# Homework Help: The form factor

1. Dec 7, 2009

### jeebs

Here is the problem:

Use the fact that the form factor, F(q), is the Fourier transform of the normalised charge
distribution $$\rho$$(r), which in the spherically symmetric case gives

$$F(q) = \int \frac{4\pi \hbar r}{q}\rho(r) sin(qr/\hbar))dr$$

to find an expression for F(q) for a simple model of the proton considered as a uniform
spherical charge distribution of radius R.

Using your calculated expression for F(q), demonstrate that in the limit
$$\frac{qR}{\hbar} << 1$$
the form factor reduces to 1.

So, what I have tried so far:

I said that for 0 < r < R, the charge density is constant, and could be taken outside the integral along with the other constants, leaving me with

$$F(q) = \frac{4\pi \hbar \rho}{q}\int^R_0 r.sin(qr/\hbar)dr$$

which when I integrate by parts leads to

$$F(q) = \frac{4\pi \hbar^3 \rho}{q^3}sin(qR/\hbar) - \frac{4\pi \hbar^2 \rho R}{q^2}cos(qR/\hbar)$$

I am certain that I have done the integration correctly.

Then I come to the part where I make the approximation that $$\frac{qR}{\hbar} << 1$$ and this happens:
$$sin(qR/\hbar) \approx qR/\hbar$$

and

$$cos(qR/\hbar) \approx 1$$

which gives me

$$F(q) = \frac{4\pi \hbar^2 \rho R}{q^2} - \frac{4\pi \hbar^2 \rho R}{q^2} = 0$$

but I am supposed to be getting F(q) = 1 when I make this approximation.

What am I doing wrong here? My only thought was that the question mentions something about normalization, and I thought that may have something to do with something, but I couln't think what. Any suggestions?

Thanks.

2. Dec 7, 2009

### teapot

I used a charge density of e/volume (3 e /4 pi R^3) integrated over all space in spherical polars & normalised it then put it in the equation. It basically just got rid of the e

But i still get 0 for the 2nd bit!

edit: Hooray, i get 1 using 2nd order approximations for sin & cos

Last edited: Dec 7, 2009