Question

If there is a MCM-41 which has wall thickness of 1 nm and pore diameter of 3 nm, where would I observe the XRD(2theta) peak?

Answer 1

XRD analysis is based on constructive interference of monochromatic X-rays and a crystalline sample: The X-rays are generated by a cathode ray tube, filtered to produce monochromatic radiation, collimated to concentrate, and directed toward the sample. The interaction of the incident rays with the sample produces constructive interference (and a diffracted ray) when conditions satisfy Bragg’s Law (nλ=2d sin θ). This law relates the wavelength of electromagnetic radiation to the diffraction angle and the lattice spacing in a crystalline sample.

The pore diameter given in question is 3 nm.

To get XRD peak , wavelength of x - rays should be comparable of pore diameter.

The wall thickness given in question ( 1 nm) is the spacing between the planes.

So In physics, Bragg's law, or Wulff–Bragg's condition, a special case of Laue diffraction, gives the angles for coherent and incoherent scattering from a crystal lattice

MCM-41 (Mobil Composition of Matter No. 41) is a mesoporous material with a hierarchical structure from a family of silicate and alumosilicate solids that were first developed by researchers at Mobil Oil Corporation and that can be used as catalysts or catalyst supports.

So using Bragg's law

2 d sin = n n= 1

2* 1 * sin = 1 * 1

sin = 1/2

   = sin^{​​​​ -1} (1/2)

= 30°

2 = 60°

We would observe XRD peak at 2 = 60 °