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Finding the potential of a 1d finite square potential well 
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#1
Feb2812, 11:15 AM

P: 1

1. The problem statement, all variables and given/known data
The deuterium nucleus (a bound state of a proton and a neutron) has one bound state. The force acting between a proton and a neutron has a strong repulsive component of range 0.4 fm and an attractive component of range ~2.4 fm. The energy needed to seperate the neutron from the proton in a deuterium nucleus is 2.2 MeV. Treat the neutron in deuterium as a particle of mass 1.67*10^27 kg in a potential well f width 2 fm. Estimate V_0 for this potential. 2. Relevant equations This potential well starts at Psi(x) = V_0 for x< a/2, then Psi(x) = 0 for x between a/2 and a/2, then back up to V_0. m = 1.67*10^27 kg E = energy hbar = reduced planck's constant k_2 = sqrt(2*m*E/hbar^2) alpha = k_2*a/2 V_0 is potential P = sqrt(m*V_0*a^2/(2*hbar^2)) alpha*tan(alpha) = sqrt(P^2alpha^2) alpha*cot(alpha) = sqrt(P^2alpha^2) 3. The attempt at a solution There's one bound state, so it'll be in the alpha*tan(alpha) = sqrt(P^2alpha^2) equation. I got it down to tan(548000sqrt(E)) = sqrt(V/E1) (in joules). V should come out to be 66 MeV according to his answers (he gives answers to some questions, just cares about work.) It's possible this was a typo too as it's happened before, but I'll assume it's right. I have no idea how to find E though. I assume it has to do with that 2.2 MeV used to remove the neutron but I'm not sure how. If my equation is wrong, that's not that big of a deal, I can fix it, I'm mostly worried about how to find E. My only idea is to relate E and V by conservation of energy but I'm not sure how. 


#2
Mar1412, 10:17 PM

P: 17

Hey there, that's an interesting question. I haven't thought this whole problem through but one thing I do notice is that if you treat the proton and neutron together as one system, then you are treating it as one particle. Therefore you must use the reduced mass between the two, which is μ = m1m2/(m1+m2) = 1/2m since the proton and neutron have basically the same mass.
http://en.wikipedia.org/wiki/Reduced_mass Then you can use the Schrodinger equation to find the energy of the standing wave solution to the problem. h^{2}/2μ*∂_{x}^{2}ψ + V(x)ψ = Eψ The function V(x) is discontinuous as you say, but the ψ must still be differentiable at the boundary points. I'm sure there would be some minimum energy state of ψ, and difference between V_{0} and E_{0} would be energy necessary to free it. I looked in a quantum mechanics book, and for a problem with zero potential energy outside the box and negative energy V inside, it says that E cannot be expressed explicitly, but must rather be numerically approximated for. Assuming a cosine function inside potential well, and an exponential well outside the potential, they get ψ = Acos(√(2m(VE))*x/h) for x< L ψ = Be^{√(2mE)*x/h} for x > L ψ = Be^{√(2mE)*x/h} for x < L Assuming that ψ is continuous and differentiable at x = L, then we get: Continuous: Acos(√(2m(VE))*L/h) = Be^{√(2mE)*L/h} Differentiable: A√(2m(VE))*L/h*sin(√(2m(VE))*L/h) = B√(2mE)*L/h*e^{√(2mE)*L/h} Dividing the two equations gets: cot(√(2m(VE))*L/h) = √(VE) / √E If you plug in hbar for h, the reduced mass for m, the potential difference for V and the distance a/2 for L, then I think should be able to approximate what E is. 


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