[Ifeffit] Ifeffit Digest, Vol 158, Issue 3
Matthew Marcus
mamarcus at lbl.gov
Sun Apr 3 20:13:43 CDT 2016
<Rant>
It shouldn't be called 'self-absorption'. That's a misnomer, which seems to have come from a 1992
paper (Troger, et. al."Full correction of the self-absorption in soft-fluorescence extended x-ray-absorption fine structure", PRB 46,3283 (1992).
The effect was described and analyzed in a 1982 paper, which called it an "attenuation factor": Goulon, et. al. "On experimental attenuation factors of the amplitude
of the EXAFS oscillations in absorption, reflectivity and luminescence measurements", J. Physique 43, 539 (1982).
</Rant>
mam
On 4/3/2016 4:56 PM, Matteo Busi wrote:
> Hi Bruce,
> Thanks for your help, it's really appreciated.
> I was trying to keep it simple but it seems I'm just messing around.
>
> What I am trying to do is to perform a new developed self-absorption correction using collected fluorescence absorption coefficient data on a CuSO4 (pentahydrate) capillar (cylinder) and spherical sample with Cu as absorber, having different values of molarity and penetration depth.
> The correction expression requires these measured quantity:
> μX(E) : the absorption coefficient due to a given core excitation of the absorbing atom
> - I used the background function for this ( bkg(E))
> μo = μ(E) : photoelectric total linear absorption coefficient of the sample at incident energy E
> - I used the xmu(E) for this
> μh = μ(E) :photoelectric total linear absorption coefficient of the sample at fluorescence emission Ef
> - I used the xmu(Ef) with Ef the K absorption edge of Cu ( 8.9789 eV)
> χ: and here I used the chi(E) values of the exported ascii .xmu file
>
> Hope it's clear now.
> Matteo
>
>
> 2016-04-02 19:00 GMT+02:00 <ifeffit-request at millenia.cars.aps.anl.gov <mailto:ifeffit-request at millenia.cars.aps.anl.gov>>:
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> 1. Re: Ifeffit Digest, Vol 158, Issue 1 (Matt Newville)
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> ----------------------------------------------------------------------
>
> Message: 1
> Date: Fri, 1 Apr 2016 15:56:29 -0500
> From: Matt Newville <newville at cars.uchicago.edu <mailto:newville at cars.uchicago.edu>>
> To: XAFS Analysis using Ifeffit <ifeffit at millenia.cars.aps.anl.gov <mailto:ifeffit at millenia.cars.aps.anl.gov>>
> Subject: Re: [Ifeffit] Ifeffit Digest, Vol 158, Issue 1
> Message-ID:
> <CA+7ESbrsbsFQGMGLwGNPJmVO=fGzksbv1SezR3M5pRkXanNyDg at mail.gmail.com <mailto:fGzksbv1SezR3M5pRkXanNyDg at mail.gmail.com>>
> Content-Type: text/plain; charset="utf-8"
>
> Matteo,
>
>
> On Fri, Apr 1, 2016 at 8:35 AM, Matteo Busi <basebush at gmail.com <mailto:basebush at gmail.com>> wrote:
>
> > Hi Bruce,
> > Now this is clear.
> > In my case the correction I have to perform requires these
> > measured/evaluated parameters: chi, mu (not sure if it's better to work
> > with normmu here) and the background function.
> > Is it physically meaningless to have the mu(k) and bkg(k) data? If that is
> > not the case I would like to have these two columns exported in the ascii
> > chi(k) file.
> > So then I can perform the correction and re-import the new corrected
> > chi(k) and make comparison with the uncorrected.
> >
> > Kind regards,
> > Matteo
> >
> >
> It would certainly help (and by that I mean "help you get the answers
> you're looking for") if you gave us more details about what you're trying
> to do rather than ask for how to what you think you want to do.
>
> Just to be clear: Converting from E to k does not involve a Fourier
> transform, just
> k = sqrt((2m*e/hbar**2)*(E-E0))
>
> for k in Ang^-1 and E in eV, that's k ~= sqrt(0.262468 <tel:%280.262468>*(E-E0))
>
> I'm not sure what you're trying, but I would image you want to use
> normalized mu(E) or at least pre-edge subtracted mu(E). That is, what we
> call mu(E) is typically really -ln(I/I0)
> where I and I0 are not the actual intensities before and after the sample,
> but counts or counts per time in some detector that samples the flux.
> These values include more or less arbitrary scale factors (amplifier gains,
> etc) included. That means that (unless you very careful) mu(E) does not
> have meaningful units, certainly not cm^-1 or cm^2/gram.
>
> Normalized mu(E) doesn't have such units either, but it's an easier place
> to start.
>
> --Matt
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