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

Semi-empirical mass formula and the pairing term

  1. May 28, 2015 #1
    We have the formula for the mass of an atom:

    uNvv565.jpg

    From our class notes I have:

    Please could someone explain how he manages to get a single parabola from the pairing term? All I understand about the pairing term is that ap<0 for Z,N even, even. ap=0 for A odd and ap>0 for Z,N odd, odd.

    At a loss if anyone can help?
     
  2. jcsd
  3. May 28, 2015 #2

    ChrisVer

    User Avatar
    Gold Member

    Simple:
    If A is odd, then N+Z is odd...
    If N+Z is odd then one of them has to be even and the other odd (summing two odds or two evens will give an even number).
    So a_P becomes zero and that term misses from the energy...
    If A is even then a_P is non-zero... at this case you can have two different Z,N that give the same A which can achieve the least energy...
     
  4. May 28, 2015 #3
    Thanks thats really helpful and clears up the a_p values. I might be being realy stupid here but I'll ask anyway, if the a_p term is zero then how is it a parabola? The only thing I can think of is that if the term is missing and you put N=A-Z into the equation you get some sort of quadratic in Z, which would be a parabola?
     
  5. May 28, 2015 #4

    ChrisVer

    User Avatar
    Gold Member

    yup, it's the quadratic equation in Z...
    you can also check :
    http://users.jyu.fi/~ptg/FYSN300/FYSN300_L3.pdf [Broken]
     
    Last edited by a moderator: May 7, 2017
  6. May 28, 2015 #5

    ChrisVer

    User Avatar
    Gold Member

    eg the figure in page 10 and its caption...
     
  7. May 28, 2015 #6
    awesome, many thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Semi-empirical mass formula and the pairing term
  1. Fermions mass terms (Replies: 5)

Loading...