# A form factor problem for high energy electrons

Assuming a uniform charge distribution within a nucleus of radius R=$$r_{0}A^{1/3}$$, show that the form factor for $$F(q)$$ for high energy electron is given by

$$F(q)=\frac{3}{q^2r_0^3A}\left[\frac{\sin(qR)}{q}-R\cos(qR)\right].$$

I started with the standard equation for the form factor. $$F(q)=\frac{4\pi}{Zeq} \int_{0}^{R} \rho(r)\sin(qr) dx,$$ solved the integral and got the $$\left[\frac{\sin(qR)}{q}-R\cos(qR)\right]$$ part. But, I'm not getting the correct constants. So, the final answer that I'm getting is

$$F(q)=\frac{3}{Zer^3_0q^2}\left[\frac{\sin(qR)}{q}-R\cos(qR)\right].$$ Can someone tell me how can I replace that $$Z$$ by $$A$$ and get rid of the $$e$$ in the denominator (which will solve the problem)?

• Are you sure the $r_0$ in your answer is the same as the $r_0$ in the problem and not what you call $R$? – jacob1729 Aug 14 at 16:55