The element curium 96Cm248 has a mean life of 1013s.. Its primary decay modes are spontaneous fission and α-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 a decay are as follows: 96Cm248=248.072220u,94Pu244=244.064100u and 2He4=4.002603u Calculate the power output from a sample of 1020Cm atoms. (1u=931MeV/c2).

Question

The element curium 96Cm248 has a mean life of 1013s.. Its primary decay modes are spontaneous fission and α-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 a decay are as follows: 96Cm248=248.072220u,94Pu244=244.064100u and 2He4=4.002603u Calculate the power output from a sample of 1020Cm atoms. (1u=931MeV/c2).

Filo · Accepted Answer

[c] The total energy released E= Energy released in fission process + energy released in α- decay process             =NF×200+Nα×(0.005517×931) =(8100×1020)×200+(92100×1020) ×(0.005517×931)             =20.725×1020MeV Power output P=E/t =20.725×1020×1.6×10−131013 =3.3×10−5W.

Question Text	The element curium 96Cm248 has a mean life of 1013s.. Its primary decay modes are spontaneous fission and α-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 a decay are as follows: 96Cm248=248.072220u,94Pu244=244.064100u and 2He4=4.002603u Calculate the power output from a sample of 1020Cm atoms. (1u=931MeV/c2).
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