The ratio of the total energy of the $2^{\text{nd}}$ orbit electron for the hydrogen atom $(_1H^1)$ to that of the helium ion $(He^+)$ $(_2^4He)$ is:

If one were to apply the Bohr model to a particle of mass $m$ and charge $q$ moving in a plane under the influence of a magnetic field $B$,the energy of the charged particle in the $n^{th}$ level will be

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).

What will be the angular momentum of an electron,if the energy of this electron in an $H$-atom is $-1.5 \, eV$ (in $J \cdot s$)?

An electron in the ground state of the hydrogen atom has an orbital radius of $5.3 \times 10^{-11} \ m$,while that for the electron in the third excited state is $8.48 \times 10^{-10} \ m$. The ratio of the de Broglie wavelengths of the electron in the ground state to that in the third excited state is:

$A$ monochromatic radiation of wavelength $\lambda$ is incident on a hydrogen sample in ground state. The sample subsequently emits radiation of six different wavelengths, then the value of $\lambda$ is. [Use $hc = 1242 \text{ eV-nm}$] (in $\text{ nm}$)