= Solution
Put $y=e^{-r/\xi}$. Differentiating the <Morse potential> gives
$$
V'(r)=\frac1\xi(-2Ay^2+By).
$$
The finite-separation minimum has $y=B/(2A)$. Thus
$$
\boxed{a=\xi\ln\frac{2A}{B}=0.36809\,\mathrm{nm},\qquad V(a)=-\frac{B^2}{4A}=-0.250\,\mathrm{eV}.}
$$
The positive <bond-dissociation energy> is $D=B^2/(4A)=0.250\,\mathrm{eV}$, measured relative to the separated-atom value $V(\infty)=0$. This distinguishes the well's negative <potential energy> from the positive <energy> required to break the bond.
For the displacement $x=r-a$, the <Taylor expansion> is
$$
V(a+x)=V(a)+\frac12\kappa x^2+O(x^3),\qquad
\kappa=V''(a)=\frac{B^2}{2A\xi^2}=\frac{2D}{\xi^2}.
$$
Consequently
$$
\boxed{\kappa=70.862\,\mathrm{eV\,nm^{-2}}=11.353\,\mathrm{N\,m^{-1}}.}
$$
The restoring <force> is $-\kappa x$ to leading order, identifying the bond's <Hooke's law> <spring constant>.
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