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Let $u=rF$. Multiplication of the radial equation by $-2mr/{\hslash }^2$ gives $u^{\prime \prime }+[A/r-\lambda (\lambda +1)/r^2-{\alpha }^2]u=0$. With $u=e^{-\alpha r}f$ the constant terms cancel, leavingThe indicial equation is $c(c-1)=\lambda (\lambda +1)$, with roots $c=\lambda +1$ and $c=-\lambda$. The usual physical Coulomb radial origin condition selects $c=\lambda +1$: it imposes the regular reduced radial boundary condition $u(0)=0$. Normalizability alone at the origin is insufficient to make this selection when an irregular branch is locally square-integrable, notably at $\lambda =0$.

Write $f=r^{\lambda +1}{\sum }_{j\ge 0}{a}_j{r}^j$. Coefficient comparison givesA nonterminating regular series has the exponentially growing asymptotic branch, suggested by $a_{j+1}/a_j\sim 2\alpha /j$: it supplies an $e^{2\alpha r}$ component, which the prefactor $e^{-\alpha r}$ cannot normalize. Equivalently the regular confluent-hypergeometric solution grows unless its first parameter is a nonpositive integer. Thus a bound state requires the numerator to vanish at some $j=k\ge 0$, givingThis is series-termination quantization of the Coulomb radial equation. It requires $A>0$ for the assumed $\alpha >0$; at $A=0$ there are no such negative-energy Coulomb bound states.

For $k=0$, $F=N{r}^{\lambda }{e}^{-\alpha r}$. The angular integral cancels when computing $⟨1/r⟩$, while the radial volume element contributes $r^{2}dr$. ConsequentlyThe ratio follows by one integration by parts and remains valid for real $\lambda \ge 0$; no integer-only factorial identity is needed. Since now $\alpha =A/[2(\lambda +1)]$,For a classical particle with that negative total energy, nonnegative kinetic energy requires $E-V={\hslash }^2(A/r-{\alpha }^2)/(2m)\ge 0$, hence $\boxed{r\le A/{\alpha }^2}$. This is a necessary radial bound; extra angular momentum can impose a stronger restriction.

At $k=\lambda =0$, $\alpha =A/2$ and the classical boundary is $r=2/\alpha$. The normalized radial probability density is $4{\alpha }^3{r}^2{e}^{-2\alpha r}$. With $z=2\alpha r$, the exterior quantum probability isThe final integration gives $(4^2+2\cdot 4+2)e^{-4}/2$. This is the nonzero quantum tail beyond the classical turning region.

Apply the Frobenius method at the origin. The lowest power in $f^{\prime \prime }-\ell (\ell +1)f/r^2$ is $r^{c-2}$, with coefficientThe terms $-2\alpha f^{\prime }$ and $2mKZf/({\hslash }^{2}r)$ start at power $r^{c-1}$ and therefore cannot cancel a nonzero $r^{c-2}$ coefficient. Because $a_0\ne 0$, the indicial equation isIts roots are $c=\ell +1$ and $c=-\ell$. Since $\psi =r^{-1}f{e}^{-\alpha r}$, their leading behaviors are respectively $\psi \sim r^{\ell }$ and $\psi \sim r^{-\ell -1}$. The physical choice isThis is the Coulomb radial origin condition. For $\ell \ge 1$, the other branch is not even locally normalizable: its radial probability integral behaves as ${\int }_0^{\epsilon }{r}^2{r}^{-2\ell -2}dr={\int }_0^{\epsilon }{r}^{-2\ell }dr$, which diverges.

For $\ell =0$, the singular behavior $1/r$ is locally square-integrable with the measure $r^{2}dr$, so local wavefunction normalization alone would not reject it. The ordinary three-dimensional Coulomb potential problem additionally requires the regular boundary condition $u(0)=0$. A $1/r$ branch violates that condition and has divergent kinetic energy integral ${\int }_0^{\epsilon }{r}^2|{\partial }_r(1/r){|}^{2}dr={\int }_0^{\epsilon }{r}^{-2}dr$. Thus it is excluded from the physical energy domain as well. The regular choice gives $u\sim r^{\ell +1}$ in every case.