 Solution

For an S-wave, put $u(r)=r\psi (r)$. A bound state has $E=-\epsilon <0$, and the radial Schrödinger equation giveswhereContinuity of $u,u^{\prime }$ at $a$ givesThe first bound state appears at threshold $\kappa \downarrow 0$, where $ka=\pi /2$. The zero-energy threshold itself is not normalizable, so a deuteron bound state exists precisely whenThe wavefunction is $\psi =u/r$, with normalization fixing the remaining constant.