5
$\begingroup$

I have a very simple question regarding the emission spectrum of a beta decay: enter image description here

Why does the $\beta^+$-spectrum start at $0$ but the $\beta^-$ doesn't? This would mean that there are a bunch of electrons emitted with no energy at all. What is the explanation?

$\endgroup$
0

3 Answers 3

4
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

I deleted my previous answer because it was based on an interpretation without a backup in theory.

This link gives the status for the beta decay spectra:

betadecay

Here is a measured energy spectrum:

beta-

It accounts for the nuclear coulomb interaction which shifts this distribution toward lower energies because of the coulomb attraction between the daughter nucleus and the emitted electron. (It shifts the distribution upward for positrons.) Q represents the energy yield of the transition and as such is the upper bound on the kinetic energy of the electron.

Note that the zero kinetic energy intersect is an extrapolation from the fit of the curve with the Fermi function (cannot measure zero momentum electrons). The Fermi function takes into account all the quantum mechanical phenomena of this complex process of a decay within a nucleus and detection outside.

It should be noted that in all expressions special relativity formulas are used.

$\endgroup$
0
$\begingroup$

"The energy spectra of electron and positron beta-decay look quite different...However this observed difference can be accounted for by the fact that the electron is attracted by the nucleus once it is created whereas the positron is repelled."(http://www.sjsu.edu/faculty/watkins/betadecay.htm)

$\endgroup$
2
  • $\begingroup$ That answer doesn't satisfy me - can you elaborate that? Do you mean that the electron loses (potentially all of its) kinetic energy because it has to leave the attractive coulomb-potential of the nucleus? $\endgroup$
    – RedLantern
    Feb 24, 2019 at 16:27
  • $\begingroup$ @RedLantern : you can find more details in the provided link $\endgroup$
    – akhmeteli
    Feb 24, 2019 at 16:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.