Using Bohr's model, calculate the ratio of the magnetic fields generated due to the motion of the electrons in the $2^{nd}$ and $4^{th}$ orbits of hydrogen atom. (in $32$ : $1$)

$A$ stationary hydrogen atom undergoes a transition from $n=5$ to $n=4$. The recoil speed of the atom is (where $R=$ Rydberg constant,$h=$ Planck's constant,and $m=$ mass of the hydrogen atom).

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:

Electrons in a certain energy level $n=n_{1}$ can emit $3$ spectral lines. When they are in another energy level $n=n_{2}$,they can emit $6$ spectral lines. The orbital speed of the electrons in these orbits are in the ratio:

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

The de-Broglie wavelength associated with the electron in the $n = 4$ level is