Estimating Nuclear Radius Through Closest Approach, Can't We Just Give The Alpha Particle Whatever Kinetic Energy We Want?

When you fire an alpha particle at a nucleus, the electrostatic forces repel each other, causing it to slow down to a stop before accelerating back in the opposite direction. So the initial kinetic energy is equal to the electrostatic potential energy it has at the point of instantaneous rest, and then you can rearrange the equation to look like this: $$r=\frac{1}{4 \pi \epsilon_0}\frac{Q_{\alpha}Q_n}{E_k}$$ The concept makes intuitive sense to me, you should be able to deduce something about a nucleus' radius by firing positively charged particles at it. But the equation has me confused, can you not just fire the alpha particle with a greater velocity, thus greater kinetic energy, and now your estimation for the radius will be smaller?

So if you choose the velocity at which the alpha particle is released carefully, you could make the estimation equal whatever you want it to equal?

• Well, one keeps increasing the $\alpha$ energy until the scattering process stops being Rutherford-like. Then you know something changed. (Of course, one keeps increasing it to fully map out the scattering process to get detail on excited nuclear states, induced nuclear reactions, and whatnot - classic nuclear physics stuff). – Jon Custer Jul 11 '18 at 13:15
• Of course once you have accelerators you can just dial the beam energy as needed (within the energy limits and precision of you machine, of course). But at the very beginning they had to settle for naturally available sources which simply come with the line energies they come with. There is no dial to turn. – dmckee --- ex-moderator kitten Dec 20 '19 at 16:30