Question

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 1

Activity (or rate ) of fission,
`A=(dN)/(dt) =lambda N =(1)/(tau)N=(10^(20))/(10^(13)) = 10^(7)`
As probability of fission is `18%` only, therefore, actual rate of fission is
`(8)/(100) xx 10^(7) =8 xx 10^(5) s^(-1)`
Energy released per fission `=200 MeV`
power output of fission `= 8 xx 10^(5) xx 200 MeV s^(-1)`
`=16 xx 10^(7) MeV s^(-1)`
Probability of `alpha`-particle decay is `92%`. Therefore, rate of decay for `alpha` -particle is
`(92)/(100) xx 10^(7) =92 xx 10^(5) s^(-1)`
For `alpha-`decay, the equation is
`._(96)^(248)Cm rarr ._(94)^(244)Pu + ._(2)^(4)He`
`(248.072220) (244.064100)(4.002603)`
Mass of decay product is
`244.064100 +4.002603 =248.06673` .
Mass defect,
`Delta m=248.072220-248.066703 =0.005517 u`
Energy released per `alpha` deacy is
`92 xx 10^(5) xx 5.1363 MeV s^(-1) =4.725 xx 10^(7) MeV s^(-1)`
`{:("Total powaer output"=16 xx 10^(7) + 4. 725xx10^(7)),(" "=20.725xx10xx1^(7)MeV s^(1)),(" "=20.725xx10^(7)xx1.6xx10^(-13)J s^(-1)),(" "=33.16xx10^(6)W=33.16 mu W):}`.