# Is the thickness of a sample related to the intensity of x-ray diffraction?

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?

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

1. normal incidence on a flat sample and
2. small scattering/diffraction angle

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).

• Why us the scattered intensity proportional to the thickness? Naively, in the regime of single scattering it should be quadratic, because the scattering amplitude is proportional with the number of scatterers which in turn increases linearly with thickness. May 25 '20 at 11:40
• For the amplitudes to add up over the whole thickness $d$ of the sample one would need both the (longitudinal) coherence length $l_c$ and the range of order to be larger than $d$. For x-rays, $l_c$ is generally much shorter than typical $d$ values. May 26 '20 at 17:03
• Well, I was thinking of $d$ in the range of a few atomic layers or a few nm. At the same time, I am more used to electron diffraction, where $l_c$ is usually well above this regime. May 28 '20 at 8:16

Each layer contributes an increase proportional to the intensity of the original beam and a decrease proportional to the intensity of the diffracted beam (from multiple diffractions and extinction). In the approximation that the original beam passes through the sample with minimal losses (and the diffracted beams escape with minimal losses), the signal increases linearly with thickness. If the sample is thick enough that the losses become significant, then you will need to use a more involved expression.