For a hydrogen atom, the ratio of the total energies of an electron...

The total energy of an electron in the $n^{\text{th}}$ orbit of a hydrogen atom is given by $E \propto -\dfrac{1}{n^2}$.

The ratio of the total energies in orbits A and B is:
$\dfrac{E_A}{E_B} = \dfrac{n_B^2}{n_A^2} = \dfrac{1}{9}$

Taking the square root of both sides gives:
$\dfrac{n_B}{n_A} = \dfrac{1}{3} \Rightarrow \dfrac{n_A}{n_B} = \dfrac{3}{1}$

According to Bohr's quantization condition, the orbital angular momentum of an electron is $L = \dfrac{nh}{2\pi}$, which implies $L \propto n$.

Therefore, the ratio of their orbital angular momenta is:
$\dfrac{L_A}{L_B} = \dfrac{n_A}{n_B} = \dfrac{3}{1}$

The ratio $L_A : L_B$ is $3:1$.

Assuming $L \propto E$ directly would incorrectly yield $1:9$, and inverting the ratio of $n$ would incorrectly yield $1:3$.

Answer: $3:1$