The figure shows standing de-Broglie waves due to the revolution of an electron in a certain orbit of a hydrogen atom. Then,the expression for the orbit radius is (All notations have their usual meanings).

The radius of the first (lowest) orbit of the hydrogen atom is $a_0$. The radius of the second (next higher) orbit will be (in $a_0$)

If Bohr's quantisation postulate (angular momentum $= n h / 2 \pi$) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantisation of orbits of planets around the sun?

In the Bohr's hydrogen atom model,the radius of the stationary orbit is directly proportional to ($n =$ principle quantum number)

$A$ free hydrogen atom after absorbing a photon of wavelength $\lambda_{a}$ gets excited from the state $n=1$ to the state $n=4$. Immediately after that,the electron jumps to $n=m$ state by emitting a photon of wavelength $\lambda_{e}$. Let the change in momentum of the atom due to the absorption and the emission be $\Delta p_{a}$ and $\Delta p_{e}$,respectively. If $\lambda_{a} / \lambda_{e} = 1/5$,which of the following options is/are correct?
[Use $hc = 1242 \text{ eV nm}$; $1 \text{ nm} = 10^{-9} \text{ m}$,$h$ and $c$ are Planck's constant and speed of light,respectively]
$(1)$ $\lambda_{e} = 418 \text{ nm}$
$(2)$ The ratio of kinetic energy of the electron in the state $n=m$ to the state $n=1$ is $1/4$
$(3)$ $m=2$
$(4)$ $\Delta p_{a} / \Delta p_{e} = 1/2$

Using Bohr's atomic model, the orbital period of an electron in a hydrogen atom in the $n^{th}$ orbit is ($\epsilon_0$ = permittivity of free space, $h$ = Planck's constant, $m$ = mass of electron, $e$ = electronic charge).