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

In a hydrogen atom,when an electron transitions from the $4^{th}$ orbit to the $2^{nd}$ orbit,the wave number is $20,397 \, cm^{-1}$. What will be the wave number when an electron in a $He^+$ ion transitions from the $4^{th}$ orbit to the $2^{nd}$ orbit?

When the electron orbiting in a hydrogen atom in its ground state moves to the third excited state,the de-Broglie wavelength associated with it

If Bohr's quantisation postulate (angular momentum $= n h / 2 \pi$) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantisation of orbits of planets around the sun?

An electron in the ground state of the hydrogen atom has an orbital radius of $5.3 \times 10^{-11} \ m$,while that for the electron in the third excited state is $8.48 \times 10^{-10} \ m$. The ratio of the de Broglie wavelengths of the electron in the ground state to that in the third excited state is:

The radius of a hydrogen atom in its ground state is $5.3 \times 10^{-11} \ m$. After collision with an electron,it is found to have a radius of $21.2 \times 10^{-11} \ m$. What is the principal quantum number $n$ of the final state of the atom?