I was reading Introduction to Nuclear Physics by Krane and stumbled on the following (page 47):
In Elastic scattering, the initial electron wave function is of the form $e^{i k_i r}$ (free particle of momentum $p_i = \hbar k_i$). The scattered electron can also be regarded as a free particle of momentum $p_f = \hbar k_f$ and wave function $e^{i k_f r}$.
The interaction $V(r)$ converts the initial wave into the scattered wave; the probability for the transition will be proportional to the square of the following quantity:
$$F(q) = \int V(r) e^{iqr}dv$$
Plugging both Coulomb's potential and charge-per-unit-volume into $F(q)$:
$$F(q) = \int e^{iqr'} \rho(r') dv'$$
Normalizing and knowing that $\rho(r')$ just depends on $r'$ (and not on $\theta'$ nor $\phi'$) we get:
$$F(q) = \frac{4\pi}{q}\int r' sin (qr') \rho(r') dr'$$
Where $q = k_i - k_f$. The scattering is elastic, so momentum is conserved ($p_i = p_f$) and $q$ is merely a function of the scattering angle $\alpha$ between $p_i$ and $p_f$.
Now a bit of vector manipulation shows:
$$q = \frac{2p}{\hbar}sin(\frac{\alpha}{2})$$
I do not know how to get the last expression
Just draw a diagram of two vectors of length $p$ with angle $\alpha$ between them, and find the difference of the vectors.