Question

6.16

The deuteron is a nucleus of heavy hydrogen consisting of ne proton and one neutron.As a simple model for this nucleos consider a single particle of mass m moving in a fixed spehreical -symmetric potencial V(r) defined by V(r)=-Vo for rro .This is called a spherical square well potential.Asssume that the particle is in a bound state with l=0

a) Fin the general solutions R(r) to the radical Schoringer euatio for rro.Use the fact that the wave function must be finite at 0 and ∞ to simpfy the solution as much as possible.(You don’t have to normalieze the solutions)

THIS PICTURE IS THE ANSWER ,BUT I DON’T UNDERSTAND WHY B1=0 AND A2=0. Could you please DRAW THE WAVE and explain it !! My book said r=0 and u(0)= 0 so B1 = 0.I don’t understand why and how the wave works here. Thank you

Fit to page Page view 6.16 (a) Taking 1 = 0, and defining u(t) = R(T), the radial Schrodinger equa na au(r) + V (r)u(r) = Eur) 2m 8r2 The potential is V = -V, for r < to and V = 0 for r>ro, so bound state wave functions correspond to - Vo <E<0. Then the general solutions to the radial Schrodinger equation are u = A, sin(kır) + B, cos(kır) for T < To and uy = Anekat + Bee-kut for T > To, where ky and k, are given by - 2m(Vo+ E and ky = 2011- But R(r) = u(r)/must be finite at r = 0, so u(0) = 0, so B = 0. Also we want R to be finite asr gives A, = 0. Then the radial wave function R(r) is R() = A, sin(kin) for r < to and Ry(t) = B, for T > to where k, and ky are given above. S