Question

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

Answer 1

The β decay of \(_{10}^{23}Ne\) may be represented as

The energy released is shared by Na nucleus and the electron-neutrino pair released. As electron-neutrino pair is much lighter the ²³Na nucleus, practically Whole of the energy released is carried by electron-neutron will carry the maximum energy. Therefore, maximum kinetic energy of the emitted electron is 4.374 MeV.