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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) |
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Answer» `4.42xx10^(-3)` W (ii)`alpha`-decay(probability 92%) The nuclear reaction is GIVEN below : `._96Cm^(248) to ._94Pu^244 + ._2He^4` `therefore` Mass defect in the reaction =`Deltam` `therefore Deltam`=mass of Cm - mass of Pu - mass of He or `Deltam`=248.072220 - 244.064100-4.002603 or `Deltam`=0.005517 u `therefore ` Energy released = (0.00517 x 931 ) MeV `E_alpha` =5.136 MeV Given : `E_f`=Each fission releases 200 MeV of energy . MEAN life of Cm = `10^13` sec. `therefore lambda = 1/"mean life"=10^(-13) s^(-1)` Again `"dN"/"DT"=lambdaN` where N is given to be `10^20`. `therefore ` Rate of decay = `(10^(-13))(10^20)` or Rate of decay = `10^7` dps Of these `10^7` dps, 8% are in fission and 92% are in `alpha`-decay process. Energy released per second due to fission. `=8/100xx10^7 xx200=16xx10^7` MeV Energy released per sec due to `alpha`-decay `=92/100xx10^7xx5.136=4.725xx10^7` MeV Total energy released per second =`(16+4.725)10^7` MeV =`20.725xx10^7` MeV `therefore ` Power output= Energy per second `=(20.725xx10^7)xx(1.6xx10^(-13))` J/s `=3.316xx10^(-5) W = 3.32xx10^(-5)` W. |
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