For the three types of beta decay: \begin{align} _z^AP&\rightarrow {}_{z+1}^AD + e^-+\bar{\nu} & Q_{\beta^-} &= (M_P-M_D)c^2 \\ _z^AP&\rightarrow {}_{z-1}^AD + e^++{\nu} & Q_{\beta^+} &= (M_P-M_D)c^2-2m_ec^2 \\ e^-+ {}_z^AP&\rightarrow{}_{z+1}^AD+\nu & Q_{EC} &= (M_P-M_D)c^2 \end{align} I'm having trouble understanding why the $Q$ equation for positron emission needs to subtract two electron masses while $\beta^-$ decay do not. The textbook says it's because the daughter nucleus has an extra negative charge because of another electron, but why don't we consider an extra positive charge in the daughter nucleus in beta-minus decay? Also, why the $Q$ value for electron capture doesn't take the electron on the left side into account?

  • $\begingroup$ My answer here possibly answers this question as well, though the questions themselves are different. $\endgroup$ Feb 22, 2022 at 19:56

1 Answer 1


The tabulated masses which are used to determine the $Q$-values are for neutral atoms. If the “spectator electrons” are undisturbed in a positron decay, a neutral parent atom $\rm P$ produces the daughter atom $\rm D$ as a negative ion,

$$ ^A\mathrm P^\text{neutral} \to {}^A\mathrm D^{1-} + e^+ + \nu $$

while electron capture has two neutral particles in the final state,

$$ ^A\mathrm P^\text{neutral} \to{} ^A\mathrm D^\text{neutral} + \nu $$

Your explicit electron on the left side of the electron-capture equation makes the arithmetic more confusing, because you have subtracted a bunch of spectator electrons whose masses and binding energies are included in your calculation.

Note that the captured electron generally comes from an inner $s$ orbital, so the neutral daughter atom has a hole deep in its electron cloud and emits x-rays as it relaxes.

For a $\beta^-$ decay, the daughter is generally created as a positive ion, unless the neutrino carries away so much of the kinetic energy that the $\beta^-$ energy is less than the binding energy for the valence electrons to the daughter ion. This is referred to as a “bound decay.”

  • $\begingroup$ Thanks for the answer! We subtract 2 electron masses in positron emission, is that because the electron pair is also considered as the product of $\beta^+$ decay, and in the reaction equation the electron is combined with the daughter ion? $\endgroup$
    – IGY
    Feb 21, 2022 at 17:59
  • 1
    $\begingroup$ Right. Count the particles in the two reaction equations. The final state for the positron emission has the extra positron that was emitted, and also an extra electron that was not captured. Then compute the $Q$-value by taking the mass difference between the initial and final states. $\endgroup$
    – rob
    Feb 21, 2022 at 18:26

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