What Causes Beta Decay to Be More Common in Heavy Elements?

pyo
Messages
3
Reaction score
0
Hey i was just wondering if anyone could help me out and shed some light into " why is B^- decay (emission of an electron) far more common among the heavy elements than the B^+ (emission of a positron) or electron capture decays?" I am looking at a plot of Z vs. N for various nuclei and believe it is the focal point of this question and how the positron of nuclei is moved by each type of decay but i still cannot see why the Beta ^- decay is the far more common type. What am i missing here? thanks!
 
Last edited:
Physics news on Phys.org
Is this homework? We can't solve your homework for you, but I think I can give you two effects to think about.

The B^- decay must always create an electron and that means that an energy gap that works for electron capture doesn't work for electron emission.

If the energy released in the decay is less than the mass of an electron, then the B^+ decay can only be an electron capture and not an emission of a positron. That means that the rate will depend on the density of the electron wave functions at the nucleus.

By the way, this last fact means that the decay rate of some atoms depends on the chemical conditions of the atom. In particular, if you strip off all the electrons, you can suppress electron absorbtion and transform a radioactive atom into a stable one.

Carl
 
Last edited:
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top