At which excited state of $Be^{3+}$ will the radius of the $e^{-}$ be the same as that of an $H$ atom in the ground state?

$A$ sample of hydrogen-like atoms produces an emission spectrum consisting of $10$ wavelengths arising from all possible transitions. During this process,the maximum angular momentum change for an electron transitioning from a higher energy level to a lower energy level is:

When the electron orbiting in a hydrogen atom in its ground state moves to the third excited state, the de-Broglie wavelength associated with it

The kinetic energy of the electron in an orbit of radius $r$ in a hydrogen atom is proportional to ($e$ = electronic charge).

The ratio of the velocity of an electron in the ground state of a hydrogen atom to the velocity of light (in $CGS$ units) is:

The radius of the first orbit of hydrogen is $r_{H}$,and the energy in the ground state is $-13.6 \text{ eV}$. Considering a $\mu^{-}$-particle with a mass $207 m_e$ revolving around a proton as in a hydrogen atom,the energy and radius of the proton and $\mu^{-}$-combination respectively in the first orbit are (assume the nucleus to be stationary):