Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Endothermic nuclear reactions

  1. Jan 14, 2010 #1
    Nuclear reactions in which the Q value is negative (endothermic reactions) imply that energy is absorbed in the process of the reaction. Is this to be interpreted as a case where the initial rest mass energy is less than the rest mass energy of the products of the reaction? If affirmative, it should be possible for an isolated proton to disintegrate as under:

    p -- n + e+ (positron) + Y(neutrino)

    However, the above is not possible because the mass of a proton is less than that of a neutron, and thus the reaction would imply a negative value of Q (mass energy on the left of the equation being less than that on the right). The reaction, otherwise, appears to be endothermic in nature because of the negative Q value. There, thus, appears to be a conflict in my understanding on this score.
    I would appreciate if the forum members could clarify the position to me, and to also provide me with one or two examples of endothermic reactions.
     
  2. jcsd
  3. Jan 14, 2010 #2

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Everything you said is correct except for this sentence:

    As you say, this is an endothermic reaction, so it requires in input of energy. Therefore it won't occur for an isolated proton.
     
  4. Jan 14, 2010 #3
    OK. Could you also indicate as to how in a beta + decay, what is the source of the input energy that triggers the above reaction, i.e., transformtion of a proton into a neutron? Thanks
     
  5. Jan 14, 2010 #4

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    There's a difference in nuclear binding energy between the initial nucleus and the final nucleus.
     
  6. Jan 15, 2010 #5
    I now understand the issue somewhat better. Should the parent mass be less than the sum of the masses of the daughter and remaining products of the hypothetical reaction, the parent is a stable nucleus and will not spontaneously decay. The reaction could, however, take place if some external energy is given to the parent nucleus. This explains why an isolated proton will not decay into a neutron plus a positron and neutrino.
    However, I am still not quite in the grip of how the proton in a nucleus decays spontaneously into a neutron plus positron and neutrino. I believe that this situation happens relatively less often, but I am trying to understand as to why this happens at all. Maybe you could explain to me using some examples. Thanks.
     
  7. Jan 15, 2010 #6

    Astronuc

    User Avatar

    Staff: Mentor

  8. Jan 15, 2010 #7

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    P->n doesn't happen less often. P->n happens for nuclei on the neutron-poor side of the line of stability, whereas n->p happens on the neutron-rich side. As to why it happens, p->n happens for the same reasons that n->p happens. The weak force treats the two processes symmetrically. As a general rule in quantum mechanics, if a process can occur without violating conservation laws, then it will actually occur.
     
  9. Jan 15, 2010 #8
    There is an interesting form of beta decay called internal conversion, where a bound [STRIKE]neutron[/STRIKE] proton in the nucleus (example: beryllium-7) absorbs an atomic electron, and becomes a bound neutron plus an anti-neutrino.

    e- + p → n + ν-bar

    This reaction is exothermic. Endothermic proton decay to a neutron has been searched for, and not found. The minimum proton decay lifetime (if it decays at all) is over 1025 years.

    Bob S
     
    Last edited: Jan 15, 2010
  10. Jan 15, 2010 #9

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I think searches for proton decay are searching for spontaneous (and therefore exothermic) decay, and it's not decay into a neutron, it's decay into something like a pion and a positron: http://en.wikipedia.org/wiki/Proton_decay
     
  11. Jan 16, 2010 #10
    Quote
    The weak force treats the two processes symmetrically. As a general rule in quantum mechanics, if a process can occur without violating conservation laws, then it will actually occur.
    Unquote

    Could you please elaborate further on the above? Does the weak force refer to force between two nucleons? If it is indeed a nuclear force such a force cannot be weak because it is supposed to be about the strongest force on earth. Hence you must be referring to some other force.
    Also in the case of B- decay, the nucleous has a surplus of neutrons and therefore one neutron decays to a proton thereby emitting an electron and an antineutrino in the process. As the mass of a neutron is a little higher than that of a proton, this reaction is energetically permitted. However, in the scenario of a nucleous being neutron deficient, one proton decays to a neutron. However, since the proton has less mass than a neutron this reaction should not be energetically permitted. This is precisely where my doubt arises. Actually in this case the nucleus has to receive some external energy so that the reaction could take place. What is the source of this external energy? Perhaps you could look into my concern from this angle. Thanks.
     
  12. Jan 16, 2010 #11

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Endothermic nuclear reactions
  1. Nuclear Reactions (Replies: 1)

  2. Nuclear Reaction (Replies: 2)

Loading...