system
Given the following post, respond as if you're commenting on Reddit:
human
Why in binding energy calculations do we include emitted neutrons but not electrons?

(not asking for help on a specific question)

In the first part of the decay a neutron is fired into a nucleus the decays and produces 2 daughter nuclei and some other emitted neutrons. 
We include these in the binding energy calculations. 

After this the two daughter nuclei decay via beta emission.  Producing emitted electrons.
We don't include these in the calculations

When I say calculations mean using binding energies and mass difference.

In both cases the particles are emitted and not part of a nucleus. 
Is it because the neutrons have nuclear forces between the quarks inside them?

Is it because the neutrons were initially a part of the nucleus? But then what about neutrons that are fired into the nucleus?

Do they count as being a part of the nucleus too?
gpt
>Producing emitted electrons. We don't include these in the calculations

We **do** include the beta particle masses in calculating beta decay Q-values. The convention in nuclear physics to use *atomic* masses (meaning the mass of the nucleus, plus masses of all electrons, minus the very small electronic binding energy). If you work out beta decay Q-values in terms of atomic masses, you find that actually the beta particle mass cancels in the beta^(-) Q-value, because the atomic mass of the daughter includes an extra electron. In the case of beta^(+), the effect is opposite, and you find that there's actually a threshold for beta^(+) decay of 2m*_e_*c^(2) because not only do you have to produce a positron, but the atomic mass of the daughter is short one electron.

Electronic binding energies are very small on nuclear physics scales, so they're often neglected when they show up explicitly in calculations.