# Calculating Q-Value for given Beta- decay

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1. Nov 29, 2014

### hasankamal007

• Missing template due to originally being posted in different forum.
Hello,
I've recently started studying Nuclear physics and would appreciate help from someone who could clarify my doubt.

Given Problem:
19O -> 19F + e + v
Calculate the Q-value in the given decay using following data:
Atomic masses:
19O 19.003576u
19F 18.998403u

The problem that I'm having is this:
In the reaction I try to apply E=Δmc2 and I take Δm = mass(19O) - [mass(19F)+mass(electron)]
But the solution of the problem, given in my textbook, ignores mass of electron.
I don't get the idea of why we have to leave the electron's mass when the electron is there as the product of the given reaction on the right hand side.

Book's solution:
Q = [mass(19O)-mass(19F)]c2

Moreover, going through other such questions given in my textbook I have inferred that book's solution always takes Δm of the NUCLEUS of the atoms involved rather than take Δm of the whole reaction that is given.
I think I'm missing out on something conceptual here.
Any help will be appreciated.

2. Nov 29, 2014

### Staff: Mentor

The atomic masses usually include the right number of electrons, 8 for oxygen and 9 for fluorine.

3. Nov 29, 2014

### hasankamal007

Yes, but at the right hand side of the equation we have a total number of 10 electrons(9of fluorine and 1emitted) and on the left 8 electrons(of oxygen). So, Δm should include the change in mass DUE to change in NUMBER of electrons from LHS and RHS. But the solutions does'nt agree with this notion.

4. Nov 29, 2014

### Staff: Mentor

You have 9 electrons at the right side. The 8 from oxygen plus the new one from the decay.
The fluorine will be an ion until it finds an electron from the environment at some point.

5. Nov 29, 2014

### hasankamal007

Shouldn't we say that there are 10 electrons on the RHS? And you mentioned something about catching electron from environment, can you elaborate on that. Does that mean we have to subtract ONLY NUCLEAR MASSES and not the whole mass defect in the reaction given to find Δm?

6. Nov 30, 2014

### Staff: Mentor

No, there are not. You start with 8 and you produce one new. Where should the 10th come from?

This process produces a fluorine ion where one electron is missing. Long after this process happened (long on the timescale of nuclear reactions), the ion will likely find an electron, catch it and become a neutral fluorine atom again - probably as part of some chemical reaction. You do not have to care about this.

No. You have to take the total mass into account.

7. Nov 30, 2014

### hasankamal007

mfb, Thanks for the reply. I think I found what the problem is.
What you're saying is indeed right! We started with 8 and produced a new one.
So there should be 9 electrons on the RHS(fluorine atom would actually be bearing a positive charge). Thus, the reaction originally given is essentially NOT 'BALANCED'.
Thus, the actual 'BALANCED' reaction should be:

19O -> 19F+ + e- + v

Now applying Δm = mass(19O) - [mass(19F+) + mass(ELECTRON)]
⇒ Δm = mass(19O) - mass(19F)

which corresponds to the correct answer!
Hence, the original reaction was essentially not balanced.

Also, I wanted to ask whether an electron can contribute to the mass defect? In all the problems I've come across, electrons have never been contributing to the mass defect in the reaction.

8. Nov 30, 2014

### Staff: Mentor

Nuclear reaction formulas simply do not care about the electrons around the nucleus (with a few exceptions).

An electron is never part of the nucleus, so it cannot contribute to the mass of the nucleus. Electrons in some very deeply bound states (like the innermost shell for heavy atoms) can have a significant binding energy, however.

9. Nov 30, 2014