Assertion $(A)$: The magnetic moment $(\mu)$ of an electron revolving around the nucleus decreases with increasing principal quantum number $(n)$.
Reason $(R)$: Magnetic moment of the revolving electron,$\mu \propto n$.

An electron in a hydrogen atom makes a transition from $n_1$ to $n_2$. If the time period of the electron in the initial state is eight times that in the final state,then:

$A$ particle of mass $m$ moves in circular orbits with potential energy $V(r) = Fr$,where $F$ is a positive constant and $r$ is its distance from the origin. Its energies are calculated using the Bohr model. If the radius of the particle's orbit is denoted by $R$ and its speed and energy are denoted by $v$ and $E$,respectively,then for the $n^{\text{th}}$ orbit (here $h$ is the Planck's constant)-
$(A)$ $R \propto n^{2/3}$ and $v \propto n^{1/3}$
$(B)$ $R \propto n^{2/3}$ and $v \propto n^{1/3}$
$(C)$ $E = \frac{3}{2} \left( \frac{n^2 h^2 F^2}{4 \pi^2 m} \right)^{1/3}$
$(D)$ $E = 2 \left( \frac{n^2 h^2 F^2}{4 \pi^2 m} \right)^{1/3}$

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:

Assume that there is no repulsive force between the electrons in an atom,but the force between positive and negative charges is given by Coulomb's law as usual. Under such circumstances,calculate the ground state energy of a $He$ atom.

The speed of the electron in a hydrogen atom in the $n=3$ level is (Planck constant $= 6.6 \times 10^{-34} \ J \ s$):