# Atomic form factor reduction from Mott Cross-section

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

I have

$$p=400 MeV/c$$

$$\theta=15$$

$$A=48$$

$$a=1.2 A^(1/3) fm$$

i have obtained

$$q= 104.42 MeV/c$$

$$<r^2> = 11.41 fm^2$$

$$x = 2.3116$$

$$F(q^2) = -0.55$$

and if i apply a different formula:

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

shouldn't they both be equal?

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