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`)

Explanation

Energy released in `alpha`decay
`._(96)^(248)Cm rarr._(94)^(244)Pu+._(2)^(4)He`
The mass defect
`Deltam = 248.072220-(244.064100+4.002603)`
`= 0.005517u`
Equivalent energy `E_(alpha) = 0.005517xx931 = 5.136MeV`
Energy released in the deacy of one atom is
`E = E_(alpha) +E_("fission")`
`= 0.92xx5.136+0.08xx200`
`= 20.725MeV`
Total energy released in `10^(20)` atoms
`= 10^(20) xx 20.725MeV`
Power output `= ("Total energy released")/("Mean life")`
`= 10^(20 xx 20.725xx1.6xx10^(-13))/(10^(13))`
`= 3.3xx10^(-5)W`