Energy and Excited States in Semi-Infinite-Well Potential

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SUMMARY

The discussion focuses on modeling a deuteron using a semi-infinite well potential with a width of 3.5 x 10-15 m and a potential energy of Vo = -2.2 MeV. The ground state energy E for the deuteron is determined using the formula E(n) = (n2 * π2 * ħ2) / (2 * m * L2), where n is the quantum number, m is the mass of the deuteron, and L is the width of the well. The excited states can be calculated similarly, with the threshold for valid states being dictated by the potential energy limit of 2.2 MeV.

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Homework Statement



Use the semi-infinite-well potential to model a deuteron, a nucleus consisting of a neutron and a proton. Let the well width be 3.5*10-15 m and Vo -(minus) E = 2.2 MeV. Determine the energy E (in MeV).

How many excited states are there? 0 1 2 3 4 5

What are their energies? (If an excited state doesn't exist, enter a value of zero.)
1st-
2nd-
3rd-
4th-
5th-


Homework Equations





The Attempt at a Solution



I assume that the ground state of the deuteron (n=1) is the E, is this correct in my thinking?

So after you find out the K value for the deuteron with L=width of the well, you use that k value to find out the E(n=1) ground state (which will be the energy we are looking for in part 1), is this correct as well?

As far as the excited states go...how do I determine when we stop using the formula En=n^2...(forgot the rest the book isn't in front of me). We can keep gooing on and on with multipule excited states.

When do we know the threshold...does it have to do with the 2.2 MeV Vo being greater than the E??
 
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Please explain the whole process of finding the energy E (in MeV) and the excited states.Thanks for your help!!
 

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