The angular momentum of an electron in the $2^{nd}$ excited state of a Helium ion $(He^+)$ is

Let $A_n$ be the area enclosed by the $n^{th}$ orbit in a hydrogen atom. The graph of $\ln (A_n/A_1)$ against $\ln (n)$:

If the radius of a hydrogen atom in its first orbit is $a_{0}$,then its radius in the third excited state is . . . . . . . (in $a_{0}$)

In the first excited state of a hydrogen atom, the energy of its electron is $-3.4 \text{ eV}$. The radial distance of the electron from the hydrogen nucleus in this case is approximately: (Take $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$, $e = 1.6 \times 10^{-19} \text{ C}$ and $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2$)

Hydrogen $(H)$,deuterium $(D)$,singly ionized helium $(He^+)$ and doubly ionized lithium $(Li^{2+})$ all have one electron around the nucleus. Consider the $n = 2$ to $n = 1$ transition. The wavelengths of emitted radiations are $\lambda_1, \lambda_2, \lambda_3$ and $\lambda_4$ respectively. Then approximately:

Which state of triply ionised Beryllium $(Be^{+++})$ has the same orbital radius as that of the ground state of hydrogen?