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`)

Options

A. `4.42xx10^(-3)` W
B. `3.32xx10^(-5)` W
C. `4.42xx10^(-5)` W
D. `3.32xx10^(-3)` W

Answer

b

Explanation

The primary decay modes of `._96Cm^248` are (i)Spontaneous fission (probability 8%)
(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.