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Hi, I'd like to demonstrate the Rutherford scattering cross section formula, assuming that the nuclei of the gold foil are not punctual charges, but uniformly charged spheres. The additional difficulty will be the fact that the electrical potential generated by a nucleus will have two expressions, one inside the nucleus and another outside. In fact, if you assume the nucleus as a uniformly charged sphere, the potential energy will be:

$$ U(r) = \left\{ \begin{array}{rl} \ k/r & \mbox{if } \; \; r > R \\ \\ kr^2/(2R^3) & \mbox{if } \; \; r < R \end{array} \right. $$

Where $k= \frac{q_1q_2}{4 \pi \varepsilon_0}$ and $R$ is the ray of the nucleus. The second expression (for $r<R$) was calculated with the Gauss' Law.

As you can see in the figure, the scattering angle $\alpha$ can be written as:

$$ \alpha= \pi - 2 (\gamma + \beta)$$

For the cross section I need the impact parameter $b$ as function of the scattering angle $\alpha$ (namely $b(\alpha)$). In fact, the cross section in terms of solid angle is:

$$ \frac{d \sigma}{d \Omega} = \frac{b(\alpha)}{\sin \alpha} \left| \frac{d \, b(\alpha)}{d \alpha} \right| $$ Now, I can try to use the energy and the angular momentum conservation ($dE/dt=0$ and $dL/dt=0$).

If I use a polar coordinate system on the plane perpendicular to the angular momentum, I have:

$$L = m \rho^2 \dot{\theta}$$

$$ E = \frac{1}{2} m \dot{\rho}^2 + U(\rho) + \frac{L^2}{2m \rho^2} $$

Hence:

$$ \frac{\dot{\theta}} {\dot{\rho}} = \frac{d \theta}{d\rho} = \pm \frac{L^2}{m \rho^2} \frac{1}{\sqrt {\frac{2}{m}(E- U_{eff}(\rho))}} $$

Where $U_{eff}(\rho)=U(\rho) + \frac{L^2}{2m \rho^2}$

In this way I can integrate $\theta$ with respect to $\rho$ and calculate $\gamma$ and $\beta$ in order to finally determinate $\alpha$ (in fact, $ \alpha= \pi - 2 (\gamma + \beta)$). Once calculating $\alpha$ I just have to hope that you can find an analytical expression for $\alpha(b)$, hence $b(\alpha)$.

In particular, using the two different expressions for $U_{eff}(\rho)$, I have:

$$ \gamma = \int_{\rho_{min}}^R \frac{1}{\sqrt {\frac{2}{m}(E- U_{eff}(\rho))}}\; d \rho $$

And

$$ \beta= \int_{R}^{+\infty} \frac{1}{\sqrt {\frac{2}{m}(E- U_{eff}(\rho))}}\; d \rho $$

Where I named $\rho_{min}$ the minimum distance reached by the particle from the center of the nucleus.

But that's the problem! How on Earth can I calculate these two integrals?!

Maybe there's an easier way with which you don't need to calculate them. What do you think?

Thank you in advance

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  • $\begingroup$ So, basically, you want to perform the calculation that Rutherford did in his original paper of 1911 , which included the term you propose? $\endgroup$ – Jon Custer Jun 21 '17 at 19:53
  • $\begingroup$ Yes. Have you got any idea? $\endgroup$ – Pietro Meloni Jun 21 '17 at 20:09
  • $\begingroup$ Well, you could start by reading Rutherford's paper - London, Edinburgh and Dublin Philosophical magazine, Vol 21, issue 125 669-688 (1911). $\endgroup$ – Jon Custer Jun 21 '17 at 20:23

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