The magnetic moment of an electron due to its orbital motion is proportional to (where $n$ is the principal quantum number).

An electron in an excited state of $Li^{2+}$ ion has angular momentum $\frac{3 h}{2 \pi}$. The de-Broglie wavelength of the electron in this state is $p \pi a_{0}$ (where, $a_{0} = \text{Bohr radius}$). The value of $p$ is

Using the formula for the radius of the $n^{th}$ orbit $r_n = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2}$,derive an expression for the total energy of an electron in the $n^{th}$ Bohr orbit.

The gravitational attraction between an electron and a proton in a hydrogen atom is weaker than the Coulomb attraction by a factor of about $10^{-40}$. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction.

If $E$ and $L$ denote the magnitudes of total energy and of angular momentum of an electron in a Bohr orbit,then the true relation between them is:

The ratio of angular momentum of an electron in $n^{\text{th}}$ orbit of a hydrogen atom to the velocity of the electron in the $n^{\text{th}}$ orbit is proportional to