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

A, B, C

Explanation

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`