 Solution

Set $u=r\psi$, choose $\alpha >0$ for a negative bound-state energy, and write $\kappa =2mK/{\hslash }^2$. Multiplying the radial equation by $r$ givesFor $u=e^{-\alpha r}f$, its second derivative is $e^{-\alpha r}(f^{\prime \prime }-2\alpha f^{\prime }+{\alpha }^{2}f)$. Substitution cancels the two ${\alpha }^2$ terms and givesFor 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. PutThe equation for the series factor $v=f/r^{\ell +1}$ is $r{v}^{\prime \prime }+(2\ell +2-2\alpha r)v^{\prime }+(\kappa -2\alpha (\ell +1))v=0$, and coefficient comparison yieldsThe 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 $q\ge 0$, namely $2\alpha (q+\ell +1)=\kappa$. Thus $n=q+\ell +1$ is an integer greater than $\ell$, andRegularity, 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 $a_0={\hslash }^2/(mK)$,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 $r^{2}dr$. Constants and the angular integral cancel in the tail ratio. The supplied integral with $\lambda =2/a_0$ gives