Really simple question, but I'm confused with it because of contradictions between sources (absence of clarity online vs lecture notes (says nuclear) and practise questions provided (uses atomic) )

In the reaction energy: $$Q=E_{R,i} - E_{R,f} = K_f - K_i$$ Does one use the atomic masses or the mass of the nucleus (like what is needed for binding energy?) when solving for this?

Thanks very much in advance!

Edit: to supply clarity, I have the reaction:

$$N^{14}_{7} \rightarrow C^{14}_{6} + e^+ + \nu_e$$

I calculated $ΔB$ as $$ΔB = (m_n - m_p - m_e + m_N - m_C) c^2$$ Where I have the masses of Neutron, proton, electron, and the atomic masses of Nitrogen and Carbon, respectively.

I had to prove $Q = ΔB + m_ec^2$ with this, So I did: $$Q = (m_n + (m_N -7m_e))c^2 - (m_p + (m_C - 6m_e))c^2$$ $$Q = (m_n - m_p - m_e + m_N - m_C)c^2 = ΔB$$ Where again, I have the atomic masses oh the Nitrogen and Carbon, take away the amount of electrons to obtain the nuclear mass.

The solution I've been provided uses only the atomic masses, rather than nuclear masses (which he says to use in his lecture notes) in the Q value to determine this relation. Any reason why using the atomic mass rather than nuclear mass misses the extra term?, or am I overlooking something completely trivial here?


1 Answer 1


Since it is essentially a nuclear phenomenon, you only consider the mass of the nucleus.

In beta minus or plus decay though, where a proton converts into neutron and positron or a neutron converts into proton and electron respectively , you should consider the mass of electron/ positron because the electron/positron is formed in the nucleus. The electron/positron is quickly emitted (in order of 10^-7 s) because of its instability.

  • $\begingroup$ I've added a question where this problem arises, could you add further comment to it if you can please? :) $\endgroup$ May 16, 2019 at 17:00
  • $\begingroup$ What you did is correct according to me. Maybe , the source has not considered the mass of electron because it’s mass/equivalent energy is very less compared to that of nucleons. $\endgroup$ May 16, 2019 at 17:11
  • $\begingroup$ So I worked it out using nuclear masses as implied by the notes, and I don't get the relation- I just get ΔB rather than this difference of electron rest-mass energy. I've clarified my text a bit more to try and explain my problem, in that I'm unsure why just atomic masses are used to calculate the Q value as opposed to nuclear masses (as in BE, where you take the electrons from the atomic mass)? $\endgroup$ May 16, 2019 at 19:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.