pedrone
Banned
- Joined
- Mar 31, 2011
- Messages
- 508
Quote:
2- However, with Z=8 and Z=20, it is taken in consideration the higgest level of energy. We can see in the graphic of the page 676 that the level 1d is lower in the level 1d5/2 , and therefore from Mayer-Jensen theory the nucleus 14Si would have to be a magic number, and not the nucleus 20Ca (this one has four protons in the highest level 1d3/2). By the same way, because 1p3/2 is the lower level, from their theory the nucleus 6C would have to be a magic number, and not the nucleus 8O (this one has two protons in the highest level 1p1/2).
You have misunderstood completely. The level scheme on the left is what you would expect without a spin-orbit potential. The level scheme on the right is what you get with a spin-orbit potential. It was the work of Mayer and Jensen (and coworkers) that showed that the spin-orbit potential allowed a correct matching of observed and theoretical magic numbers. This is precisely what they won the Nobel prize for.
No, actualy you are tying to fool everbogy in here.
What you say is just as it's written bellow the graphic in Eisberg-Resnick book.
What I said in my previous post is NOT regarding to the left side of the graphic, as you smartly are suggesting that should be my interpretation.
What I said is concerning the right side, just where it's considered the interaction WITH a spin-orbit potential.
As it can be seen easily, the Mayer-Jensen theory is disagree to the RIGHT side of the graphic (with spin-orbit potential) for the nuclei 20Ca and 8O.
They could not be magic numbers.
Instead of, 14Si and 6C would have to be, according to the Mayer-Jensen theory.
Tubbythin, stop trying to fool us with your lies.
