The nucleus `""_(10)^(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) = 22.994466 u`
`m (""_(11)^(23)Na) = 22.989770 u`.

Explanation

The equation is `""_(10)^(23)Ne to ""_(11)^(23)Na + beta + barupsilon +Q`
`Q = [m_n(""_(10)^(23)Ne) - m_(n) ""_11^23Na - m_e]c^2`
(Neglect the rest mass of `barupsilon)`
`= {{m_n ""_(10^(23)Ne + 10 m_e} - {m_n(""_11^23Na) + 11m_e}]c^2`
`= [m ""_(10)^(23)Ne - m ""_(11)^(23)Na]c^2`
`=[22.994466 - 22.989770]931`
`= 4.374 MeV`.