# Why is Bragg diffraction small-angle for thicker layers and/or larger molecules or unit cells? And wide-angle for small or shallow crystals?

It is easy to find equations that quantify this on the internet, but not an explanation as to why...

Also, does this apply to electrons and neutrons? Or just X-ray reflection/diffraction?

• you know diffraction of light on a grid or double slit? the coarser the grid, or the larger the distance of the slits, the smaller the angle,. This is exactly the same. Aug 21, 2020 at 20:32
• @trula that's an answer Aug 21, 2020 at 22:32
• I noticed that you haven't voted on, commented on, or otherwise interacted with the answer that's there already. If you need an answer with a different approach, what would that be?
– uhoh
Sep 12, 2020 at 3:57

It has to do with optical laws, regardless of the wave you use.

Bragg's law gives you the angles in which there is a maximum, but it doesn't say anything about the intensity you will detect on them.

If you perform the calculations, you'll find out that the intensity of the maxima rises with $$N^2$$, but the width decreases as $$1/N$$.

You can check this in a simple 1D model. Consider a chain of $$N$$ equally spaced atoms at a distance $$a$$. They're indexed from $$0$$ to $$N-1$$. Any lattice vector is $$R=n\cdot a$$

The intensity would be the sum $$S$$ squared.

$$S=\sum_{n=0}^{N-1} e^{-i \Delta k\cdot R}=\sum_{n=0}^{N-1} e^{-i \Delta k\cdot (na)}=\sum_{n=0}^{N-1} \left(e^{-i \Delta k\cdot a}\right)^n=\dfrac{e^{-i \Delta k\cdot Na}-1}{e^{-i \Delta k\cdot a}-1}$$

$$S=e^{-i \Delta k\cdot (N-1)a/2} \cdot \dfrac{\sin(N\Delta k a/2)}{\sin(\Delta k a/2)}$$

Taylor to First order yields

$$I\propto |S|^2 = 1\cdot \left| \frac{N k a/2}{k a /2 } \right|^2 \propto N^2$$

On the otehr hand, the distance between the two first minima is proportional to $$1/N$$, so the width of the peak decreasses with $$1/N$$.