# Nuclear shell model - finite square well

I am trying to make a simplified approximation and solve Schrodinger equation in the finite square well to model the nucleus of Ca (shell nuclear model). The potential is $V(r) = -V_0$ for $0<r<a$ and $V(r)=0$ for $r>a$. So $\psi_{nlm}(r,\theta,\phi) = R_{nl}(r)Y_{lm}(\theta,\phi)$. Now I am solving the equation for $R$: $$\frac{d^2 R}{d \rho^2}+\frac{2}{\rho}\frac{dR}{d\rho}+ \left[1-\frac{l(l+1)}{\rho^2} \right]R=0$$ Where $\rho = \alpha r$ if $r<a$ and $\rho = \beta r$ if $r>a$. And $$\alpha = \left[ \frac{2m(V_0-|E|)}{\hbar^2} \right]^{1/2} \\ \beta = \left[ \frac{2m|E|}{\hbar^2} \right]^{1/2}$$ To calculate $E$, I am using the Coulomb potential given by: $$E = \frac{3}{5}\frac{e^2}{4\pi \epsilon_0}\frac{Z(Z-1)}{R'}$$ Where $R'$ is the nuclear radius and $R' = R_0 A^{1/3}$, with $R_0=1.2$ fm and $A$ is the mass number (=48 for Calcium). Plugging these numbers in and making $Z=20$ I found $E\approx 75$ MeV. But I am supposed to use $V_0=45$ MeV, which does not make sense since I expected $E<V_0$. What could be the problem here?

• I can confirm that your $E$ is around 75 MeV (assuming "eV" was a typo). I observe that calcium-48 has a mass excess of $-44$ MeV (though that's different from the binding energy). But I don't understand how you're using both a square well and a Coulomb potential. Shouldn't your $E$ be the eigenvalue of $\hat H$ once you've found your wavefunction? Are you treating protons and neutrons differently? Modeling $\beta$ or $2\beta$ decay?
– rob
May 30, 2014 at 3:43
• The Coulomb potential was used in order to estimate the energy of the proton. Yes, I am treating them differently. Only protons will have this energy $E$. Page 140 of this pdf shows something similar to what I am trying to do physics.umd.edu/courses/Phys741/xji/chap8_12.pdf May 30, 2014 at 3:59
• Shouldn't you then have a deep square well for the neutrons and a shallower square well for the protons?
– rob
May 30, 2014 at 4:08
• I think so. But I am solving only for protons in this model. May 30, 2014 at 4:15

You can't have both a square well and a Coulomb potential — they are mutually exclusive. In Gamow's model for $\alpha$ decay you see people talking about a square well for the $\alpha$ while it's inside the nucleus, and a $1/r$ potential while it's outside the nucleus. Sometimes the Gamow potential is said to have a "Coulomb barrier." But that's not the potential that you describe: you have $V=0$ outside the nucleus.
In that case your electrostatic energy $E$ is a red herring. The energy that matters in the Schrödinger equation is the eigenvalue of the Hamiltonian operator $H$, once you have found a suitable wavefunction.
• You are completely correct. After thinking a bit more about this, I realised I should find the energies $E$ by applying the condition of continuity at $r=a$. May 30, 2014 at 6:06