# Introductory Particle Physics - Form factor, charge distribution?

1. Oct 5, 2009

### jeebs

Here is the problem. I've been messing around with it for a while but I'm not sure if what I'm trying to do is the right way to approach this.

The form factor F(q) = $$\int\rho(\vec{r})e^{i\vec{q}.\vec{r}/\hbar}d^{3}\vec{r}$$ is the 3D Fourier Transform of the normalised charge distribution $$\rho$$($$\vec{r}$$).

For a simplified model of a proton's charge distribution, $$\rho(r)\propto$$ (e$$^{-r/R}$$)/r.

R can be considered as some characteristic "size" of the proton, setting the rate at which the charge dies away, but does not constitute a hard edge to the proton.

i) Find the constant of proportionality required to normalise $$\rho$$ correctly.
ii) something else that presumably needs the answer to i) first.

I am new to all this particle physics business, so I am in unfamiliar territory and I'm not sure how to approach this question. I've so far just aimlessly waded into this and ended up with a couple of sides of mindless mathematical messing around. This could be a simple question or a complicated one for all I know, so I thought i'd post it here before I bother my busy lecturer...

Anyway, my closest attempted solution:

-what I thought was that I should assume the proton has its highest charge density at its centre and it gradually fades away, uniformly in all directions.

-I'm also thinking that r must be the distance from the centre of the proton, so that as r tends to infinity, the charge density $$\rho(r)$$ approaches zero.

-I'm trying to find some constant of proportionality here, lets call it A, so that $$\rho(r) =$$ A(e$$^{-r/R}$$)/r.

-I'm thinking that if I do $$\int \rho(r) dV = 1$$ then I can solve for A, but this is as far as I have got, I'm struggling with how to take this integral any further.

Am I on the right lines, has anyone got any suggestions that would make my life easier?

thanks.

2. Oct 5, 2009

### gabbagabbahey

Looks fine so far, now just compute the volume integral using spherical coordinates, centered on the proton's center (r=0)....you should know how to do that.