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

The first four spectral lines in the Lyman series of a $H$ atom are $\lambda = 1218 \, \mathring{A}, 1028 \, \mathring{A}, 974.3 \, \mathring{A}, 951.4 \, \mathring{A}$. If instead of Hydrogen,we consider Deuterium,calculate the shift in the wavelength of these lines.

In the Bohr model of the hydrogen atom,the centripetal force is provided by the Coulomb attraction between the proton and the electron. If $r_0$ is the radius of the ground state orbit,$m$ is the mass,$e$ is the charge on the electron,and $\varepsilon_0$ is the permittivity of vacuum,the speed of the electron is:

Would the Bohr formula for the $H-$ atom remain unchanged if the proton had a charge of $+ \frac{4e}{3}$ and the electron a charge of $- \frac{3e}{4}$ (where $e = 1.6 \times 10^{-19} \ C$)? Give reasons for your answer.

$A$ sample of hydrogen atoms in its ground state is radiated with photons of $10.2 eV$ energy. The radiation emitted from the sample is absorbed by excited ionized $He^{+}$ ions. Which of the following statement$(s)$ is/are true?

According to de-Broglie,the de-Broglie wavelength for an electron in an orbit of radius $5.3 \times 10^{-11} \ m$ of a hydrogen atom is $10^{-10} \ m$. The principal quantum number for this electron is: