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

`3.32 xx 10^(-5)` watt

Explanation

(Take a sample of `10^(20)` Cm atoms)
`alpha`-decay `._(96)Cm^(248) rarr ,_(96)Pu^(244)+._(2)He^(4)`
Here `Deltam=(248. 072220-244.064100-4.002603)u=0.005517 u`
`implies E_(alpha)=Deltamc^(2)=0.005517xx931=5.136 MeV`
Energy released in the decay of one atom
`E=E_("fission")+E_(alpha)=0.08xx200+0.92xx5.136`
`=20.725 MeV`
Total energy released from the decay of all `10^(20)` atoms=`20.725xx10^(20)MeV=3.316xx10^(8)J`
Power output
`=("Total energy released")/("mean life")=(3.316xx10^(-8))/10^(-13)`
`=3.3xx10^(-5)` watts