Question

The nucleus ${}_{11}^{23}Ne$ decays by ${\beta }^{-}$ emission. Write down the $\beta$ - decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
$m{(}_{10}^{23}Ne)=22.994466u$
$m{(}_{11}^{23}Na)=22.089770u$.

Answer 1

The beta decay equation is as shown.
${}_{10}^{23}Ne\to {}_{11}^{23}Na+e^{-}+Q$

$Q=\Delta m\times$ 931.5 MeV

Here, $\Delta m=m_N{(}_{10}^{23}Ne)-m_N{(}_{11}^{23}Na)-m_e=$ mass defect.

$\Delta m=\left[m_N{(}_{10}^{23}Ne\right)-m_N{(}_{11}^{23}Na)-m_e]=22.994466-22.089770=0.904696$ amu

Here, the masses used are masses of nuclei and not of atoms. Note: $m_e$ has been cancelled.

$Q=\Delta m\times$ 931.5 MeV $=0.904696\times 931.5=4.37$MeV.
The maximum kinetic energy of the electron (max $E_e$) = Q = 4.37 MeV.