Semi-empirical mass formula and the pairing term

In summary, the student found that if the atom's atomic number is odd, then the pairing term (which is a function of the element's atomic number and the nuclear charge) will miss one term from the atomic energy.
  • #1
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
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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  • #3
ChrisVer said:
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 that's really helpful and clears up the a_p values. I might be being really 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
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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  • #5
eg the figure in page 10 and its caption...
 
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  • #6
awesome, many thanks!
 

1. What is the Semi-empirical mass formula?

The Semi-empirical mass formula is a mathematical formula used to estimate the nuclear binding energy of atomic nuclei. It takes into account the number of protons and neutrons in the nucleus, as well as other factors such as the pairing energy and the asymmetry energy.

2. How does the pairing term affect the Semi-empirical mass formula?

The pairing term in the Semi-empirical mass formula takes into account the additional binding energy that is present when protons and neutrons are paired in the atomic nucleus. This pairing energy is stronger for even numbers of protons and neutrons, leading to a more stable nucleus.

3. What is the role of the pairing term in nuclear stability?

The pairing term plays a crucial role in determining the stability of atomic nuclei. It contributes to the overall binding energy of the nucleus, making it more stable. Without the pairing term, many nuclei would be unstable and undergo radioactive decay.

4. How does the pairing term affect the mass of the nucleus?

The pairing term affects the mass of the nucleus by adding an additional binding energy, which in turn affects the overall mass of the nucleus. This additional binding energy reduces the mass defect, making the nucleus slightly more massive than it would be without the pairing term.

5. Are there any limitations to the Semi-empirical mass formula and the pairing term?

Yes, there are limitations to the Semi-empirical mass formula and the pairing term. It is based on empirical data and is not accurate for all nuclei, especially those with extreme proton-neutron ratios. Additionally, it does not take into account other factors such as nuclear spin and nuclear deformation, which can also affect the stability of nuclei.

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