The beta‑minus decay ${}^{60}\text{Co}{\to }^{60}\text{Ni}+e^{-}+{\bar{u}}_e$ occurs with the following atomic masses: $m{(}^{60}\text{Co})=59.933822\text{u}$, $m{(}^{60}\text{Ni})=59.930786\text{u}$, and the electron mass $m_e=0.00054858\text{u}$. (a) Calculate the Q‑value of the decay in MeV. (b) Assuming the neutrino carries negligible kinetic energy, find the maximum kinetic energy of the emitted electron (in MeV).

Step-by-step solution

1. Write the mass balance for ${\beta }^{-}$ decay: initial mass = $m{(}^{60}\text{Co})$, final mass = $m{(}^{60}\text{Ni})+m_e+m_{\nu }$; neutrino mass is negligible.
2. Compute the mass difference $\Delta m=m{(}^{60}\text{Co})-[m{(}^{60}\text{Ni})+m_e]$.
3. $$\begin{gathered} Deltam=59.933822-(59.930786+0.00054858)=0.00248742 \\ , \\ textu \end{gathered}$$.
4. Convert to energy: $Q=\Delta m\times 931.5\text{MeV/u}=0.00248742\times 931.5\approx 2.32\text{MeV}$.
5. In ${\beta }^{-}$ decay, the released energy is shared between the electron and the antineutrino. If the neutrino takes negligible kinetic energy, the electron receives essentially all of $Q$.
6. Thus $$\begin{gathered} K_e^{ \\ max}=Q \\ approx2.32 \\ , \\ textMeV \end{gathered}$$.
7. Both (a) and (b) give the same numerical value.

Concept tested: Q‑value of $\beta^{-}$ decay

Published 17 September 2026 · How this solution was prepared and checked