If the ground state of a particle in a spherical square well is just barely bound, show that the well depth and radius parameters $V_0$ and $a$ are related to the binding energy by the expansion
$$
\frac{2 m V_0 a^2}{\hbar^2}=\frac{\pi^2}{4}+2 \kappa a+\left(1-\frac{4}{\pi^2}\right)(\kappa a)^2+\cdot \cdot
$$
where
$$
\hbar \kappa=\sqrt{-2 m E}
$$
The deuteron is bound with an energy of $2.226 \mathrm{MeV}$ and has no discrete excited states. If the deuteron is represented by a nucleon, with reduced mass, moving in a square well with $a=1.5$ fermi, estimate the depth of the potential.