The deuteron is a nucleus of "heavy hydrogen" consisting of one proton and one neutron. As a simple model for this nucleus, consider a single particle of mass m moving in a fixed spherically-symmetric potential V(r), defined by V(r)=-V0 for r<r0 and V(r)=0 for r>r0. This is called a spherical square-well potential. Assume that the particle is in a bound state with l=0.
a) Find the general solution R(r) to the radial Schrodinger equation for r<r0 and r>r0. Use the fact that the wave function must be finite at 0 and infinity to simplify the solution as much as possible. (You do not have to normalize the solutions)
b) The deuteron is only just bound; i.e., E is nearly equal to 0. Take m to be the proton mass, m=1.67x10^(-27) kg, and take r0 to be a typical nuclear radius, r0=1x10^(-15) m. Find the value of V0 (the depth of the potential well) in MeV (1 MeV = 1.6x10^(-13) J). (Hint: The continuity conditions at r0 must be used. The radial wave function R(r) and its derivative R'(r) must both be continuous at r0; this is equivalent to requiring that u(r) and u'(r) must both be continuous at r0, where u(r)=rR(r). The resulting equations cannot be solved exactly but can be used to derive the value for V0.)
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