Nuclei of `.^(64)Cu` can decay be electron capture (probability 61%) or by `beta^(+)` decay (39%). Half life of `.^(64)Cu` is 12.7 hour. Find the partial half life for electron capture decay process,

Answer

`._(30)^(64) Zn,._(28)^(64)Ni, (t_(1//2))_(3)= 29.76 hr `

Explanation

`beta^(-) "emission"`
`._(29)Cu^(64)rarr ._(30)Zn^(64) +._(-1)e^(0)`
`beta^(+) "emission"`
`._(29)Cu^(64) rarr ._(28)Ni^(64)+._(+1)e^(0)`
Electron capture
`.(29) Cu^(64)+_(-1)e^(0) rarr._(28)Ni^(64)`
`% "of I product" = 38%=(lambda_(1))/(lambda_(1)+lambda_(2)+lambda_(3)) xx 100 =(lambda_(1))/(lambda)xx100=((0.693)/((t_(1//2))_(1)))/((0.693)/(t_(1//2)))xx100`
`lambda rArr` overall disintegration constant
`(38)/(100)=((t_(1//2)))/((t_(1//2))_(1))`
`:. (t_(1//2))=(t_(1//2))xx(100)/(38)=12.8 xx(100)/(38)=33.68 "hour"^(-1)`
similarly
`(t_(1//2))_(II)=(t_(1//2))xx(100)/(19)=12.8 xx(100)/(19)=67.36 "hour"^(-1)`
`(t_(1//2))_(III)=(t_(1//2))xx(100)/(43)=12.8xx(100)/(43)=29.776 "hour"^(-1)`