The angular momentum of an electron in Bohr's hydrogen atom having energy $(-0.544) \text{ eV}$ is ($h$ = Planck's constant)

The kinetic energy of an electron in the first Bohr orbit of the hydrogen atom is $...... eV$.

Kinetic energy of an electron in one of the orbits of a hydrogen atom is $x$. Then, its total energy is . . . . . . .

If $\lambda_{1}$ and $\lambda_{2}$ are the wavelengths of de-Broglie waves for electrons in the first and second Bohr orbits in a hydrogen atom,then the ratio $\left(\frac{\lambda_{1}}{\lambda_{2}}\right)$ is equal to:

The muon has the same charge as an electron but a mass that is $207$ times greater. The negatively charged muon can bind to a proton to form a new type of hydrogen atom. How does the binding energy $E_{B\mu}$ of the muon in the ground state of a muonic hydrogen atom compare with the binding energy $E_{Be}$ of an electron in the ground state of a conventional hydrogen atom?

Which state of triply ionised Beryllium $(Be^{+++})$ has the same orbital radius as that of the ground state of hydrogen?