# Classical Rutherford scattering (partial) derivation

I am having trouble answering the following question, please could you help! Thank you in advance for any assistance you can give.

Consider classical Rutherford scattering of a particle with mass $m$ and initial speed $v_0$ from a potential $$V(r) = \frac{\alpha}{r}$$ where $\alpha$ is some constant and $r$ is the location of the particle from the origin. Starting from Newton's second law, show that $$|\Delta {\bf{p}}| = \frac{2 \alpha}{v_0 b} \cos \Big(\frac{\Theta}{2} \Big).$$ Note that $b$ is the impact parameter and $\Theta$ is the scattering angle.

Please note that I have already shown that "from geometry" the change in momentum is $|\Delta {\bf{p}}| = 2 p \sin(\Theta / 2)$, and that $b v_0 = r^2 \frac{d \theta}{dt}$ where $t$ is time and $\theta$ is the angle $\angle({\bf{r}},{\bf{r^*}})$ where ${\bf{r^*}}$ is the point of closest approach. I am unsure however if the the above two equations will be of assistance.

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I think it best to go to the source, Ernest Rutherford, "The Scattering of $\alpha$ and $\beta$ Particles by Matter and the Structure of the Atom", London, Edinburgh and Dublin Philosophical Magazine and Journal of Science, Volume 21, Issue 125, pages 669-688 (1911). This is the paper in which he details what is now known as Rutherford scattering, with the full derivation of the cross section to be expected from an atom with a nucleus. A nice read.

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The full text for Rutherford's article is available online, too. – Alex Nelson Jul 30 '14 at 19:19
@AlexNelson - thanks for the link. Be aware that access may depend on your institution's agreements with Taylor and Francis (sigh). Annoying that access to an article well past copyright remains limited for many people. – Jon Custer Jul 30 '14 at 19:42

Certainly late for your homework problem, but figure below shows the schematic of elastic scattering. The scattering angle is $\Theta$. The momentum transfer vector is $\Delta \bf{p}$.

Since your question starts with potential, we shall obtain the force first:

\begin{align} F(r) &= \frac{dV}{dr} \\ &= -\frac{\alpha}{r^2} \end{align}

Since the force is central, at any time the force in the direction of $\Delta \bf{p}$ is $F \cos\phi$, so the total momentum transfer is \begin{align} \int_{\infty}^{\infty} F \cos{\phi}\,dt \end{align}

Using one of your information $b_0=r^2\frac{d\phi}{dt}$ (which is obtainable from conservation of angular momentum) we make the change of variable

\begin{align} dt = \frac{r^2}{bv_0}d\phi \end{align}

and the integration becomes

\begin{align} \int_{-\frac{\pi-\Theta}{2}}^{\frac{\pi-\Theta}{2}} \left(-\frac{\alpha}{r^2}\right)cos{\phi}\,\frac{r^2}{bv_0}d\phi &= \frac{\alpha}{bv_0}\left[\sin\phi\right]^{\frac{\pi-\Theta}{2}}_{-\frac{\pi-\Theta}{2}} \\ &= \frac{2\alpha}{bv_0}\cos{\left(\frac{\Theta}{2}\right)} \end{align}

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