How can I calculate the orbital angular momentum of an electron?

Given below are two statements :
Statement I : Rutherford's gold foil experiment cannot explain the line spectrum of hydrogen atom.

Statement II : Bohr's model of hydrogen atom contradicts Heisenberg's uncertainty principle.

In the light of the above statements, choose the most appropriate answer from the options given below:

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is equal to $\frac{{\mathrm{h}}^2}{\mathrm{x} {\mathrm{ma}}_0^2}.$ The value of $10 \mathrm{x}$ is (${\mathrm{a}}_0$ is radius of Bohr's orbit)

(Nearest integer)

[Given: $\pi =3.14$]

Given below are two statements:

Statement I : Bohr's theory accounts for the stability and line spectrum of ${\mathrm{Li}}^{+}$  ion.

Statement II : Bohr's theory was unable to explain the splitting of spectral lines in the presence of a magnetic field.

In the light of the above statements, choose the most appropriate answer from the options given below:

The emission series of hydrogen atoms is given by, $\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$, where, $R$ is the Rydberg's constant. For a transition from $n_2$ to $n_1$, the relative change $\frac{\mathrm{\Delta }\lambda }{\lambda }$ in the emission wavelength, if hydrogen is replaced by deuterium (assume that, the mass of proton and neutron are the same and approximately $2000$ times larger than that of electrons) is

The magnitude of acceleration of the electron in the $n^{\text{th}}$ orbit of hydrogen atom is $a_{\mathrm{H}}$ and that of singly ionised helium atom is $a_{\mathrm{He}}$. The ratio $a_{\mathrm{H}}:a_{\mathrm{He}}$ is,