 Solution

The general time-independent Schrödinger equation is $[-{\hslash }^2{\nabla }^2/(2m)+V]\psi =E\psi$. For a spherically symmetric wave function, ${\nabla }^2\psi ={\psi }^{\prime \prime }+2{\psi }^{\prime }/r$; the Coulomb attraction between the proton and electron is $V=-q^2/(4\pi {ϵ}_{0}r)$. These supply respectively the kinetic and potential terms in the displayed radial equation. Rearranging identifiesas the Bohr radius.

Substitute $\psi =(A+Br)e^{-\lambda r}$ into the rearranged equation and put $ϵ=2mE/{\hslash }^2$. Equating the coefficients of $r,1,r^{-1}$ after cancelling the exponential givesA nonzero state has $A\ne 0$. If $B=0$, these identities give $\lambda =1/a$ and $ϵ=-1/a^2$. If $B\ne 0$, they give $\lambda =1/(2a)$, $B=-A/(2a)$, and $ϵ=-1/(4a^2)$. Thus the two independent eigenfunctions and their corresponding energies areThey are eigenfunctions at different energies, rather than two independent solutions at one fixed energy.

For the higher-energy state, radial probabilities include the spherical volume factor $4\pi r^2$. The normalization constant and $4\pi$ cancel in the ratio. Using the supplied integral at $b=a$ and at infinity,HenceThis calculation fixes the denominator as $8e$ in the source's unparenthesized expression; interpreting it as a product with $e$ would give a negative probability.