Question

The radio nuclide ${}^{11}C$ decays according to
${}_6^{11}C\to {}_5^{11}B+e^{+}+\nu$: $T_{1/2}$ $=20.3min$
The maximum energy of the emitted positron is $0.960MeV$.
Given the mass values:
$m{(}_6^{11}C)=11.011434u$ and $m{(}_6^{11}B)=11.009305u$
Calculate $Q$ and compare it with the maximum energy of the positron emitted.

Answer 1

${}_6^{11}C\to {}_6^{11}B+e^{+}+\nu +Q$
$Q=[m_N\left({}_6^{11}C\right)-m_N\left({}_6^{11}B\right)-m_e]c^2$
Where, the masses used are those of nuclei and not of atoms. If we use atomic masses, we have to add $6m_e$ in case of ${}^{11}C$ and $5m_e$ in case of ${}^{11}B$. Hence,
$Q=[m\left({}_6^{11}C\right)-m\left({}_6^{11}B\right)-2m_e]c^2$ (Note $m_e$ has been doubled)
Using given masses, Q = 0.961 MeV.
Q = $E_d+E_e+E_{\nu }$
The daughter nucleus is too heavy compared to $e^{+}$ and $\nu$, so it carries negligible energy ($E_d$ $\approx$ 0). If the kinetic energy ($E_{\nu }$) carried by the neutrino is minimum (i.e., zero), the positron carries maximum energy, and this is practically all energy Q, hence, maximum $E^e$ $\approx$ Q.