013×.
I model the 77
Ir near-resonance brutally with electrons 0.91×10-30
kg heavy and 8.7MN/m stiff to resonate at 492PHz=2040eV. Well below the resonance, 1V/m move the electron by 1.8×10-26
Ir contains 117kmol for 7.0×1029
3d electrons, so
1V/m induces 2.1fC×m displacement. The resulting permittivity is ε=(1+2.3×10-4
As a quick check, electrons in optical materials resonate at 5eV instead of 2keV, they are 4002
times less stiff, so their permittivity would be 38 - compare with 3 usually, not bad for such a model. Maybe only one 3d electron from Ir moves per atom. More probably, few valence electrons per atom move at optical materials.
With F=2040eV/2014eV=1.013, the near-resonance multiplies the movement amplitude by 1/(F-1/F)=39 so the permittivity becomes ε=(1+9.1×10-3)×ε0
. The near-resonating 10 electrons are more brilliant
than the 67 others. This improves the diffraction by crystals and, at lower energies, by gratings.X-ray lenses
were reported using this near-resonance. Here a Fresnel lens made by semiconductor process needs steps 68nm thick, nicely transparent, and its 30nm to 300nm wide rings as in the previous message fit the thickness. More transparent than a zoned lens but limited by the coherence time too and narrowband. 30% loss allow only 156nm thickness, no good alternative. Better matched element pairs would improve. 39
Y is more transparent in this use.
An X-ray mirror
must provide atomic planes spaced by multiples of the half-wave on the path of the ray. Fcc Ir spaces the planes by 192pm along <100>, less in the others. That's too small for 616pm wavelength, so some iridium compound is needed, with lighter atoms for transparency, maybe O, F, Cl, B, C, a metal... The incident and reflected angle makes the fine tuning. The ray shall encounter a dense iridium plane every 308pm in some adequate crystal.
I treat this as a dichroic mirror. The 3d shells are small and of higher index, but for the reflection they can be taken as 308/2=154pm layers of permittivity 1+11×10-3
that alternate with 154pm layers of vacuum. The index is 1+55×10-4
and the field reflection amplitude 28×10-4
at each interface.
Over 0.2µm coherence time, 62% iridium attenuates ×0.75, nice. The ray encounters 325 Ir layers and as many on the return leg. The 650 interfaces reflect 28×10-4
field each so the reflection is strong
Marc Schaefer, aka Enthalpy