The element curium `._96^248 Cm` has a mean life of `10^13s`. Its primary decay modes are spontaneous fission and `alpha`-decay, the former with a probability of `8%` and the later with a probability of `92%`, each fission releases `200 MeV` of energy. The masses involved in decay are as follows
`._96^248 Cm=248.072220 u`,
`._94^244 P_u=244.064100 u` and `._2^4 He=4.002603u`. Calculate the power output from a sample of `10^20` Cm atoms. (`1u=931 MeV//c^2`)

Answer

`3.32xx10^(-5)Js^(-1)`

Explanation

Activity of fission `=A=(dN)/(dt)=lambdaN=(1)/(tau)N=(10^(20))/(10^(13))=10^(7)`
as probability is `8%` so reactivity `=(8)/(100)xx10^(7)=8xx10^(5)Mev//sec`.
Rate of decay of `alpha`-particle `=(92)/(100)xx10^(7)=92xx10^(5)`
`Delta M` for `alpha` decay `=0.00517 "amu"`
Energy released per `alpha`-decay `=92xx10^(5)xx5.1363 Mev//sec`
Total power output `=16xx10^(7)+4.725xx10^(7)=33.16 mu W`