The ratio of the energies of the hydrogen atom in its first to second excited state is

As the $n$ (number of orbit) increases,the difference of energy between the consecutive energy levels

$A$ hydrogen atom at the ground level absorbs a photon and is excited to $n=4$ level. The potential energy of the electron in the excited state is

The total energy of an electron in the $n^{th}$ stationary orbit of the hydrogen atom can be obtained by

An electron in a hydrogen atom undergoes a transition from an orbit with quantum number $n_i$ to another with quantum number $n_f$. $V_i$ and $V_f$ are respectively the initial and final potential energies of the electron. If $\frac{V_i}{V_f} = 6.25$,then the smallest possible $n_f$ is:

An electron of a hydrogen-like atom,having $Z=4$,jumps from the $4^{\text{th}}$ energy state to the $2^{\text{nd}}$ energy state. The energy released in this process will be $......... \text{eV}$.
(Given $Rch = 13.6 \text{ eV}$)
Where $R =$ Rydberg constant,
$c =$ Speed of light in vacuum,
$h =$ Planck's constant.