Beta decay fundamental question

Smed
Messages
36
Reaction score
1
Can someone please clarify beta decay for me. As I understand it, in B- decay, a neutron turns into a proton, electron, and anti-neutrino, then the electron and anti-neutrino are ejected. But from this example, it looks like there are two additional electrons as a result of decay.

19/8 O -> 19/9 F + 0/-1 e + 0/0 v

You start with 8 electrons in the Oxygen atom, then a new electron is created from beta decay which would give 9 electrons. On the right side of the equation Fluorine has 9 electrons plus the beta minus electron = 10. Where is the 10th electron on the right hand side coming from?

Thanks
 
Physics news on Phys.org
This notation refers to nuclei, not atoms. The 9 means there are 9 protons.
 
When the oxygen-19 (neutral atom with 8 bound electrons) decayed to fluorine-19, the fluorine atom still had only 8 bound electrons.
 
Adding to Avodyne's comment, the number of electrons in a neutral atom is proportional the atomic number, Z, which gives the number of protons (+charges) in the nucleus. The 19 in those equations is the atomic mass in integer atomic mass units (amus). The electron mass to proton mass is ~1/1836, so it's departure from the nucleus does not substantially change the atomic mass.

A beta particle would slow down near (within cm) the original nucleus, and there would be a cascade of electrons along the ionziation path. Change in Z is balanced by that free electron so the net charge between O-19 and F-19 doesn't change on the atomic level.

See this regarding beta decay - http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/beta.html#c2
 
This helps, but I'm still a little confused. I thought that the electrons were always equal to the atomic number Z, otherwise how can you know how many electrons are in the atom? In other words, if the Fluorine atom from this reaction only has 8 electrons, then how is it distinguishable from 19/9 F which has 9 electrons?
 
Smed said:
This helps, but I'm still a little confused. I thought that the electrons were always equal to the atomic number Z, otherwise how can you know how many electrons are in the atom? In other words, if the Fluorine atom from this reaction only has 8 electrons, then how is it distinguishable from 19/9 F which has 9 electrons?
In a neutral state, there are Z electrons (of negative charge) balancing Z protons (positive charge) in the nucleus. With beta decay, the charge of the nucleus increases by 1, so the number of atomic electrons must also increase by 1, to remain neutral.

Usually what happens is the ejected electron (beta particle) slows down by ionization and excitation (by virtue of collisions with atomic electrons). Meanwhile the original atom (from where the beta particle originated) 'steals' an a electron from a neighbor, and there is a cascade until that all the electrons along the ionzation path, including the beta particle, have been reabsorbed into the neighboring atoms.

Similar, when an alpha particle (+2) is ejected, the new nucleus has a new charge (Z(orig)-2), so two atomic electrons are lost. The alpha particle passes through other atoms ionizing and exciting them, until it comes to rest, absorbs two electrons and become a He atom.


Also in the case of positron emission (+ beta), the original nucleus loses a + charge, so it also loses 1 atomic electron. Eventually, the positron annihilates with an electron, again maintaining neutrality. There is a cascade of electrons which migrate to fill the 'holes' in the atoms from the atom which lost its electron in the annhilation process back toward the original atom that emitted the positron.
 
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