Am I visualizing/understanding atomic form factor correctly?

Question

I'm having difficulty visualizing the concept of atomic form factor or atomic scattering factor as it has been presented in a chemistry course on X-ray crystallography that I'm taking. At the moment, I visualize it as such:

(Please note that the question marks in the diagram above are included to communicate my confusion.)

Am I visualizing this correctly? Is there any other way for me to interpret these diagrams? The course doesn't dive into the mathematics of atomic form factor (and neither does our accompanying text), so I don't have much to go on in developing a mental image of the phenomenon at play here--and I strongly suspect that the mental image that I do have is wrong since I can't find any evidence to support it.

Answer 1

In your illustration, there seems to be a little «almost step». But more importantly, if displaying a heavy scatterer (iron, Fe) next to a weaker one (C), you should put a bit more emphasis on the descent for iron. This is one of the reasons why the atomic position of the heavier atoms are easier to attribute in the Fourier map, than the lighter ones.

Take a look at IUCr's illustration from their freely accessible on-line dictionary:

(credit)

Maybe your school has the printed or / and electronic version of the IT (the International Tables of Crystallography) which goes into further detail (e.g., listing $f$ and $f'$ for the elements, obviously dependent on the wavelength of radiation used during the diffraction experiment).

Because currently many schools' libraries may restrict access and borrow the books in person, note that IUCr moderates a dedicated section of teaching pamphlets, too. Topics include e.g., crystal packing and symmetry.

Answer 2

I agree with Ivan, your coloured circles are misleading. The factor is a ratio, a number, and is scattering relative to a single electron's scattering and as such is independent of any radial distance $r$ in the (spherical) atom. It depends on the angle that scattering is observed at and also on the total number of electrons. $f=\int_o^\infty U(r)\sin(\mu r)/(\mu r) dr$ where $U(r)=4\pi r^2\rho(r)$ and $\mu=4\pi\sin(\theta)/\lambda$ making $f$ after integration only a function of $\mu$ which as you see depends on the scattering angle. The integration removes any dependence on radial distance; $\rho(r)$ is the electron density at $r$. As $f$ depends on the number of electrons in an atom, it is larger for heavier atoms, i.e. more electrons more scattering.

Answer 3

There are two major differences between carbon and iron in a diffraction experiment. First, iron has more electrons. Second, most electrons in iron (18 of 26) are tightly bound inner electrons, so they are close to the nucleus. On the other hand, most of the electrons in carbon are valence electrons, so its electron density is more diffuse.

If you need an analogy or a mental image, iron is like a peach and carbon is like a strawberry. You should not have different mental images of the objects depending on wavelength - it is the same object. If you need some hard data, you can Fourier-transform the form factor to get the radial electron density of the atoms of the two elements.