I was reading Nuclear Physics and the author mentioned something about the atomic form factor, something in relation to the Fourier transform of the spatial distribution of the electric charge. but I don't know what it means. What is the physical interpretation of the atomic form factor?
Answer
There are a variety of ways one could answer this. As you note, the form factor is the Fourier transform of the spacial distribution of the electric charge density. The subtlety with this definition is that for a composite system such as a nucleus (or atom, or nucleon), defining a "net charge density" that is just a function of spacial coordinates, i.e. $\rho(x, y, z)$, is an approximation. The exact description of an atom or nucleus is given by a many-body wavefunction.
That aside, $\rho(x, y, z)$ for an atom or nucleus can be roughly interpreted as the charge density associated with an effective potential that is arrived at by averaging over the motion of the constituent particles. The form factor $f(Q)$ is Fourier transform of this.
Consequently, one should expect the form factor associated with an atom to deviate from zero around $1/(10^{-10} \textrm{m})$ and $1/(10^{-15} \textrm{m})$. The former corresponds to electronic degrees of freedom (e.g. the "wavefunction" of the electrons) and the latter corresponds to nuclear degrees of freedom.
The form factor $f(Q)$ will provide a measure of the interaction of between the atom and an incident photon with momentum $Q$. Consequently, photons with a wavelength $\sim 10^{-10} \textrm{m}$ (i.e. visible or UV light) or $\sim 10^{-15} \textrm{m}$ (i.e. gamma rays) will interact strongly with the atom by coupling to the electronic or nuclear degrees of freedom, respectively.
One can also define other kinds of form factors that correspond to different sorts of interactions (e.g. magnetic, the strong force, etc.).
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