Not exactly a homework problem in the real sense of the word but a way to understand how I can make arrive at the $Q$-values algebraically without relying on formula sheets or memorising.

Let's begin with the positive beta-decay:

$$_{Z}^{A}\textrm{X}_{N}\rightarrow {}_{Z-1}^{~~~~~A}\textrm{X'}_{N+1}+e^{+}+\nu $$ Let M represents Atomic mass.

$$Q=\left[ \left( M\!\!\left(_{Z}^{A}\textrm{X}\right) - ZM_{e}\right) -\left( \left(M\!\!\left({}_{Z-1}^{~~~~~A}\textrm{X'}\right)-(Z-1)M_{e}\right)-M_{e^+}-\nu \right) \right]c^{2}$$ (The nuclear mass of an atom with Z proton is the atomic mass of the atom sans Z electrons since the number of protons and electrons must coincide in the case of a neutral element)

Which reduces to

$$Q=\left[ M\!\!\left({}_{Z}^{A}\textrm{X}\right) -M\!\!\left(_{Z+1}^{~~~~~A}\textrm{X'}\right) -2M_{e} \right]c^2$$

My questions are:

  1. Is the conversion done correctly?

  2. The Q value represents the energy of the positron and the neutrino but in my final equation, if it is correct, does not appear convincingly to me that this is the case. In my final equation, I have subtracted away the energy of the positron so the Q-value seems like some other kind of energy. I really appreciate a sound explanation with regards to the above.

  • 1
    $\begingroup$ I don't understand your question...first $Q = M(_Z^AX) - M(_{Z - 1}^A X') - M_{e^+} - M_{\nu}$. Secondly, yes the $Q$ value represents either 1) the energy necessary or 2) the energy in excess released and thus if it's the second case, the $Q$ value will be split between the kinetic energy of the original atom plus that of the positron and neutrino (since $E \propto \frac{p^2}{m}$ most of the excess energy goes to the lighter particles). $\endgroup$
    – Jared
    May 31, 2015 at 7:33
  • $\begingroup$ It can be tough to post a clear question when I myself am confused over the text that I'm using. So Q is derived correctly. What my text claim is that the Q value represents the kinetic energy of the positron after a positive beta decay. (actually positron and neutrino but since neutrino is massless according to the standard model of particle physics, we discount the energy of the neutrino) It's confusing. $\endgroup$
    – Physkid
    May 31, 2015 at 7:39
  • $\begingroup$ So are you asking why is it that the majority of the energy released goes towards the emitted positron? As opposed to the atom or neutrino? $\endgroup$
    – Jared
    May 31, 2015 at 7:48
  • $\begingroup$ Partially so. But the core of the question is, if you look at my final question above, it is obvious that I have moved everything(including the positron's mass) into the square bracket. There is no other terms left. I'm puzzled about the existence of there being some kinetic energy of a positron 'hidden' in the Q-value. The Math isn't quite telling me the same thing. For if Q consists of the positron's KE, then using backwards deduction, we must have $$_{Z}^{A}\textrm{X}_{N}\rightarrow _{Z-1}^{A}\textrm{X'}_{N+1}+e^{+}+\nu + ( e^+) $$ $\endgroup$
    – Physkid
    May 31, 2015 at 7:53
  • $\begingroup$ I think you are misunderstanding the meaning of this $Q$ value then. Initially you have an atom, then afterwards you have a lighter atom, a positron, and a neutrino plus energy (the difference in mass). The $Q$ value is the added energy which must go towards the added kinetic energy of each of the resulting pieces. You don't subtract the kinetic energies from $Q$--you find the kinetic energies (and momenta) from $Q$. $\endgroup$
    – Jared
    May 31, 2015 at 7:57

1 Answer 1


Your simplification is correct: a nuclear beta decay in general has $A-1$ "spectator nucleons" and $Z$ "spectator electrons." The difference in final-state mass is the difference in the nuclear masses plus the mass of the created leptons (and the neutrino mass is so miniscule that it's buried in the experimental noise in the measurement of the $Q$-value.

In positron decay you have the slight extra wrinkle that the daughter nucleus will usually still have $Z$ electrons, since the electrons in the parent nucleus are spectators to the weak decay. Thus if you consider the measured masses of the neutral atoms (which is generally what's tabulated) rather than the nuclei, then your final state is heavier by $2m_\text e$: one $m_\text e$ for the decay positron and one $m_\text e$ for the extra spectator electron in the daughter ion.

This has the consequence that weak decays by electron capture have higher $Q$-value by $1.022\rm\,MeV$ than equivalent decays by positron emission: the positron in the final state is replaced by an electron in the initial state, and the daughter atom is born neutral. In the case of beryllium-7, electron capture decays are energetically allowed, but positron decays are energetically forbidden.

Be warned that it's hard to measure the mass of a bare nucleus, so most tabulated masses are for neutral atoms.

As to where the kinetic energy goes: it is shared among all three constituents. To find the distribution of kinetic energies, you assume the decay happens at rest (always true in some reference frame); the momenta of the three final-state products must add up to zero, so the daughter nucleus, electron, and neutrino will have order-of-magnitude the same momentum. If the lepton momenta are MeV-scale then both leptons are relativistic, with $E^2\approx p^2c^2$; the daughter nucleus is always nonrelativistic, with $E\approx p^2/2 m_X$. So the statement that the $Q$-value gives the energy of the electron and neutrino is only approximately correct.


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