Since we have made a transformation to spherical coordinates it means that \( r\in [0,\infty) \). The quantum number \( l \) is the orbital momentum of the nucleon. Then we substitute \( R(r) = (1/r) u(r) \) and obtain $$ -\frac{\hbar^2}{2 m} \frac{d^2}{dr^2} u(r) + \left ( V(r) + \frac{l (l + 1)}{r^2}\frac{\hbar^2}{2 m} \right ) u(r) = E u(r) . $$ The boundary conditions are \( u(0)=0 \) and \( u(\infty)=0 \).