What is the ratio of the de Broglie wavelength of an electron...

According to Bohr's quantization rule, the angular momentum is $mvr = \dfrac{nh}{2\pi}$.

The de Broglie wavelength of an electron is given by $\lambda = \dfrac{h}{mv}$.

Substituting $mv$ from the first equation, we get $\lambda = \dfrac{2\pi r}{n}$.

The radius of the $n$-th orbit of a hydrogen-like atom is $r \propto \dfrac{n^2}{Z}$.

Therefore, the de Broglie wavelength is proportional to $\lambda \propto \dfrac{n}{Z}$.

For the electron in the third orbit of $He^+$, $n_1 = 3$ and $Z_1 = 2$.

$\lambda_1 \propto \dfrac{3}{2}$

For the electron in the second orbit of $Li^{2+}$, $n_2 = 2$ and $Z_2 = 3$.

$\lambda_2 \propto \dfrac{2}{3}$

The ratio of their de Broglie wavelengths is $\dfrac{\lambda_1}{\lambda_2} = \dfrac{\dfrac{3}{2}}{\dfrac{2}{3}} = \dfrac{9}{4}$.

Answer: $9 : 4$