- #1

jeebs

- 325

- 4

Use the fact that the form factor, F(q), is the Fourier transform of the normalised charge

distribution [tex]\rho[/tex](r), which in the spherically symmetric case gives

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

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

[tex]\frac{qR}{\hbar} << 1[/tex]

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

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

which when I integrate by parts leads to

[tex] 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) [/tex]

I am certain that I have done the integration correctly.

Then I come to the part where I make the approximation that [tex]\frac{qR}{\hbar} << 1[/tex] and this happens:

[tex] sin(qR/\hbar) \approx qR/\hbar [/tex]

and

[tex] cos(qR/\hbar) \approx 1 [/tex]

which gives me

[tex] F(q) = \frac{4\pi \hbar^2 \rho R}{q^2} - \frac{4\pi \hbar^2 \rho R}{q^2} = 0[/tex]

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.