Distribution of released energy in nuclear fusion

  • #1
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Main Question or Discussion Point

Hi,

On Wikipedia (here), we can find that in following channels of nuclear fusion reactions:
H-2 + H-3 -> He-4 (3.5 MeV) + n (14.1 Mev)
H-2 + H-2 -> H-3 (1.01 MeV) + H-1 (3.02 MeV)
H-2 + H-2 -> He-3 (0.82 MeV) + n (2.45 MeV)
H-2 + He-3 -> He-4 (3.6 MeV) + H-1 (14.7 MeV)
The released energy is always distributed between products.
But I have a few questions regarding above and other reaction channels:

1) Does this energy always manifests as kinetic energy of products?
2) If no, does the product nuclide can be created in excited state (and consume some of kinetic energy)?
3) or may a gamma photon be emitted consuming some the energy?
4) and finally, if only one nuclide is produced (e.g. He-4), does whole energy is transferred to that product?

Oh, I am talking about reactions in low energies, up to ± 30 MeV.

Many Thanks,
Toreno
 

Answers and Replies

  • #2
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1) Does this energy always manifests as kinetic energy of products?
Most of the time. All those reactions can happen with the emission of gamma rays (directly), but that is a rare process. I don't think there are nuclear excitations that don't decay via proton or neutron emission (which effectively looks like a different reaction then).
4) and finally, if only one nuclide is produced (e.g. He-4), does whole energy is transferred to that product?
That would violate energy-momentum conservation.
 
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  • #3
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Hi,

Thank you for fast response.
Could you please expand a little more that energy-momentum violation?

Thanks,
Toreno
 
  • #4
jtbell
Mentor
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Could you please expand a little more that energy-momentum violation?
Suppose for the sake of discussion that the fusion reaction X + Y --> 4He exists, and that there are no other products (e.g. gammas).

Recall that in a fusion reaction that releases energy, the sum of the masses of the reactants must be greater than the sum of the masses of the products. In this case, we must have

m(X) + m(Y) > m(4He).

Consider this reaction in the reference frame in which the final 4He is at rest. That is, its kinetic energy K(4He) = 0. The total energy must be conserved. Therefore we must have

m(X)c2 + K(X) + m(Y)c2 + K(Y) = m(4He)c2

The previous condition then implies that either K(X) or K(Y) or both must be negative. But kinetic energy can't be negative!
 
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  • #5
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Ok, I got this now. Thank you for. Thank you for explanation!
 
  • #6
The CNO cycle emits 3 gamma rays in each turn of the cycle, and 2 positrons, which, upon annihilation, generate 4 additional gamma rays, bringing the total to 7.
 

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