Binding Energy in U-235 and daughter atoms

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SUMMARY

The discussion centers on the binding energy of Uranium-235 (U-235) and its daughter nuclei during fission. When U-235 decays, it releases energy due to the higher binding energy per nucleon of the resulting daughter nuclei, which is calculated to be approximately 200 MeV. The initial binding energy of U-235 is -1786 MeV, while the final state of the daughter nuclei is -1942 MeV, resulting in a net release of 256 MeV. This energy release occurs despite the negative binding energy, as the process adheres to the principles of energy conservation.

PREREQUISITES
  • Understanding of nuclear fission processes
  • Familiarity with binding energy concepts in nuclear physics
  • Knowledge of energy conservation principles
  • Basic calculations involving MeV (mega-electronvolts)
NEXT STEPS
  • Study the principles of nuclear fission in detail
  • Learn about the binding energy calculations for different isotopes
  • Explore the concept of spontaneous fission and its implications
  • Investigate the relationship between binding energy and nuclear stability
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Students and professionals in nuclear physics, nuclear engineers, and anyone interested in understanding the energy dynamics of nuclear fission reactions.

physmurf
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I am a little bit confused.
If U235 decays 2 neutrons, and the two daughter atoms posses a larger amount of binding energy per nucleon, then why is there excess energy? Why doesn't this process require the input of additional energy? I know I am missing something fairly simple.
 
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Because binding energy is conted as "negative".
 
malawi_glenn said:
Because binding energy is conted as "negative".

So, wouldn't that imply that this would require energy from the surroundings in order to make the reaction to take place? I know that we can safely assume that the kinetic energy of the incoming neutron to be negligible. So, how can this happen without the addition of external energy?
 
Well now there is many different ways for the U-235 to decay and being fissioned. Just tell the specific reaction that you have questions about and I will try to explain it.

If the daugher nucleis has greater binding energy per nucleon (i.e. more negative energy), then energy is released (in general, depends on the neutrons also etc but basically this is what happens)

Mother nucleus A' has 20 nucleons, 5MeV / nucleon; total energy: -100MeV

Decays to daughter nuclei B' and C', they have binding energy 6MeV / nucleon; total energy: -120MeV so here you have less energy, so in order to obey energy conservation; 20 MeV is then kinetic energy of daugher nucleis (NB! this is only a crude model for explanation)
 
Okay, from lecture, the professor uses the following example. U-235 decays to two daughter nuclei X and Y each with A=117 with the release of two free neutrons. He says the change in binding energy = (7.6 Mev * 235 - 8.3 Mev * 234) = (approximately equals) (0.7)(235) = (approximately) 200 MeV.

The above equation will yield a negative energy. Where does the energy come from that allows more overall binding energy between the two daughter nuclei?
 
Spontaneous fission is that reaction called.

Binding energy is counted as negative, so You have to be aware of this and count backwards.

U-235; 7.6MeV / nucleon
X; Y : 8.3MeV / nucleon respectevly

gives:

235*(-7.6) = -1786MeV initally
2*117*(-8.3) = -1942MeV final

You have more energy in the beginning then in the final state, so you must add energy to obey energy conservation.

So there is relased 256MeV, do the comparison with a hydrogen atom, that is excited and the electron has 1.51eV binding energy; deexcites to ground state that has binding energy 13.6eV; there is realeased 12.89eV as a photon (and recoil energy of atom). So then things is rather clear, cos' binding energy is counted negative, but in nuclear physics, you have to take into account that the initial particle can be divided into two etc, but basically it is the same principle.
 
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