`""^(64)Cu(T""_(50)=12.8 year)` decays `beta-`emission (38%), `beta""^(+)` decayproducts and calculate partial half-lives for each of the decay process.

Explanation

`""_(29)Cu""^(64)` undergoes
(i) `B""^(-)` emission (ii) `B""^(+)` emission (iii) electron capture

Let the rate constant for `beta ""^(-)` emission = `k""_(1)`
Let the rate constant for `beta""^(+)`emission = `k""_(3)`
Let the rate constant for electron capture emission = `k""_(3)`
Overall rate constant
`K=K""_(1)+K""_(2)+K""_(3)=(0.693)/(t""_(1//2) )=(0.693)/(12.8)h""^(-1)`
`t""_(1//2)` of `""_(29)Cu""^(64)=12.8h`
`K""_(1)` = % of its decay `xx` Rate constant (overall)
`(38)/(100)xx k=(0.38xx0.693)/12.8h""^(-1)`
`t""_(1//2)` of `beta""^(-)` emission = `(0.693)/(K)=(0.693xx12.8)/(0.38xx0.693)`
`=(12.8)/(0.38)`
(partial half life) = 33.68 hr.
`K""_(2)` = % of `beta""^(oplus)` decay `xxK` (overall)
`=(19)/(10)xx(0.693)/(12.8)xx(0.19xx0.693)/(12.8)h""^(-1)`
Let `(t""_(2))""_(1//2)` be the half life of `beta""^oplus` emission
`=(0.693)/(K""_(2))=(0.693xx12.8)/(0.693xx0.19)=(12.8)/(0.19)`
= 67.37 hr