When an electron orbiting in a hydrogen atom in its ground state jumps to a higher excited state, the de-Broglie wavelength associated with it

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).

For an electron moving in the $n^{\text{th}}$ Bohr orbit,the de Broglie wavelength of the electron is:

Angular momentum of an electron in a hydrogen atom is $\frac{3h}{2\pi}$ ($h$ is the Planck's constant). The kinetic energy $(KE)$ of the electron is (in $\text{ eV}$)

Imagine an atom made up of a proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr model to this atom. The longest wavelength photon that will be emitted has wavelength $\lambda$ (given in terms of the Rydberg constant $R$ for the hydrogen atom) equal to:

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