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as derived it is: \begin{equation}F(q^2) = \frac{3}{x^3} (\sin x-x \cos x),\quad x=\frac{qa}{\hbar}, \quad q=2p\sin(\theta/2)\end{equation}

I have

\begin{equation} p=400 MeV/c\end{equation}

\begin{equation}\theta=15\end{equation}

\begin{equation}A=48\end{equation}

\begin{equation}a=1.2 A^(1/3) fm\end{equation}

i have obtained

\begin{equation}q= 104.42 MeV/c\end{equation}

\begin{equation}<r^2> = 11.41 fm^2\end{equation}

\begin{equation}x = 2.3116\end{equation}

\begin{equation}F(q^2) = -0.55 \end{equation}

and if i apply a different formula:

\begin{equation}F(q^2) = 1- \frac{q^2<r^2>}{6\hbar^2} ,\quad \end{equation} \begin{equation}F(q^2) = 0.47 \end{equation}

shouldn't they both be equal?

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  • $\begingroup$ Form-factor can be of any sign. $\endgroup$ – Vladimir Kalitvianski Feb 1 '15 at 17:20
  • $\begingroup$ Your derivation was wrong, check it. $\endgroup$ – Vladimir Kalitvianski Feb 1 '15 at 17:56
  • $\begingroup$ Hello vladimir. these derivations were taken from the text though. $\endgroup$ – rebc Feb 1 '15 at 17:59
  • $\begingroup$ Then, maybe, you have two different formulas describing just different cases. $\endgroup$ – Vladimir Kalitvianski Feb 1 '15 at 18:05

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