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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2024

Paper 1

For fixed $l$, the radial equation is
$$
-\frac{\hbar^2}{2\mu r^2}\frac{d}{dr}
\left(r^2\frac{dR}{dr}\right)
+\frac{\hbar^2l(l+1)}{2\mu r^2}R
-\frac{e^2}{4\pi\epsilon_0r}R
=ER.
$$
Substitute
$$
R(r)=Cr^l e^{-r/b}.
$$
Cancellation of the coefficient of $1/r$ requires
$$
b=(l+1)a,
\qquad
\boxed{a=\frac{4\pi\epsilon_0\hbar^2}{\mu e^2}},
$$
the reduced-mass <Bohr radius>. The remaining constant term gives
$$
\boxed{E_l=-\frac{\hbar^2}{2\mu a^2(l+1)^2}
=-\frac{e^2}{8\pi\epsilon_0a(l+1)^2}}.
$$
Each such no-radial-node state is a <circular Coulomb bound state> with principal quantum number $n=l+1$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 1 / 14A / ii / Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 1 / 14A / ii / Solution

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Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 1 / 14A / ii / Solution

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