If an electron jumps from the $1^{st}$ orbital to the $3^{rd}$ orbital,then it will:

When an electron falls from a higher energy level to a lower energy level,the difference in energy is emitted as electromagnetic radiation. Why can it not be emitted as other forms of energy?

$A$ diatomic molecule has moment of inertia $I$. By applying Bohr's quantization condition,its rotational energy in the $n^{\text{th}}$ level is $[n \geq 1]$ $(h = \text{Planck's constant})$

$A$ 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 $1/4$
$(3)$ $m=2$
$(4)$ $\Delta p_{a} / \Delta p_{e} = 1/2$

Hydrogen $(H)$,deuterium $(D)$,singly ionized helium $(He^+)$ and doubly ionized lithium $(Li^{2+})$ all have one electron around the nucleus. Consider the $n = 2$ to $n = 1$ transition. The wavelengths of emitted radiations are $\lambda_1, \lambda_2, \lambda_3$ and $\lambda_4$ respectively. Then approximately:

Assuming the atom in the ground state,the expression for the magnetic field at a point (nucleus) in a hydrogen atom due to the circular motion of the electron is [$\mu_{0} =$ permeability of free space,$\epsilon_{0} =$ permittivity of free space,$m =$ mass of electron,$e =$ electronic charge,$h =$ Planck's constant].