# I Why use the mass of 2 electrons in calculation of Q for beta

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1. Feb 15, 2017

### Imolopa

When dealing with a beta ^+ decay in the calculation of the disintegration energy Q, one includes the mass of 2 electrons.

But the output result is 1 electron in the fundamental reaction:
p => n + ( e^+ ) + v

Where the neutron n has a neutral charge and the neutrino v has Z = 0.

So why does one use 2 electron masses in the Q calculations?

2. Feb 15, 2017

### ChrisVer

Hi, you may have a look here:
from/after Eq9.3

Last edited by a moderator: May 8, 2017
3. Feb 16, 2017

### gleem

In case you have not yet waded through the article ChrisVer suggested I offer this. As expected in nuclear decay the energy available (Q) for the emitted particle is determined by the difference in the nuclear masses of the parent and the daughter nuclei.

However, the masses that are provided in tables are the neutral atomic masses not nuclear. So for β- decay it turns out that the difference in the masses of the parent and daughter nuclei are the same as the difference in the masses of the parent and daughter neutral atoms. That is not true for β+ decay. So it you take the Q equation for β+ decay

Q = ZMAn - [ Z-1MAn + me] where the subscript "n" refers to the nucleus.

and add and subtract Zme and convert the equation to atomic masses you get

Q = ZMA - Z-1MA + 2me

4. Feb 16, 2017

### Imolopa

Thanks for the explanation, I tried reading the linked document, but can't seem to find an explanation/understand the reasoning for the lowermost question in this post.
However it's not intuitively clear to me yet.

As I understand it so far:
- During β- decay a neutron n from the nucleus is converted to a proton p (thus increasing the atomic number Z by 1 in the daughter), an electron e- and an approximately massless antineutrino.

So from the previous I further deduce that the total number of electrons increases by 1, since it came from the converted neutron n.

So the mass of the mother atom (nucleus including electrons surrounding it):
Mmother = mmothernucleus + Zmelectron [Z equals the number of electrons since the mother is neutral]

Then we can rearrange to get the mass for the nucleus:
Mmothernucleus = mmother - Zmelectron

However when it comes to calculating the mass for the daughter atom I see in textbooks that is written as:
• Mdaughter = mdaughternucleus + (Z+1) * melectron

In the case of a β+ decay, where a proton p converts to a neutron n (thus decreasing the atomic number Z by 1 in the daughter) , an positron e+ and an approximately massless neutrino. Which in textbooks yields an expression for the mass of the daughter atom of the form:
• Mdaughter = mdaughternucleus + (Z-1) * melectron

In the expression for calculating the disintegration energy as we can see the mass for the extra electron is included allready:
Q = mmothernucleus - ( mdaughternucleus + melectron)

Now I see how we end up with 2 masses for electrons when inserting for the nuclear masses from the above expressions in the β+ decay version.
However I don't understand why (Z+1) * melectron and (Z-1) * melectron are used instead of: (Z) * melectron and (Z) * melectron in the bulletpoints above. In my mind the extra electron has allready been accounted the last expression for Q.

5. Feb 16, 2017

### gleem

Starting with the nuclear masses

Q = ZMAn -[ Z-1MAn+ me]

Now and and subtract Zme

Q = ZMAn + Zme - Zme - [ Z-1MAn+ me]
= ZMAn + Zme - [ Z-1MAn+ Zme + me]
=ZMAn + Zme - [ Z-1MAn+ (Z-1)me +me+ me]

Convert to atomic mass notation

=ZMA + [ Z-1MA +2 me]

6. Feb 16, 2017

### ChrisVer

Intuitively, the extra electron from the beta- is going to be part of the daughter atom [which got an extra proton+that electron = atom neutral].... the extra positron of the beta+ is going to find an electron and be annihilated giving photons of energy $2m_e$ [the number of protons decreased, together with taking out an electron, so again going to a neutral atom + you had the annihilation which sends away the 2me]... I am open to corrections on this.

7. Feb 16, 2017

### gleem

The future annihilation of the positron should not and does not enter into this discussion. We do not know when or with what it will annihilate. There is no physics involved in the Q computation it is just a matter of book keeping. because of the data that is used in the energy balance equation.

As an example take the decay of C11 to B11+ The C11 on decay looses the one me The B11 produced at that instant end up with one extra orbital me which it discards to become neutral.. Since we use the mass of the neutral atom in this equation we must adjust the equation it with one extra me.

8. Feb 16, 2017

### ChrisVer

the energy conservation is only taken into account in my intuitive description....

9. Feb 16, 2017

### gleem

correct. That is why the use of atomic mass works.

I see what you are doing but I think the reasoning is not a good approach. The positron does not have to annihilate. The positron is ejected from the system for one me and an extra orbital electron is discarded to make the daughter neutral for the other me.

10. Feb 17, 2017

### Imolopa

Ok so as I can see you are adding and subtracting Zm twice, for simplicity we can say that it is the same as adding 1 and removing 1, and as a result not actually changing anything when it comes to the total sum/quantity as it remains the same. This leads me to guess that the method is used so that we can "convert the format" from the nuclear masses (excluding electrons since they are not part of the nuclei) and into the atomic masses (the whole atom which includes the nuclei and each of their corresponding surrounding electrons in the orbits). In short words nothing is changed in our "total" system, rather just the perspective.

Aah ok so this is where the extra electron mass comes from in the Q [disintegration energy], (for beta+) the one extra positron attracts an electron from the daughter cause of the attractive nature of opposite charges. Beeing antiparticles of each other they annihilate when coming in contact. Since we are calculating "energy" in Q and not masses, the vanishing 2 "e" return the energy equal to their masses, the output are the photons that are approximately massless and their energy is received from the actual masses (E = mc^2).
And for the beta-, since the extra electron becomes part of the daughter, its mass thus vanishes from Q expression, since the neutral atomic mass "notation" has it included within itself.

Ok so in other words we can disregard it's corresponding energies as they are not considered part of our system when not attached to the atoms. In short one can imagine these particles floating around outside our 3D-room of reference, a place in space considered irrelevant for our calculations.

Last edited: Feb 17, 2017
11. Feb 17, 2017

### ChrisVer

My explanation applies only as a rough way to think how the factor of 2 appears, it doesn't necessarily mean that the positron will be annihilated with an electron of the daughter atom. The only reason it works is mainly because of the conservation of energy (now when or how an electron+positron will annihilate is irrelevant). Even if the positron won't be annihilated with an electron of the daughter nucleus, it will be annihilated with some other electron elsewhere - and the electron that gets released from the daughter atom will be captured from the other atom that lost the electron (to become neutral again). In the end, it depends when you want to apply energy conservation, and of course all this works because we don't consider the differencies in binding energies (Which you also neglect in the determination of Q).

But, for example, you can see a beta+ decay of a nucleus using a Photomultiplier, and the signal appears as a large peak on the mass of the electron. Eg the spectrum of Na22, as detected by a NaI(TI), undergoing beta+ can be seen in the following link:
https://image.slidesharecdn.com/gam...using-na-itldetector-13-728.jpg?cb=1358093358

It means that the mass of the electron+proton is enough to give you the mass of the atom you have in the end.

The difference is as stated above by @gleem that the final atom in beta- has a free energy level (or energetically favorable) for the electron to occupy, while in beta+ you are losing one (e- gets freed and a e+ gets emitted by the nucleus).