The nucleus `._(10)^(23)Ne` decays by `beta` - emission. Write down the `beta` - decay. 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)=22.089770 u`.

Explanation

The `beta` decay of `._(10)Ne^(23)` may be represented as
`._(10)Ne^(23)rarr ._(11)Na^(23)-._(1)e^(0)+ v + Q`
Ignoring the rest mass of antineutrino and v electron
Mass defecr, `Delta m = m(._(10)Ne^(23))-m(._(11)Na^(23))`
` = 22.994466-22.989770`
`= 0.004696` amu
`Q = 0.004696xx931 MeV = 4.372 MeV`
As `._(11)Na^(23)` is very massive, this energy of 4.3792 MeV is shared by `e^(-)` v pair. The max K.E of `e^(-)=4.372` MeV when energy carried by v is zero.