Kinetic energy required for alpha particle to 'stick' to nucleus of a silver atom

Question

I am relatively new to nuclear physics and have just picked it up as a hobby. I found this question online and require some help.

"An alpha particle is incident on a silver nucleus, what is the minimum kinetic energy required in order to become stuck in the silver nucleus? Take the diameter of the nucleus to be 10^-15 m"

I have directly copied this from the review sheet (hence the quotations). I'm unsure if the question mistakenly refers to the silver atom as just its nucleus, as both the nucleus and the alpha particle are positively charged. Could someone please provide working to calculate the kinetic energy, for both the silver atom and the nucleus?

Additionally, would I be correct in saying that if the question is considering simply the nucleus, it would never become 'stuck', rather an external force is being applied against the alpha particle in it's collision with the nucleus.

I'm only 13 so go easy on me if i've made a mistake :)

Answer 1

The question is really about just the nucleus. Including the electron hull would make things much more complicated (because then you also have to include its effect on the electric field of the nucleus).

The reason that the alpha particle can stick to the nucleus are the short ranged nuclear forces (that keep the nucleus itself bound as well). At short ranges those forces are much stronger than the repellent electrostatic force. But for those forces to act, the alpha particle must get very close to the nucleus.

The question simplifies the specifics by assuming that the nucleus has a certain radius and that the alpha particle gets stuck when it reaches that radius. (Also, in the real world other things than sticking could happen depending on the specific nucleus, e.g. the binding energy released by the alpha particle can be enough to cause the nucleus to fission or to eject neutrons or small nucleon clusters).

With those assumptions the question reduces to finding the energy necessary to reach a distance of $10^{-15}\,\text{m}$ against the electrostatic forces between the nucleus and the alpha particle.

Ususally, such questions let you silently assume further things, e.g. that the mass ratio between the alpha particle and the nucleus is so low, that you can ignore the acceleration of the nucleus. (You could also visualize that, by imagining the nucleus pinned fixed in space). After all, such a calculation will be only a rough estimation in any case (because the real processes are much more complicated and require quantum mechanical treatment).