In the Bohr's hydrogen atom model,the radius of the stationary orbit is directly proportional to ($n =$ principle quantum number)

As per the Bohr model,the minimum energy (in $eV$) required to remove an electron from the ground state of a doubly ionized $Li$ atom $(Z = 3)$ is:

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?

The energy required to excite an electron from the $1^{st}$ to the $3^{rd}$ Bohr orbit in $Li^{++}$ is ...... $eV$.

The kinetic energy of an electron revolving around a nucleus will be

An electron rotates in a circle around a nucleus having positive charge $Ze$. The correct relation between the total energy $(E)$ of the electron and its potential energy $(U)$ is: