Mass for solving the nuclear energy level

In summary, the person is trying to solve a problem involving the finite square well for ##s## states using graphical methods. They have successfully found the energy levels and wavefunctions for a deuteron using the reduced mass, but are unsure of which mass to use for a larger nucleus like ##^{48}Ca##. They have tried using the reduced mass from literature, but their results do not match with realistic Skyrme Hartree Fock calculations. The person then shares their graphical solutions and asks for help in understanding their calculations and why they do not match with the Skyrme Hartree Fock results. They also mention that they are following guides from a textbook on Quantum Mechanics.
  • #1
just_mb
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0
Hi all,
I'm trying to solve a problem of finite square well for the ##s## states graphically. The task is to find energy levels and wavefunctions of proton in a spherically symmetric potential, first for deuteron then ##^{48}Ca##. What makes me confused is the mass. For deuteron, the mass used is the reduced mass because it's a two-masses system. Using that, the energy from my calculation is similar to that from literature. But what about the ##^{48}Ca##, what mass should I use? When I use reduced mass of ##m_{proton}(19 \times m_{proton} + 28 \times m_{neutron})/(20 \times m_{proton} + 28 \times m_{neutron})## which I found from a literature, the energies don't match with realistic Skyrme Hartree Fock calculations (figure attached). When I use the total mass ##(20 \times m_{proton} + 28 \times m_{neutron})##, I found 7 ##s## energy levels as oppose to 2, as shown in the figure. Please help me, I have been searching through tons of books and journals with no luck.

Screen Shot 2019-07-07 at 11.54.03 am.png
 
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  • #2
Use the reduced mass, or the proton mass - they are nearly the same for larger nuclei.
It is hard to tell what went wrong if you don't show your results.
 
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  • #3
mfb said:
Use the reduced mass, or the proton mass - they are nearly the same for larger nuclei.
It is hard to tell what went wrong if you don't show your results.
Thanks. That's what I used and this is the graphical solutions that I got.
246252

where ##\xi## is given by $$\xi = \sqrt{\frac{2m(V_0 - |E|)}{\hbar^2}}a$$ where ##V_0 = 45 \, MeV##, and ##a = ½A^⅓ = 4.36\, fm##. As shown in the graph, the solutions are ##\xi = 2.702## and ##5.298## which according to the equation above correspond to ##E = 1.04\, MeV## and ##13.73\, MeV##. The coulomb energy for ##^{48}Ca## is ##75\, MeV##. If I substract those energies by Coulomb energy, the results are nowhere near those in Skyrme Hartree Fock calculations. What's wrong with my calculation?
 

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  • #4
It is still not very clear what you did, and it is surprising that you only found two solutions.

Is this part of some textbook problem (so you know this approach should give a useful answer)?
 
  • #5
mfb said:
It is still not very clear what you did, and it is surprising that you only found two solutions.
Only the ##s## states need to be solved so I think there are only two solutions, as suggested by Skyrme Hartree Fock calculations. I am also confused with that solution. In the picture, for example, the ##0s1## state of protons only differ by about 1 MeV from that of neutrons. The difference is caused by Coulomb energy. But from my calculation, the Coulomb energy is 75 MeV. Am I understanding it correctly? Can you please explain this to me?

mfb said:
Is this part of some textbook problem (so you know this approach should give a useful answer)?
No, it's not. But we follow the guides from Quantum Mechanics by Schiff.
 
  • #6

1. What is the purpose of mass for solving nuclear energy levels?

The mass of a nucleus is important for understanding nuclear energy levels because it determines the stability and binding energy of the nucleus. This information is crucial for designing and controlling nuclear reactions.

2. How is mass used to calculate nuclear energy levels?

The mass of a nucleus is used in the famous equation E=mc², where E represents energy, m represents mass, and c represents the speed of light. This equation allows us to calculate the amount of energy released or absorbed in a nuclear reaction.

3. How does mass affect the stability of a nucleus?

The more massive a nucleus is, the more stable it is. This is because the strong nuclear force, which holds the nucleus together, is stronger than the repulsive force between positively charged protons. A more massive nucleus has more neutrons, which help to balance out the repulsive forces and keep the nucleus stable.

4. Can mass be changed to alter nuclear energy levels?

Yes, mass can be changed through nuclear reactions such as fission and fusion. In fission, a nucleus is split into smaller fragments, releasing energy. In fusion, two smaller nuclei combine to form a larger one, also releasing energy. These reactions change the mass of the nuclei involved, which in turn affects the nuclear energy levels.

5. How is mass measured in nuclear energy level calculations?

Mass is measured in atomic mass units (amu) in nuclear energy level calculations. One amu is equivalent to 1/12 the mass of a carbon-12 atom. The mass of a nucleus is typically expressed in terms of its atomic mass number, which is the total number of protons and neutrons in the nucleus.

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