Beta decay is presented in this lecture. The neutrino hypothesis and its relationship with beta decay is discussed. A review of Q value calculations for beta decay is provided. The importance of spin and parity, and how it can be used to assess beta decay, is discussed. Modeling beta decay through the weak force is provided.. The impact of Coulomb interactions on positron and electron spectral shape is presented. The use of Kurie plots in understanding beta decay is introduced. Selection rules in beta decay and beta transitions are explained. Calculating logft and its relation to spin and parity are presented. Double beta decay is discussed.
I had a question about your slide on comparative half-lives. The example there uses 14O -> 14N and I can't seem to get the same Q as you have listed there. I'm using the eq from the earlier slide for positron decay and I keep getting around 4.1 Mev where you have Q listed as 1.81 Mev.
ReplyDeleteMy numbers are Q = [14.008595285 - (14.003074005 + 2*0.000548579909)]*931.5 = 4.121 Mev
I get about the same thing using mass excess values. I'm getting the numbers from http://nucleardata.nuclear.lu.se/database/masses/
Any help would be much appreciated!
The Q value you have is correct. I will make this correction. Thanks
DeleteCorrection in lectures uploaded.
DeleteI never knew there was so much quantum mechanics involved in Beta Decay. Most of my nuclear engineering curriculum just skirts over that stuff. It was nice getting to see it in depth.
ReplyDeletethanks for the comment. We are looking to expresses some fundamental understanding of the material.
DeleteWhat is considered a high enough A where a nucleus will decay via a Gamow-Teller transition instead of Fermi transition?
ReplyDeletethe transition occurs in nuclei that are no longer Z and N symmetric. This change start around Cr, Z= 24. Above Cr you have higher probability of a nucleon transition during beta decay involving a change in shell states from the shell model. 67Cu has the unpaired 29th proton in p3/2, 3/2-. The daughter 67Zn has the unpaired 37 neutron transition from this state to f5/2, 5/2-. This is an allowed transition with no parity change, so it is Gamow-Teller.
DeleteWhat are the fundamental differences of beta decay versus double beta decay?
ReplyDeletelook at the A=100 isobar in the table of the isotope. 100Mo decays by double beta decay. A single beta would yield 100Tc, which is at a higher energy state. This transition is not spontaneous. However if 100Mo were to undergo double beta decay it would go to the lower energy state 100Ru. This occurs but with a longer half life. The fundamental different is the relative energy state of the next neighbor. If the next neighbor is lower in energy then a single beta decay occurs. If the next neighbor is higher energy, but the one after that lower, then double beta may occur.
DeleteSpeaking about how neutrinos were "discovered" to balance conservation laws reminded me that I actually had a lab on neutrino detection when I was taking quantum physics; it was very speculative.
ReplyDeleteNeutrinos are elusive. Thanks for the comment!
DeleteI learned a lot of new information relating to B-decay. I appreciate the fact that we are now getting more in depth into each decay type.
ReplyDeleteThanks. The idea is to provide some fundamental understand of why the decay occurs.
DeleteOn the topic of Positron Emission Tomography, it seems that areas with the highest metabolic activity (or most mitochondria) would theoretically yield the clearest 3D images, due to the electron transport occurring at these sites, and the availability of more electrons for the positrons to annihilate.
ReplyDeleteMaybe. You are more likely to enhance imaging from having higher concentration of radiopharmaceutical.
ReplyDeleteMe Ed and Carlie attempted to calculate number one on the quiz with the equation, we understand the answer but couldn't calculate it can you explain this?
ReplyDeleteAs Breanna stated above, we worked a while on number one, doing many calculations, mostly Q value, only to come to the conclusion that we don't necessarily have to do any calculations, this is because of the annihilation and conservation of rest mass, correct? Also, if we did want to use calculations to find that a positron is 511KeV in positron decay, could we use the Q value to calculate this?
ReplyDeleteYou understand it. The photon energy must come from the mass of the electron and positron. Since the photon has no mass all the energy comes from the particle annihilation. Two photons at 511 KeV from two particles. This is also discussed somewhat is pair production.
DeleteComments and answers to PDF quiz 5 posted.
ReplyDelete