Calculate the maximum kinetic energy of the beta particle emitted in the following decay scheme:
`N^12 rarr C^12 + e^(+)+v`
`C^12 rarr C^12 + gamma(4.43 MeV).`
The atomic mass of `N^12` is `12.018612 u.`

Answer

A

Explanation

`^(12) N rarr ^(12 C^(**)+ e^-)+ v`
` ^(12)C^(**) + e^(-) + v`
` ^(12)C^(**) rarr ^(12)C + V (4.43 MeV ) `
Net reaction:
` ^(12)N rarr ^(12)C + e^(+)+ v+ v (4.43MeV) `
Energy of
` (e^(**) + v) = N^(12) - (C^(12) + v) `
` = 12.0018613 u - (12) u - 4.43 `
` = 0.018613 u - 4.43 u `
` = 17. 43 u = 12.89 MeV`
Maximum energy of `e^(-)`
(Assuming 0 energy for v)
` = 12.89 MeV`