I understand that in general if we're adding more planes of atoms (increasing thickness of sample) then the intensity would increase because we have more constructive interference. But isn't there a breaking point for this? Shouldn't there be a finite thickness past which the intensity decreases?

# [Physics] Is the thickness of a sample related to the intensity of x-ray diffraction

condensed-matterdiffractionsolid-state-physicsx-ray-crystallographyx-rays

#### Related Solutions

You are mistaking the mechanism by which part of the wavefront is used selectively for the phenomena in which the selected bits interfere with one one another and create a pattern.

I have a (optical) teaching spectrometer that uses a *reflective* diffraction grating, but it is also common to use diffraction gratings in transmission mode.

There are some nice pictures of reflective diffraction devices in the Wikipedia article on diffraction gratings.

Selective reflection or transmission is how you *partition the initial wavefront*. Diffraction is what happens when the selected bits interfere resulting in a macroscopic pattern in the angles of outgoing signal.

Also possibly helpful on Physics: Why is it difficult to differentiate between interference and diffraction?

Both descriptions are correct; some people prefer the geometric description: the lattice of atoms is replace by a collection of planes, with different orientations. This corresponds to the Bragg model of partially reflective mirrors, and the K-vectors give the directions for the reflections which form the diffraction pattern.

The description given by Ashcroft & Mermin simply puts the diffracted waves first, and the planes follow. Since it is the waves that one measures, this is more physical to some experimentalists.

However, if you actually work with crystals, the abstract planes from the geometric definition correspond to the cleavage planes.

The Bragg formula gives the paths of constructive interference; these are the bright spots in the diffraction pattern. You could also find the dark places, but the usual thing is to map out the bright spots, as this is sufficient to determine the crystal structure.

The Ewald sphere is a map of the reciprocal lattice, and corresponds to rotations as you describe. For x-rays it is a valuable tool, for it tells you which spots are accessible. With electron diffraction the wavelengths are so short that essentially all of the diffraction spots are accessible at once.

Powder diffraction provides an average across all of the rotations, because the powder already contains all of the possible orientations simultaneously.

## Best Answer

Yes, there is such a point. The precise formula varies as a function of the scattering geometry, but if we consider a special case:

it is quite simple: the scattered intensity is proportional with the sample thickness $d$ but it gets attenuated as $\exp(-\mu d)$ (the Beer-Lambert law with $\mu$ the absorbance).

The strongest signal is thus obtained at the maximum of $I_S \sim d \exp(-\mu d)$, which occurs at $d = 1/\mu$, i.e. a transmission of $1/\text{e}$. Beyond this value, the attenuation (an exponential effect) dominates the increase of scattering volume (a linear effect).