Rutherford Alpha Scattering

Question

Why is the orbit of an alpha particle in the Rutherford gold foil scattering experiment a hyperbola?

The equation of orbit of a two-body system under the influence of gravity ($F_g = -\gamma/r^2$) is given by

$$r(\phi) = \frac{c}{1+\epsilon \cos\phi} $$

where

$$c = \frac{\ell ^2}{\gamma\mu}$$ with $\ell$ the angular momentum and $\mu$ the reduced mass of the system. $\epsilon$ turns out to be the eccentricity of the orbit

For a hyperbolic orbit, $\epsilon > 1$ or $E > 0$ where the energy $E$ is

$$E = \frac{\gamma ^2 \mu}{2\ell ^2}(\epsilon ^2 -1)$$

All these equations were derived given that the force of interaction is $-\gamma/r^2$. In Rutherford's case, the force of interaction is electrostatic between the positive nucleus and positive alpha, so $F = +\gamma/r^2$.

From this alone can we use the above equations to see that the orbit of the alpha particle is a hyperbola? If you sketch the alpha particle and the nucleus, it certainly looks like the orbit is hyperbolic, but how do we know it's not parabolic ($\epsilon = 1$ for parabola). How do we know $\epsilon > 1$?

Answer 1

Simple!

1. If $\epsilon < 1$ then it is a circle or a ellipse. In this case, happened no scattering. On the contrary, the particle were absorbed.

2. If $\epsilon = 1$ it is a parabola. There's only one value of energy (for a given angular momentum) for the orbit to be a parabola. Therefore, mathematically, the probability of the orbit is a parabola it is 0%. Therefore, almost surely a parabolic orbit event won't happen.

Besides, from a practical point of view, if you want to do scattering experiments you want to avoid absorption. Therefore, you want the energy to be far away from the limit between scattering and absorption. And since the limit is a parabola, you will want to be away from it.

Answer 2

When the alpha particle is far from the nucleus the potential energy of the system is zero, so the total energy $E$ is the kinetic energy of the alpha particle, which is greater than zero ($E > 0$). From your equation this requires that $\epsilon > 1$, therefore the trajectory is a hyperbola.