# Nuclear shell model

1. Jun 15, 2010

### roshan2004

I am trying to explain nuclear energy levels in terms of Shell model and I hope that you guys would assist me through this-
1. The potential which exists inside the nucleus is assumed to be harmonic oscillator potential and it's value is obviously given by $$V(r)=\frac{1}{2}m\omega ^2r^2$$ However in my friend's note it is given as$$V(r)= -V_{o}+\frac{1}{2}m\omega ^2r^2$$ I don't know whether my friend's note is correct or not as I couldnot understand his equation at all.

Last edited: Jun 15, 2010
2. Jun 15, 2010

### Meir Achuz

The constant V_0 is correct in that simple model, but it doesn't affect the order of the energy levels.
It is there because the potential is ot necessarily zero at the origin.

3. Jun 15, 2010

### Orion1

independent particle nuclear shell model...

That potential is part of the independent particle nuclear shell model.

Harmonic oscillator potential:
$$V(r) = \frac{1}{2} m \omega^2 r^2$$

According to Wikipedia, a 'more realistic' potential is the Woods–Saxon potential.

Woods–Saxon potential:
$$V(r) = -\frac{V_0}{1 + \exp(\frac{r - R}{a})}$$

Reference:
http://en.wikipedia.org/wiki/Nuclear_shell_model#Deforming_the_potential"
http://en.wikipedia.org/wiki/Woods_Saxon_potential" [Broken]
http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/shell.html#c3"

Last edited by a moderator: May 4, 2017
4. Jun 15, 2010

### roshan2004

So, which one will be better for me to use, the first one or the second one?

5. Jun 16, 2010

### Meir Achuz

The HO potential has simple wave functions and energy levels, helpful in identifying shell model states. The W-S potential has to be treated numerically. You do have to add a spin-orbit term (L.S)
to the HO to get the correct energy levels and states.

6. Jun 16, 2010

### roshan2004

Now, I am trying to find the nuclear energy levels without spin orbit coupling which goes like this-(correct me if I have made some mistakes here-)
The Schrodinger's equation for the Quantum harmonic oscillator is given by-
$$-\frac{\hbar^2}{2m}\triangledown ^2\psi (r)+\frac{1}{2}m\omega ^2r^2\psi(r)=E\psi(r)$$
Solving this equation we have the quantum energy levels of the harmonic oscillator given by
$$E_{n}=(n+\frac{3}{2})\hbar\omega$$
On solving the radial and angular parts of this equation we get the quantum numbers $$n_{r}$$ and l which are related to n as $$n=2n_{r}+l-2$$ where $$n_{r}=1,2,3......$$ and l=0,1,2.... are known as s,p,d,f,g........ (orbitals of the nucleon)
The occupancy of the nucleons in nucleus is given by 2(2l+1) and I calculated different states and occupancy as
for n=0 I got 1s state with occupancy 2
for n=1 I got states 1s and 1p with occupancy 6
for n=2 I got states 1d and 2s with occupancy 12
for n=3 I got states 1f and 2p with occupancy 20
for n=4 I got states 1g,2d and 3s with occupancy 30
Now the problem is I just couldnot draw these energy levels like the one we have in the case of energy levels of nucleons after spin orbit coupling

Last edited: Jun 16, 2010
7. Jun 17, 2010

### Meir Achuz

The L.S coupling makes the J=L+S state lie lower than the J=L-S state. This changes the numbers because the ten 2d states split into 6 and then 4. By n= 2 (or maybe n=3), the 3d_5/2 lies lower than the 2p_1/2.
The common counting of states lists them so that what you call
1s and 1p for your n=1 are called 2s and 2p, and so on.

8. Jun 17, 2010

### Orion1

nuclear shell model quantum energy level...

The equation that I derived including spin orbit coupling is...

Nuclear spin quantum number:
$$s = \frac{1}{2}$$

Nuclear shell model quantum energy level:
$$\boxed{E_{nljs} = \hbar \omega \left[ \left( (2n_r + l - 2) + \frac{3}{2} \right) + C_1 \cdot l(l + 1) + \frac{C_2}{2} \left( j(j + 1) - l(l + 1) - s \left( s + 1 \right) \right) \right]}$$

Nuclear shell model constants:
$$C_1 = -0.0225$$ - spherical harmonic constant
$$C_2 = -0.1$$ - spin orbit coupling constant

Reference:
http://en.wikipedia.org/wiki/Principal_quantum_number" [Broken]
http://en.wikipedia.org/wiki/Angular_momentum#Relation_to_spherical_harmonics"
http://en.wikipedia.org/wiki/Total_angular_momentum" [Broken]
http://en.wikipedia.org/wiki/Spin_quantum_number" [Broken]
http://en.wikipedia.org/wiki/Spin–orbit_interaction#Evaluating_the_energy_shift"

Last edited by a moderator: May 4, 2017
9. Jun 17, 2010

### roshan2004

Thanks,but the suggestions given by you guys is regarding spin orbit coupling. I am trying to draw only the energy levels without spin orbit coupling from the states and occupancy I have calculated.

10. Jun 17, 2010

### Meir Achuz

Then your post #4 is it, but it won't give the right magic numbers.

11. Jun 17, 2010

### Orion1

Even without spin orbit coupling there is a spherical harmonic constant resulting from degeneracy of states with different orbital angular_momentum quantum numbers.

Principal quantum number:
$$N = (2n_r + l - 2)$$

$$n_r$$ - radial quantum number
$$l$$ - orbital angular_momentum quantum number

Nuclear shell model quantum energy level without spin orbit coupling:
$$\boxed{E_{nl} = \hbar \omega \left[ \left( (2n_r + l - 2) + \frac{3}{2} \right) + C_1 \cdot l(l + 1) \right]}$$

$$C_1 = -0.0225$$ - spherical harmonic constant

Integration via substitution:
$$N_N = (N + 1)(N + 2) = ((2n_r + l - 2) + 1)((2n_r + l - 2) + 2) = (2n_r + l)(2n_r + l - 1)$$

Number of occupied quantum states:
$$\boxed{N_N = (2n_r + l)(2n_r + l - 1)}$$

Reference: