These information are availble in the CXRO database. Link given below-

https://henke.lbl.gov/optical_constants/getdb2.html

Talina Terrazas Monje

Dear Talina Terrazas Monje ,

just in addition to the answer of Mangalika Sinha ,

the critical angle of total reflections has to be calculated from delta via:

critical angle= sqrt( 2*delta); with delta given by the link above.

According to : L.G. Parrat, Phys. Rev. 95; pp 359ff (1954) delta is proportional to the density of the material:

delta = A*lambda²*Z with A = N/2pi * e²/(mec²)* rho/M = 2,7019*1010cm*rho/M.

So play around with this formula (e.g. via the link above) and go for high density material.

I do not know for what purposes you need the critical angle.

It is just a number describing the angular dependence of the x-ray reflectivity in cooperation with the x-ray attenuation.

For the meaning of the 'critical angle' one often expects a steep cut off of reflectivity with increasing glancing angle. But for high attenuation this 'cut off' is often a gradual decrease of reflectivity without any peculiar characteristic feature.

Good luck

Gold at 8kev should be pretty much the best you may get. You are below the L-edges and gold has one of the highest densities of all elements (19.3 g/cm^3). Good luck, Dirk

I don't think you'll find a single material with larger critical angle compared to gold. However a multilayer may provide a higher reflectivity at shallow angles around few degrees. But you need to give some more details on why you need large reflectivity and critical angle.