According to my textbook, for all the beta decays, it is required that the mass of the original atom to be heavier than the mass of the final atom.

Is this due to the fact that all the beta decays involve the production of neutrinos and these neutrinos take away the energy of the atoms?

It is further required that for beta-plus decay the mass of the original atom must be at least 2 electron mass heavier than the mass of the final atom. Why?

  • $\begingroup$ The radioactive decay label as a beta decay requires that an electron (or a positron) be emitted from the nucleus. That is one electron mass. $\endgroup$
    – LDC3
    Apr 16, 2014 at 5:18
  • $\begingroup$ In realm of these nuclear reactions mass conservation does not exactly apply. en.wikipedia.org/wiki/Mass%E2%80%93energy_equivalence $\endgroup$
    – evil999man
    Apr 16, 2014 at 5:20

2 Answers 2


In analyzing mass-energy calculations involving beta decay, it is important to avoid simple bookkeeping errors. You look at the balanced nuclear reaction, and use tables of isotope masses to calculate the loss of mass (and its conversion into energy)

The problem comes from the two terms "mass of the final atom" and "tabulated mass of the final atom"

Consider this example: $$ _{12}^{23}\mathrm{Mg} \longrightarrow \, _{11}^{23}\mathrm{Na} + e^{+} + \nu $$ It is basic in all nuclear reactions that the sum of all the mass and energy on the LHS must equal the sum of all mass and energy on the RHS. The mass of the parent isotope must be greater (perhaps only slightly) than the combined mass of the daughter isotope plus the mass of the positron, with the missing mass appearing as energy split between the positron and neutrino.


When you sit down to do the calculation, there's a problem: a bookkeeping problem.

You started with a normal, run-of-the-mill magnesium atom, with a surrounding cloud of 12 electrons. That's the mass you find listed for that isotope of magnesium, and that's what you write down to start your calculation.

Now you look at the daughter sodium isotope. It's sitting in the reaction vessel, still with the 12 orbital electrons it had a picosecond ago, when it was a magnesium atom.. But when you look up the mass of the sodium isotope, the tabulated value only includes the mass of the 11 orbital electrons that all normal sodium atoms have.

So to replicate the masses in the actual reaction, you need to add an electron mass to the tabulated sodium isotope mass.

Finally, you can look at the positron produced and look up its tabulated mass for the calculation. There's your required mass difference of two positron masses...


First part of your question: part of the mass is used as kinetic energy for the electron/positron and (anti-)neutrino to leave the core. Therefore mass can't be conserved.

For the beta plus decay: I don't know your textbook, but assume the mass of the atom includes the surrounding electrons. Then the core emits an positron (1st half of the mass loss), and losses an electron from the atomic orbitals, since the charge of the core decreases (2nd half).


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