# Infinite Square Well

1. May 1, 2010

### Slayer537

I've been working at this problem for about an hour and can't seem to make any progress. Any help would greatly be appreciated.

1. The problem statement, all variables and given/known data
Estimate the ground state energy level of a proton in the Al nucleus which has a potential energy of 100 MeV. Compare your answer to that calculated from the infinite square model. The radius of the Al nucleus is 5 fm.

2. The attempt at a solution

I thought that for the first part of the question this equation should be used

En = n2*h2/(8*m*L2)

However, I was getting nowhere close to the answer of 1.72 MeV. For the second part I figure that it would involve Schrodinger's equation and and this equation:

$$\psi$$ = (2/L)1/2*sin(n*pi*x/L)

Oddly enough using the first equation and using the diameter instead of the radius I got the right answer for the second part of the question of 2.05 Mev; however, I don't think that I solved it correctly.

2. May 1, 2010

### nickjer

The first part isn't an infinite well. The 2nd part is an infinite well, the first equation you listed is the energy levels for an infinite well, that is why it worked. Also, "L" is the width of the well which is the diameter and not the radius (that is why you got the right answer using diameter).

The first part sounds like you will be using a finite potential well. Unless you are learning some other method like the shell model.

3. May 2, 2010

### Slayer537

Thanks, that explains the second part of the question. Still can't figure out how to do the first part. We have done finite potential well, but not shell. I looked up the equations in my book and think that I should use Schrodinger's time independent equation:

-(ħ/2m)(d2/dx2)Ψ(x)+U(x)Ψ(x)=EΨ(x)

Where I would solve for E. Could I then use this for Ψ(x) :

Ψ(x)=(2/L)1/2sin(pi*x/L)

? If so what value would I use for x, or am I still missing something?

4. May 2, 2010

### Slayer537

Never mind. I just figured it out.

First solve for δ:

δ=ħ/(2*m*U)1/2

Then use δ to solve for Energy, making sure to use diameter, not radius:

E=pi22/(2*m*L2)

---> E = 1.72 MeV