β− Decay Energy from Masses

As in the preceding example, we must first find $\mathrm{\Delta }m$, the difference in mass between the parent nucleus and the products of the decay, using masses given in Appendix A. Then the emitted energy is calculated as before, using $E=(\mathrm{\Delta }m)c^2$. The initial mass is just that of the parent nucleus, and the final mass is that of the daughter nucleus and the electron created in the decay. The neutrino is massless, or nearly so. However, since the masses given in Appendix A are for neutral atoms, the daughter nucleus has one more electron than the parent, and so the extra electron mass that corresponds to the ${\beta }^{–}$ is included in the atomic mass of Ni. Thus,

The ${\beta }^{-}$ decay equation for ${}^{\text{60}}\text{Co}$ is

As noticed,

Entering the masses found in Appendix A gives

Thus,

Using $\mathrm{1\; u}=\mathrm{931.5\; MeV}/c^2$, we obtain

Perhaps the most difficult thing about this example is convincing yourself that the ${\beta }^{-}$ mass is included in the atomic mass of ${}^{\text{60}}\text{Ni}$. Beyond that are other implications. Again the decay energy is in the MeV range. This energy is shared by all of the products of the decay. In many ${}^{\text{60}}\text{Co}$ decays, the daughter nucleus ${}^{\text{60}}\text{Ni}$ is left in an excited state and emits photons ( $\gamma$ rays). Most of the remaining energy goes to the electron and neutrino, since the recoil kinetic energy of the daughter nucleus is small. One final note: the electron emitted in ${\beta }^{-}$ decay is created in the nucleus at the time of decay.