The nucleus ${^{23}_{10}Ne}$ decays by $β^-$ emission. The maximum kinetic energy of the electrons emitted is:
Given that: $m({^{23}_{10}Ne}) = 22.994466 u, m({^{23}_{11}Na}) = 22.989770 u$ and $1u ≡ 931 MeV/c^2$
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 4.372 MeV
For $\beta^-$ decay:
$Q = [m(^{23}_{10}Ne) - m(^{23}_{11}Na)]c^2$
Given: $m(^{23}_{10}Ne) = 22.994466 \, u$, $m(^{23}_{11}Na) = 22.989770 \, u$
Mass difference $= 22.994466 - 22.989770 = 0.004696 \, u$
$Q = 0.004696 \times 931 \, \text{MeV} = 4.37 \, \text{MeV}$
In $\beta^-$ decay, the maximum kinetic energy of the emitted electron $\approx Q$ (neglecting neutrino rest mass).
Final Answer: $K_{max} = 4.37 \, \text{MeV}$