What is the angular momentum of an electron in Bohr's hydrogen atom whose energy is $-0.544 \ eV$?

In a hydrogen atom,the electron and proton are bound at a distance of about $0.53 \; \mathring{A}$.
$(a)$ Estimate the potential energy of the system in $eV$,taking the zero of the potential energy at infinite separation of the electron from the proton.
$(b)$ What is the minimum work required to free the electron,given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in $(a)$?
$(c)$ What are the answers to $(a)$ and $(b)$ above if the zero of potential energy is taken at $1.06 \; \mathring{A}$ separation?

The de-Broglie wavelength of an electron in the $3^{rd}$ orbit of a $He^{+1}$ ion is approximately (in $\mathring{A}$):

A hydrogen atom, initially in the ground state, is excited by absorbing a photon of wavelength $980 \ \text{\AA}$. The radius of the atom in the excited state, in terms of Bohr radius $a_0$, will be (given $hc = 12500 \ \text{eV-\AA}$). (in $a_0$)

$A$ hydrogen atom in the ground state absorbs $\Delta E$ amount of energy. If the orbital angular momentum of the electron is increased by $\frac{h}{2 \pi}$ ($h=$ Planck constant), then the magnitude of $\Delta E$ is (in $eV$)

The radius of the first orbit of a hydrogen atom is $0.5 \, \mathring{A}$ and the speed of the electron is $2.2 \times 10^6 \, m/s$. Find the magnetic induction at the proton due to the motion of the electron in $Tesla$.