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so this is bothering me a tad but why during calculations, do we subtract the masses of all electrons from both the original atom and the "new atom" and then subtract the mass for the emitted electron?

Let's say we have the beta decay:

$$\rm ^{14}_{6}C \to {}^{14}_{7}N + \beta^- + anti-neutrino$$

We neglect the mass for the anti-neutrino, so in the solution we have then the masses of both the C and N atoms and the electron emitted, and we subtract N and e^- from the C mass, but they also subtract all electron masses involved for both atoms,that is 6 electrons for C and 7 for N, why? And if so why doesn't the formula for beta decays show this too?

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  • $\begingroup$ I mean, you can stick 6 electrons on each side of the equation if you like, but they don't really do anything. Their rest masses are always going cancel and their energies change by, at most, a few hundred eV, while the rest mass of the $\beta^-$ is around 500 keV, so its pretty negligible. (The Nitrogen is created as an $N^+$, ion, with one missing electron. It does not gain a complete set of electrons from nowhere. This would violate charge conservation) $\endgroup$
    – By Symmetry
    Jan 9, 2018 at 17:21
  • $\begingroup$ So far I understand the concept at least and why, having more trouble setting it up mathematically now... $\endgroup$
    – Myzanthros
    Jan 9, 2018 at 19:31
  • $\begingroup$ Do you have a specific question, as opposed to simply stating what "they" do, then asking "why"? That's not a good physics question without explaining your expectation. When you say "in the solution" what solution are you talking about? $\endgroup$
    – Bill N
    Jan 9, 2018 at 22:54
  • $\begingroup$ I simply wanted to know why my book included the electron mass in the calculations only to ignore them at the same time and then I was also wondering how one can calculate this mass defect without the mass given for Nitrogen but since every tells me that they should give me the mass for N then I'm guessing my book wants me to look it up which is just stupid since you don't get to do that during the exam $\endgroup$
    – Myzanthros
    Jan 10, 2018 at 11:40

2 Answers 2

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The full equation is as follows

$$\rm ^{14}_{6}C \to {}^{14}_{7}N^+ + {}^{\;\;0}_{-1}\beta^- + {}^{0}_{0}\bar \nu$$

Now $\rm {}^{14}_{7}N^+ + {}^{\;\;0}_{-1}\beta^-$ is the same as $\rm {}^{14}_{7}N^+ + {}^{\;\;0}_{-1}e^-$ which is the same as $\rm {}^{14}_{7}N$ except for the binding energy of the electron in the $\rm {}^{14}_{7}N$ atom.

The tables are for atomic masses and the assumption which is made is that the binding energy of the electrons orbiting the nucleus is very much smaller than the binding energy of the nucleons within the nucleus.
There is thus no need to do any subtraction of electrons.

If you do your counting of particle this decay is equivalent to $$\rm ^{1}_{0}n \to {}^{1}_{1}p^+ + {}^{\;\;0}_{-1}\beta^- + {}^{0}_{0}\bar \nu$$

ie a neutron decaying into a proton etc

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  • $\begingroup$ How can I calculate this if I don't get the atomic mass for N? I only got it for C, and like Jasper said below, the masses just don't add up. This is what my solutions book writes to calculate (Write it below) $\endgroup$
    – Myzanthros
    Jan 9, 2018 at 19:24
  • $\begingroup$ @Myzanthros You do need the atomic mass of nitrogen to find the mass defect and hence the energy released for this reaction. $\endgroup$
    – Farcher
    Jan 10, 2018 at 6:13
  • $\begingroup$ Yeah I understand it now but that is what made me confused on that exercise since they don't give me the mass for Nitrogen so I was a bit lost on how I would solve it. $\endgroup$
    – Myzanthros
    Jan 10, 2018 at 11:41
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The notation

$$\rm ^{14}_{6}C \to {}^{14}_{7}N^+ + {}^{\;\;0}_{-1}\beta^- + {}^{0}_{0}\bar \nu$$

is not about masses but rather (relevant) particles. You start with one $\rm{C}$ atom that brings an appropriate number of electrons with it and end up with a $\rm{N}$ atom with the proper number of electrons attached plus the emitted electron and neutrino.

The masses don't even add up:

  • Carbon-14: $14.003241 \rm{~u}$
  • Nitrogen-14: $14.0030740048(6) \rm{~u}$
  • (source: wikipedia)

And they shouldn't, because the "missing" is converted to energy (excitation of the nitrogen atom and kinetic energy of the electron and maybe the neutrino).

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  • $\begingroup$ Then how can I find the mass defect if I don't have the mass given for Nitrogen? This is my biggest issue... $\endgroup$
    – Myzanthros
    Jan 9, 2018 at 20:14
  • $\begingroup$ I don't understand that comment (or the question) completely, but does this physics.stackexchange.com/questions/308617/… help? $\endgroup$
    – Jasper
    Jan 9, 2018 at 20:39
  • $\begingroup$ Yes this is basically what my question was, the problem was in my case that I did not receive the mass for Nitrogen hence the confusion about how to solve it. $\endgroup$
    – Myzanthros
    Jan 10, 2018 at 11:42

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