The ratio of the radius of the first Bohr orbit to that of the second Bohr orbit of the orbital electron 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 $v \propto n^{1/3}$
$(C)$ $E = \frac{3}{2} \left( \frac{n^2 h^2 F^2}{4 \pi^2 m} \right)^{1/3}$
$(D)$ $E = 2 \left( \frac{n^2 h^2 F^2}{4 \pi^2 m} \right)^{1/3}$

$A$ light of energy $12.75 \; eV$ is incident on a hydrogen atom in its ground state. The atom absorbs the radiation and reaches to one of its excited states. The angular momentum of the atom in the excited state is $\frac{x}{\pi} \times 10^{-17} \; eVs$. The value of $x$ is $........$ (use $h=4.14 \times 10^{-15} \; eVs$)

If the radius of the first orbit of an $H$ atom is $a_0$, then the de Broglie wavelength of an electron in the third orbit is: (in $\pi a_0$)

When the electron orbiting in a hydrogen atom goes from one orbit to another orbit (principal quantum number $= n$),the de-Broglie wavelength $(\lambda)$ associated with it is related to $n$ as:

The $21\, cm$ radio wave emitted by hydrogen in interstellar space is due to the interaction called the hyperfine interaction in atomic hydrogen. The energy of the emitted wave is nearly