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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)?

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

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