How is the linear velocity $v$ of an electron in a $Bohr$ orbit related to its principal quantum number $n$?

In a hydrogen atom,if $r_n$ is the radius of the $n^{th}$ orbit and $L_n$ is the orbital angular momentum,then which of the following relations is correct?

The electron in a hydrogen atom makes a transition $n_1 \rightarrow n_2$,where $n_1$ and $n_2$ are the principal quantum numbers of the two states. Assume the Bohr model to be valid. The frequency of orbital motion of the electron in the initial state is $1/27$ of that in the final state. The possible values of $n_1$ and $n_2$ are

Match the following List-$I$ with List-$II$ in connection with Bohr's atomic model.

$A$. Speed of revolution of electron	$i$. $\frac{1}{4 \pi \varepsilon_0} \frac{2 \pi Z e^2}{n h}$
$B$. Kinetic energy	$ii$. $-\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$
$C$. Total energy	$iii$. $\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$
$D$. Frequency	$iv$. $\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{4 \pi^2 Z^2 e^4 m}{n^3 h^3}$

In a hydrogen atom in its ground state,the first Bohr orbit has radius $r_1$. When the atom is raised to one of its excited states,the electron's orbital velocity becomes one-third of its initial value. The radius of that orbit is: (in $r_1$)

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 $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2$)