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The element curium .96^248 Cm has a mean...

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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The elecron curium _(96)^(248) On has a mean life of 10^(13) second ist pirmary dacay mode with a probilly of 8% and the letter with a probillty of 92% Each fission released 200 MeV of energy . The masses involved in a - dacay are as follows _(96)^(248) Cm = 248.072220 u, _(94_^(244) Pu = 244.064100 u and _(2)^(4) He = 4 .002603 u calculate the power output from a sample of 10^(20) Cm atom (1 u = 931 MeV//e^(2))

Find the kinetic energy of the alpha - particle emitted in the decay ^238 Pu rarr ^234 U + alpha . The atomic masses needed are as following: ^238 Pu 238.04955 u ^234 U 234.04095 u ^4 He 4.002603 u . Neglect any recoil of the residual nucleus.

Find energy released in the alpha decay, Given _92^238Urarr_90^234Th+_2^4He M( _92^238U)=238.050784u M( _90^234Th)=234.043593u M( _2^4He)=4.002602u

Calculate the amount of ""_(92)U^(235) required to release energy of 1 kWh. Given energy released during fission of one atom is 200 MeV.

200 MeV of energy may be obtained per fission of U^235 . A reactor is generating 1000 kW of power. The rate of nuclear fission in the reactor is.

The energy liberated on complete fission of 1 kg of ._92 U^235 is (Assume 200 MeV energy is liberated on fission of 1 nucleus).

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