= Solution
The <scattering length> is $a_0=-\lim_{k\to0}\tan\delta_0/k$. Since $\sin(kR)\sim kR$ and $\cos(kR)\sim1$, the exact <phase shift> gives, away from the pole,
$$
\boxed{a_0=R\frac{g}{1+g},\qquad g=\frac{2mV_0R}{\hbar^2}}.
$$
Thus plot $a_0/R$ against $g$, so that both axes are <dimensionless>.
\Image[/past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2018-delta-shell-scattering-length.png]
{title=Delta-shell scattering length versus dimensionless coupling, with the bound-state threshold at g = −1}
For repulsion, $g>0$, the curve rises from zero to the horizontal asymptote $a_0/R=1$, staying between zero and one. For weak attraction, $-1<g<0$, the <scattering length> is negative and tends to $-\infty$ as $g\to-1^+$. For stronger attraction, $g<-1$, it is positive and greater than $R$, tending to $+\infty$ as $g\to-1^-$ and to $R$ as $g\to-\infty$. Its only zero is $g=0$.
The pole at $g=-1$ is the zero-energy threshold of an S-wave <bound state>. To see this directly, use $u\propto\sinh(\kappa r)$ inside and $u\propto e^{-\kappa r}$ outside for energy $-\hbar^2\kappa^2/(2m)$. The same derivative jump gives
$$
g=-\kappa R[1+\coth(\kappa R)]
=-\frac{2\kappa R}{1-e^{-2\kappa R}}.
$$
This approaches $-1$ as $\kappa\downarrow0$ and admits a positive $\kappa$ for $g<-1$. At the threshold the <scattering length> diverges; this is a <zero-energy scattering resonance>, not a divergence of the finite-energy cross-section beyond its <partial-wave unitarity> bound. These are the main features of the <scattering length of a delta-shell potential>.
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