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`)

The element curium `_96^248 Cm` has a mean life of `10^13s`. Its prima

1.	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» Correct Answer - A::B::C The reaction involved in `alpha`-decay is `_96^248Cmrarr_94^244Pu+_2^4He` Mass defect, `Deltam=mass of _96^248Cm-mass of _94^244Pu-massof_2^4He` `=(248.072220-244.064100-4.002603)u` `=0.005517u` Therefore, energy released in `alpha`-decay will be `E_alpha=(0.005517xx931)MeV=5.136MeV` Similarly, `E_(fission)=200MeV` (given) Mean life is given as `t_(mean)=10^13s=1//lambda` :. Disintegration constant `lambda=10^-13s^-1` Rate of decay at the moment when number of nuclei are `10^20` `=lambdaN=(10^-13)(10^20)` `=10^7` disintegration per second Of these, 8% are in fission and 92% are in `alpha`-decay. Therefore, energy released per second `=(0.08xx10^7xx200+0.92xx10^7xx5.136)MeV` `=2.074xx10^8MeV` :.Power output (in watt) = energy released per second`(J//s)`. `=(2.074xx10^8)(1.6xx10^-13)` `=3.32xx10^-5Js^-1` :. Power output `=3.32xx10^-5 W`

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