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Beta ray maximum energy

Physics Asked by Frogfire on November 26, 2020

I recently participated in a classical experiment that can calculate the maximum energy of a beta radiation emission (for Tl-204). When I searched the internet for other results for comparison, I found results ranging from 0.7 to 3 MeV. Why is the range so big? Is the maximum energy dependent on time?

Thanks in advance.

2 Answers

I think the difference has got to do with the different modes of beta decay. There are two types:

  1. β− decay (electron emission) with associated reaction: $n → p + {e}^-+{nu}_{el}$

This process is mediated by the weak interaction. The neutron turns into a proton through the emission of a virtual W− boson. At the quark level, W− emission turns a down quark into an up quark, turning a neutron (one up quark and two down quarks) into a proton (two up quarks and one down quark). The virtual W− boson then decays into an electron and an antineutrino.

enter image description here

  1. β+ decay (positron emission) with associated reaction: $prightarrow n+ e^+ +bar{{nu}}_{el}$

In the Tl-204 experiment, we have to do with the first mode of beta decay. Assuming the uncertainty in energy isn't due to instrumental uncertainties (which of course they always are but not to such an extent as mentioned), the energy variation has to have the energy of the neutron.ve its cause in the decay mechanism. Obviously, the energy of the electron is dependent on the energy of the neutron and the proton. The higher the energy of the neutron (which depends on which nuclear orbital the neutron finds itself: see the shell model) the higher the energy of the proton and/or electron, after the decay. I don't have the knowledge of the precise mechanism (I don't know if the decay occurs when the neutron falls back to a lower shell with less energy, or if the decay is initiated by other means, but I think it's the former; don't blame me for that please). The more energy is transferred to the proton, the less to the electron. As @PM2Ring (sounds like an M5 spy) rightly observed one also has to take into consideration the fact that there will be various energy differences between the electron and its accompanying neutrino.
According to the former answer, the energy will have a maximum of about $764keV$.
See also this Wikipedia article.

Answered by Deschele Schilder on November 26, 2020

The two main sources I use are the Evaluated Nuclear Structure Data Files and the Evaluated Nuclear Data Files. For this problem, use ENSDF.

Search on 'Nuclide or Mass' = 204.

Scrolling down the long list of different data available, one finds three:

Under $^{204}$Hg there is '204TL EC DECAY'

Under $^{204}$Tl there is '204TL IT DECAY (61.7 US)'

Under $^{204}$Pb there is '204TL B- DECAY'

Tick the box next to the ones you want to look at (here really only the $beta -$ decay), grab the PDF, and you will see:

enter image description here

Looking at the Q-value, the energy (for the 3-body result) is 763.76keV.

There are no options getting anywhere close to 3MeV, so I don't know where that number is coming from (the electron capture path to $^{204}$Hg has a Q of 344keV).

Answered by Jon Custer on November 26, 2020

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