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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2014

Paper 2

The normalized <spherical harmonics> separate the angular variables. The <Radial Schrodinger equation for the hydrogen atom> is
$$
-\frac{\hbar^2}{2m}\left(R''+\frac2rR'-\frac{l(l+1)}{r^2}R\right)
-\frac{e^2}{4\pi\epsilon_0r}R=ER.
$$
Write $R=Ar^le^{-\kappa r}$. Then
$$
\frac{R''+2R'/r-l(l+1)R/r^2}{R}
=\kappa^2-\frac{2\kappa(l+1)}r.
$$
Substitution separates a constant term and a $1/r$ term. The latter vanishes precisely when $\hbar^2\kappa(l+1)/m=e^2/(4\pi\epsilon_0)$. With $\kappa=1/[a(l+1)]$ this determines the <Bohr radius>
$$
\boxed{a=\frac{4\pi\epsilon_0\hbar^2}{me^2}.}
$$
The constant term determines
$$
\boxed{E=-\frac{\hbar^2}{2ma^2(l+1)^2}
=-\frac{e^2}{8\pi\epsilon_0a(l+1)^2},\qquad D=8(l+1)^2.}
$$
These are the <maximal-angular-momentum hydrogen states>, with principal quantum number $n=l+1$.

Because the angular integral is one, normalization requires
$$
1=|A|^2\int_0^\infty r^{2l+2}e^{-2\kappa r}\,dr
=|A|^2\frac{(2l+2)!}{(2\kappa)^{2l+3}}.
$$
Therefore
$$
\boxed{|A|^2=\frac{2^{2l+3}}{a^{2l+3}(2l+2)!(l+1)^{2l+3}},
\quad F=2l+3,\ G=2l+2,\ H=2l+3.}
$$
The phase of $A$ remains arbitrary.

Solution

Past exam of the mathematics course of the University of Cambridge / 2014 / ib / Paper 2 / 17A / i / Solution

Past exam of the mathematics course of the University of Cambridge / 2014 / ib / Paper 2 / 17A / i / Solution

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