I How can the e- escape by overcome the suck force in β- decay

kiwaho
Messages
71
Reaction score
0
Not like decay of β+ or EC(electron capture), in β- decay, the nucleus positive charges definitely exerts strong suck force to negative charged electron, and the heavier the nuclide, the stronger the force!

So, no coulomb barrier for β+/EC, but does for β- decay.

How can the electron in a β- radioactive isotope can overcome that super strong attractive force even in just a few energy Q(β-) of single digital KeV?

Also wondering why no significant influence between light nuclides and heavy nuclides for the said factor.

I guess it is the magic tunneling effect?

And supposedly in β-, the initial velocity at the exit point should be smaller than the velocity at a little far distance to exit point, because the suck force is inversely proportional to the square of distance. But in fact, it seems not like that.

Deducedly, in same energy Q(β), β+/EC should be easier than β-, that is why our universe is not symmetric in the abundance of quasi-stable isotope. For example, you can see the β- potential Rb-87 with abundance 28%, In-115 with abundance 96%, Re-187 with abundance 63% and Q(β-) only 2467eV, and so on, but nobody can enumerate high abundance of β+ potential nuclides. The God is really left-handed!

There is electron capture, why is not there positive electron capture, or say positron capture decay? I boldly predict it should have e+ capture along with β- decay. Let's work hard to prove it.
 
Last edited:
Physics news on Phys.org
It can.
The electron energy/momentum distribution follows from the Fermi Theory [which takes into account the coulomb interactions through the Fermi function, and results to different spectra for electrons and positrons].
 
kiwaho said:
How can the electron in a β- radioactive isotope can overcome that super strong attractive force even in just a few energy Q(β-) of single digital KeV?
The attraction between electron and nucleus is already taken into account in that energy. It is the remaining energy that is left after the electron escaped "to infinity".
If the energy is not sufficient for the electron to escape, the decay does not happen.

This has nothing to do with tunneling - there is no region of lower potential energy outside.
kiwaho said:
And supposedly the initial velocity at the exit point should be smaller than the velocity at a little far distance to exit point, because the suck force is inversely proportional to the square of distance. But in fact, it seems not like that.
That does not make sense.
 
  • Like
Likes kiwaho
mfb said:
The attraction between electron and nucleus is already taken into account in that energy. It is the remaining energy that is left after the electron escaped "to infinity".
If the energy is not sufficient for the electron to escape, the decay does not happen.

This has nothing to do with tunneling - there is no region of lower potential energy outside.
That does not make sense.
I see now. Thank you mfb.
But for e+ capture, there should exist coulomb barrier. Maybe that is why no e+ capture decay?
 
e+ capture would need e+ around...
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top