What is the numerical value of angular momentum for an electron in the first orbit of a hydrogen atom? $(h = 6.626 \times 10^{-34} \ J \ s)$

If the radius of the electron orbit in the excited state of a hydrogen atom is $476.1 \ pm$, the energy of the electron in that excited state in $J$ is (Radius and energy of the electron in the first orbit of a hydrogen atom are $52.9 \ pm$ and $-2.18 \times 10^{-18} \ J$ respectively).

An electron initially present in an excited state of $H$ atom is further excited to another energy level by an incident photon. It releases $10$ photons while coming back to the ground state,out of which $7$ have higher energy than the incident photon. The electron was initially present in:

Given below are two statements.
Statement $I:$ According to Bohr's model of an atom,qualitatively the magnitude of velocity of electron increases with decrease in positive charges on the nucleus as there is no strong hold on the electron by the nucleus.
Statement $II:$ According to Bohr's model of an atom,qualitatively the magnitude of velocity of electron increases with decrease in principal quantum number.
In the light of the above statements,choose the most appropriate answer from the options given below:

For an emission line of atomic hydrogen from $n_i = 8$ to $n_f = n$,the plot of wave number $(\bar{\nu})$ against $(\frac{1}{n^2})$ will be (The Rydberg constant,$R_H$ is in wave number unit).

The frequency of radiation emitted when the electron falls from $n = 4$ to $n = 1$ in a hydrogen atom will be (Given ionization energy of $H = 2.18 \times 10^{-18} \ J \ atom^{-1}$ and $h = 6.625 \times 10^{-34} \ Js$)