InterviewSolution
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InterviewSolution
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The element Curium `_(96)^(248Cm` has a mean life of `10^(13)` second. Its primary decay mode are spontaneous fission and a `alpha-decay` , the former with a probability of `8%` and the latter with a probability of `92%` Each fission released `200 MeV` of energy . The masses involved in `alpha - decay` are as follows `_(96)^(248) Cm = 248.072220 u, _(94^(244) Pu = 244.064100 u and _(2)^(4) He = 4 .002603 u ` calculate the power output from a sample of `10^(20)` Cm atom `(1 u = 931 MeV//e^(2))` |
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Answer» Correct Answer - `3.32xx10^(-5)Js^(-1)` 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` |
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