238
92U decays spontaneously by α emission to 234
90Th. Calculate the total energy released by
this process and the recoil velocity of the 234
90Th nucleus. The atomic masses are 238.050783
u for 238
92U and 234.043595 u for 234
90Th.
1
Expert's answer
2020-04-09T09:35:04-0400
The decay energy Q is given by conservation of mass-energy as
c2Q=MU−MTh−Mα.
We get
c2Q=238.050783−234.043595−4.002603=0.004585u
If multiplied by the conversion factor 931.5MeV/c2 , then Q≈4.27MeV.
So, the total energy released by this process Q≈4.27MeV.
If the parent nucleus is at rest when it decays, the daughter nucleus and the particle must have equal and opposite momenta. If p is the magnitude of the momentum of either particle, the decay energy is
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