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

In $(i)$ a hydrogen atom and $(ii)$ a singly ionized helium atom,an electron makes a transition from the same excited state $n$ to the ground state. The ratio of the wavelengths of the emitted photons in the two cases will be:

In any Bohr orbit of a hydrogen atom,the ratio of $K.E.$ to $P.E.$ of a revolving electron at a distance $r$ from the nucleus is:

The emission series of hydrogen atom is given by $\frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$ where,$R$ is the Rydberg constant. For a transition from $n_{2}$ to $n_{1}$,the relative change $\Delta \lambda / \lambda$ 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 number of revolutions per second made by an electron in the first Bohr orbit of a hydrogen atom is of the order of

The de-Broglie wavelength of the electron in the ground state of the hydrogen atom is $...... \text{ } \mathring{A}$ (radius of the first orbit of hydrogen atom $= 0.53 \text{ } \mathring{A}$).