Hello All;
I see these graphs (See image) in the papers, where the X-rays attenuation coefficient is plotted against the thickness of various materials. I know, these are not experimental but they are plotting these using some online software. They have mentioned the NIST platform but I found only the absorption vs energy plotting facility on the website. Could anyone please help me where can I plot the X-rays attenuation coefficient vs the thickness of any material/compound using online software?
Thank you so much.
Dear Nasir Ali ,
you can used any plot program which is able to handle some basic formulae...
The x-ray transmission T = I/Io is given via the Lamber-Beer law:
T = I/Io = exp[-(µ/rho)*rho*d]
with 'µ/rho' being the x-ray mass attenuation coefficient*), 'rho' the material density and 'd' the thickness of the sample.
You see, that there is no necessity to use some online tools for that...
*) 'µ/rho' is energy dependent (you noted that already) and can be taken for preset energies from
https://physics.nist.gov/PhysRefData/Xcom/html/xcom1.html
even in the case of a compound.
Please note that the energies are given in MeV here...
Good luck and
best regards
G.M.
Dear Nasir Ali ,
If youd don't want to make the calculation yourself as suggested by Dr. Gerhard Martens and you want to get an accurate calculation plot, you can just use this link to the Center of X-Ray Optics (CXRO):
https://henke.lbl.gov/optical_constants/atten2.html
In this page, you can enter your sample and beam paramters, and get the requested output plot.
Best regards,
Elhoucine HADDEN
Dear @Nasir Ali ,
What you show in the plots is the attenuation efficiency (the absorption) vs the thickness. The attenuation coefficient depends on the material and the wavelength, not on the thickness.
Best regards,
Elhoucine HADDEN
Dear Elhoucine Hadden ,
Nasir Ali wants to plot the transmission vs the thickness of the sample, but not the energy dependence of the coefficient µ(E).
Your link mainly provides the penetration depth (D1/e) dependence on the incident angle*) phi and the attenuation coefficient µ; but not the transmission:
D1/e = 1/µ(E) * sin(phi); (for phi >> phic)
*) phi taken beween the sample surface and the x-ray beam
Best regards
G.M.
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