Question

The nucleus ${}^{23}Ne$ decays by $\beta -$ emission into the nucleus ${}^{23}Na$. Write down the $\beta -$ decay equation and determine the maximum kinetic energy of the electrons emitted. Given, $m{(}_{10}^{23}Ne)=22.994466amu$ and $m{(}_{11}^{23}Na)=22.989770amu$. Ignore the mass of antineutrino $\left(\overline{v}\right)$:

Answer 1

${}_{10}^{23}Ne{+}_{+1}^0\beta {\to }_{11}^{23}Na$

The given equation is of ${\beta }^{+}$ decay.

Maximum kinetic energy = $△m{c}^2$

$△m=m\left(Ne\right)-m\left(Na\right)=22.994466-22.989770=0.004696amuE=△m{c}^2=0.004696\times 931MeV=4.371976MeV$