Why is the alpha particle in alpha decay considered to be in a potential well? I understand that when modelling alpha decay, it is useful to consider the $\alpha$ particle as being preformed, in a region confined to the daughter nuclei. I also understand that the term $V_{0}$ comes from the strong nuclear force.
My question is why this problem is treated like a potential well problem. The $\alpha$ particle's energy, $Q$, is stated to be less than $B$. But I don't understand why? Why does the Coulomb force drop off at the nucleus? Also since the alpha particles are positively charged, why wouldn't they spontaneously leave the nucleus once at the distance $a$? Why would they have to quantum tunnel to get across?
 A: The constant $V_0$ term is only an approximation to the strong nuclear potential, all that you need to know is that it is attractive (and thus the potential is negative) and this attraction is restricted to a finite range. The nuclear force is actually repulsive at very close distances and falls off with distance, but those are extra details to the problem that aren't relevant so we exclude them. It's not that the Coulomb force suddenly vanishes, it's that we are approximating the combined Coulomb and Strong interactions at that nuclear distance as just being flat, for simplicity's sake.
The particle's energy is $Q$ because it just is. An alpha particle in a nucleus has some non-zero energy, and that energy is less than the energy it takes to get another similar alpha particle to that same distance. You are correct in thinking that if the alpha particle had energy $B$ instead of energy $Q$, it would not need to quantum tunnel, it would simply just have the energy to escape. It is simply the case that the alpha particle just does not have this energy. If it did, well then it wouldn't be confined to being in the nucleus in the first place.
