Assertion : Bohr had to postulate that the electrons in stationary orbits around the nucleus do not radiate.
Reason : According to classical physics all moving electrons radiate.

The gyromagnetic ratio of an electron in a hydrogen atom,according to the Bohr model,is

The inverse square law in electrostatics is $|\vec F| = \frac{{{e^2}}}{{4\pi { \in _0}{r^2}}}$ for the force between an electron and a proton. The $\frac{1}{r^2}$ dependence of $|\vec F|$ can be understood in quantum theory as being due to the fact that the particle of light (photon) is massless. If photons had a mass $m_p$,the force would be modified to $|\vec F| = \frac{{{e^2}}}{{4\pi { \in _0}}}\left( {\frac{1}{{{r^2}}} + \frac{\lambda }{r}} \right)\left( {{e^{ - \lambda r}}} \right)$ where $\lambda = \frac{{{m_p}c}}{\hbar }$ and $\hbar = \frac{h}{{2\pi }}$. Estimate the change in the ground state energy of a $H$-atom if $m_p$ were $10^{-6}$ times the mass of an electron.

The frequency of the light emitted when an electron transitions from the $n=4$ to $n=2$ level in a hydrogen atom is $\frac{3}{7}$ times the frequency of a transition in a $Li^{2+}$ ion. Which transition in the $Li^{2+}$ ion corresponds to this?

The ratio of the velocity of an electron in the ground state of a hydrogen atom to the velocity of light (in $CGS$ units) is:

$A$ 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 $v \propto n^{1/3}$
$(C)$ $E = \frac{3}{2} \left( \frac{n^2 h^2 F^2}{4 \pi^2 m} \right)^{1/3}$
$(D)$ $E = 2 \left( \frac{n^2 h^2 F^2}{4 \pi^2 m} \right)^{1/3}$