= Solution
<Set> $u=r\psi$, choose $\alpha>0$ for a negative bound-state energy, and write $\kappa=2mK/\hbar^2$. Multiplying the radial equation by $r$ gives
$$
u''+\left[-\alpha^2+\frac\kappa r-\frac{\ell(\ell+1)}{r^2}\right]u=0.
$$
For $u=e^{-\alpha r}f$, its <second derivative> is $e^{-\alpha r}(f''-2\alpha f'+\alpha^2f)$. Substitution cancels the two $\alpha^2$ terms and gives
$$
\boxed{f''-2\alpha f'-\frac{\ell(\ell+1)}{r^2}f+\frac\kappa r f=0.}
$$
For the physical hydrogen problem, the angular eigenvalue has $\ell$ a <nonnegative integer>. Regularity at the origin selects the Frobenius exponent $u\sim r^{\ell+1}$, rather than the singular branch. Put
$$
f=r^{\ell+1}\sum_{j=0}^\infty c_jr^j.
$$
The equation for the <series> factor $v=f/r^{\ell+1}$ is $rv''+(2\ell+2-2\alpha r)v'+(\kappa-2\alpha(\ell+1))v=0$, and coefficient comparison yields
$$
c_{j+1}=\frac{2\alpha(j+\ell+1)-\kappa}{(j+1)(j+2\ell+2)}c_j.
$$
The regular confluent-hypergeometric solution generically grows like $e^{2\alpha r}$ in this <series> factor; after the extracted $e^{-\alpha r}$ it is not <normalizable>. A <bound state> requires termination at some degree $q\ge0$, namely $2\alpha(q+\ell+1)=\kappa$. Thus $n=q+\ell+1$ is an <integer> greater than $\ell$, and
$$
\boxed{\alpha=\frac{mK}{\hbar^2n},\qquad E_n=-\frac{mK^2}{2\hbar^2n^2}.}
$$
Regularity, normalizability and the <integer> angular index are the physical conditions selecting this spectrum; the <differential equation> without those conditions would admit many other solutions.
For $n=1$, necessarily $\ell=0$ and $q=0$. With $a_0=\hbar^2/(mK)$,
$$
\boxed{f_1(r)=Cr,\qquad\psi_1(r)=Ce^{-r/a_0}.}
$$
A normalized spatial <ground state> is $\Psi_1=(\pi a_0^3)^{-1/2}e^{-r/a_0}$. In <spherical coordinates> the radial <probability> has weight $r^2dr$. Constants and the angular <integral> cancel in the tail ratio. The supplied <integral> with $\lambda=2/a_0$ gives
$$
\Pr(r>a_0)=\frac{\int_{a_0}^\infty r^2e^{-2r/a_0}\,dr}{\int_0^\infty r^2e^{-2r/a_0}\,dr}
=e^{-2}(1+2+2)=\boxed{5e^{-2}}.
$$
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