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Need clarification about asymmetry term in Bethe and Weizsäcker formula

  1. Oct 31, 2015 #1
    well ...
    we are studying bethe and weiszacker semi empiric formula , and i am confused . specifically about the third term the asymmetry term and how it lowers the binding energy .
    so just so i know that i am getting this straight , originally the binding energy no matter the nucleus is constant but there variable that change the effective energy used for binding , like the second term which is the Colombian term , part of the energy is wasted fighting the repulsion between the protons , for the asymmetry term , it is explained with the pauli exclusion , saying that the surplus of neutrons have a higher energy than protons , yeah so what ?

    i tried using how an electron being in a higher level of energy is less bound to the nucleus as an analogy but it doesn t make complete sens either.
    since the binding energy given to each nucleus is constante , and the energy actually used for binding is less than it should be , then where does the energy go ? what is it used for ? is it used to make the protons have a higher energy to bind with the neutrons ? or is it what it gives the neutrons higher energy in the first place and pauli s exclusion principle just means that they could do that ?

    i am so confused , and i am sorry if i am not being clear enough , thank you very much in advance
     
  2. jcsd
  3. Oct 31, 2015 #2

    mfb

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    The binding energy per nucleon is constant if you neglect various effects that are negligible.

    Both protons and neutrons fill their energy levels in ascending order of energy. The lowest states get filled first. If you have many neutrons but not many protons, some neutrons are forced to occupy relatively high energetic levels. Same with an asymmetry in the other direction. A higher energy means a weaker bond, similar to the electrons.
    They have to do it due to the Pauli exclusion principle.
     
  4. Oct 31, 2015 #3
    thank you , thank you . i kind of had to go to the bathroom , and while i was there i understood what the formula does , it simply calculates the total internal energy of the nucleus , and then eliminating all the elements that contributes to the internal energy so that we have the binding energy left. the volume and surface terms are the total internal energy , and then we take away the electric energy which is the third term , and then we take away the energy that the neutrons get by pauli's principle ,and then the energy that come from the parity of protons and neutrons , and so what s left is the energy that binds the nucleus together . am i right ?
     
  5. Oct 31, 2015 #4

    mfb

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    The binding energy is negative - more binding energy means the nucleus is lighter. The sign can be a bit confusing, you are actually adding all contributions, but some are negative, some are positive.
     
  6. Oct 31, 2015 #5
    in my mind i understand when you say that the binding energy is negative , since isolated nucleons go to a more stable state which means lower energy and so it should be negative . but when we calculate with the formula we find positive results , were the signs flipped for the sake of making things easier ?
     
  7. Oct 31, 2015 #6

    mfb

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    Yes, binding energy is the energy released when a nucleus forms, so it is positive (for bound nuclei).
     
  8. Oct 31, 2015 #7
    i am so sorry to be bother , but i think i finally understand now .
    state 1 , isolated nucleons . state 2 a nucleus
    going from state 1 to 2 :
    lowers the energy of the system by the first two terms
    increases the total energy by the third term ( coulomb energy )
    increases the energy by the fourth term ( asymmetry )
    increases or lowers the total energy by the fifth term ( pairing energy )
    the sum of all these energy is the energy that the system loses by going from 1 to 2 ,also referred to by the binding energy and it is negative .
    it is also the energy that i must provide / that the system receives , to cancel out the advantages of being in state 2 thus forcing him to returned to state 1 .
    is everything i am saying now is finally correct ? , this day has been a journey of enlightenment , sorry if i bothered but please tell me if i got anything wrong ?
     
  9. Oct 31, 2015 #8

    mfb

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    Right.

    (I guess "second term" is the surface term: it goes in the opposite direction as the volume term)
     
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