Spectrum of beta decay

Nice question!

You don't say what $$\beta^+$$ and $$\beta^-$$ spectra these are (and, BTW, if this is from a book, you should credit the author). If these are from a single odd-odd nucleus that undergoes both $$\beta^+$$ and $$\beta^-$$, then there is one very simple reason why the two spectra should look different, and that is that the amount of binding energy released in the two cases will be a little different. But that would actually rescale the energy axes relative to one another rather than producing this kind of additive effect.

For comparison, let's consider the neutrino, which is uncharged. The spectrum of neutrinos emitted in beta decay is a bell curve that goes to zero at zero energy, and this is basically because of the number of available states, which is maximized if the beta and the neutrino share the available kinetic energy roughly equally.

If we model beta decay in purely classical terms as a process in which the beta and neutrino are released in a pointlike event inside the nucleus, then we don't get a prediction that matches the data. In such a description, we would expect the $$\beta^+$$ and $$\beta^-$$ to be shifted to the right and left by an amount equal to the electrical potential energy. Since a couple of answers have linked to the page by Watkins at SJSU, let's use his example of 64Cu, which exhibits both $$\beta^+$$ and $$\beta^-$$ decay. The radius $$r$$ of this nucleus is on the order of 4 fm ($$4\times 10^{-15}$$ m). Therefore the potential energy lost or gained by the $$\beta^+$$ or $$\beta^-$$ on the way out would be $$U\sim kZe^2/r\sim10\ \text{MeV}$$. This is totally incompatible with observation. For one thing, it predicts that $$\beta^-$$ decay could never happen, since the $$\beta^-$$ can't be created with this much energy. It also predicts that $$\beta^+$$'s would be detected with huge energies, which is not the case.

To understand what's really going on, we need quantum mechanics, not just classical physics. The amount of kinetic energy that is available to a beta based on conservation of energy is on the order of 0.3 MeV. This gives it a de Broglie wavelength of about 2000 fm, which is hundreds of times greater than the size of the nucleus. Therefore the beta cannot be much better localized than that when it is emitted. This means that we should really use something more like $$r\sim 500\ \text{fm}$$ (a quarter of a wavelength) in our calculation of the electrical energy. This gives $$U\sim0.08\ \text{MeV}$$, which is about the right order of magnitude compared to observation.

A byproduct of this analysis is that a $$\beta^+$$ is always emitted within the classically forbidden region, and then has to tunnel out through the barrier. It's a counterintuitive fact about quantum mechanics that a repulsive force can hinder the escape of a particle. This is also observed in alpha decay.