Cu^(64)(" half life" =12.8 "hours" ) decay by beta^(c-)- emission (38%), beta^(o+)- emission(19%), and electron capture (43%). Write the decay products and calculate partial half lives for each of the decay processes.

Cu^(64)(" half life" =12.8 "hours" ) decay by beta^(c-)- emission (38%

1.	Cu^(64)(" half life" =12.8 "hours" ) decay by beta^(c-)- emission (38%), beta^(o+)- emission(19%), and electron capture (43%). Write the decay products and calculate partial half lives for each of the decay processes.
Answer» SOLUTION :The nuclear reactions are `:` `._(29)Cu^(64) rarr ._(30)Zn^(64)+._(-1)E^(0)(._(-1)beta^(0))` `._(29)Cu^(64) rarr ._(28)Ni^(64)+._(+1)e^(0)(._(+1)beta^(0))` `._(29)Cu^(64)+._(-1)e^(0) rarr ._(28)Ni^(64)` Given, `lambda_(avg)=(0.693)/(12.8)hr^(-1)` `:. lambda_(1)+lambda_(2)+lambda_(3)=lambda_(avg)=(0.693)/(12.8)=5.41xx10^(-2)hr^(-1)......(i)` Also for parallel PATH decay `lambda_(1)=` Fractional yield of`._(30)Zn^(64)xxlambda_(avg)` `lambda_(2)=` Fractional yield of `._(28)Ni^(64)xxlambda_(avg)` `lambda_(3)=` Fractional yield of `_(28)Ni^(64^(**))xxlambda_(avg)` `(lambda_(1))/(lambda_(2))=(38)/(19)....(ii)` and `(lambda_(1))/(lambda_(3))=(38)/(43) ....(iii)` From Eqs. `(i), (ii),` and `(iii), lambda_(1)=2.056xx10^(-2)hr^(-1)` `lambda_(2)=1.028xx10^(-2)hr^(-1),lambda_(3)=2.327xx10^(-2)hr^(-1)` `t_(1//2)` for `beta^(c-)-` emission `=(0.693)/(2.056xx10^(-2))=33.70hr` `t_(1//2)` for `beta^(o+)-` emission`=(0.693)/(1.028xx10^(-2)0=67.41hr` `t_(1//2)` for electron capture `=(0.693)/(2.327xx10^(-2))=29.78hr`