Semi-empirical mass formula and the pairing term

  • Thread starter rwooduk
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We have the formula for the mass of an atom:

uNvv565.jpg


From our class notes I have:

By keeping A constant and varying Z there is generally only one stable nuclide for each odd value of A. We can show this by looking at the pairing term to show that odd A gives a single parabola with a single minimum.
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?
 

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  • #2
ChrisVer
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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...
 
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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...
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?
 
  • #4
ChrisVer
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yup, it's the quadratic equation in Z...
you can also check :
http://users.jyu.fi/~ptg/FYSN300/FYSN300_L3.pdf [Broken]
 
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ChrisVer
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eg the figure in page 10 and its caption...
 
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awesome, many thanks!
 

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