In atomic units the <Time-independent Schrodinger equation> is
$$
\left(-\frac12\nabla^2-\frac1r\right)\psi=E\psi.
$$
The displayed equation is the <Radial Schrodinger equation for the hydrogen atom>. Its radial derivative terms come from the <Laplacian in spherical coordinates>, the Coulomb term is the potential $-1/r$, and separation into <spherical harmonics> uses the <orbital angular momentum> eigenvalue $l(l+1)$, producing the centrifugal term $-l(l+1)R/r^2$.

For $E=-1/(2n^2)$, substitute $R=r^\alpha e^{-r/n}$. Dividing the equation by $R$ and comparing powers of $r$ gives
$$
\alpha(\alpha+1)=l(l+1),
\qquad
\frac{\alpha+1}{n}=1.
$$
The solution regular at the origin has $\alpha=l$, and when $l=n-1$ both equations agree. This is the <circular Coulomb bound state>. Thus
$$
\boxed{\alpha=n-1.}
$$
The radial probability measure is $|R(r)|^2r^2\,dr$. With $a=2/n$ and the <Gamma integral>, the <mean radius of a circular Coulomb bound state> is
$$
\langle r\rangle
=\frac{\int_0^\infty r^{2n+1}e^{-ar}\,dr}
{\int_0^\infty r^{2n}e^{-ar}\,dr}
=\frac{2n+1}{a}
=\boxed{\frac{n(2n+1)}2}.
$$

At fixed <principal quantum number> $n$, the allowed <orbital angular momentum> quantum numbers are $l=0,\ldots,n-1$, and each has <magnetic quantum number> $m=-l,\ldots,l$. Hence the orbital <quantum degeneracy> is
$$
\boxed{\sum_{l=0}^{n-1}(2l+1)=n^2.}
$$
Including electron spin would double this to $2n^2$, but spin is absent from the stated wavefunctions.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ib / Paper 2 / 17B / Solution

{title2=$n$}
{wiki}

For a hydrogenic bound state, the principal quantum number is the positive integer $n$ that determines the unperturbed energy. Its allowed orbital quantum numbers are $\ell=0,\ldots,n-1$.


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 Incoming links: Principal quantum number