The ratio of the energies of the electron in the hydrogen atom in the first and second excited states is

The following diagram indicates the energy levels of a certain atom. When the system moves from the $4E$ level to the $E$ level,a photon of wavelength ${\lambda _1}$ is emitted. The wavelength of the photon produced during its transition from the $\frac{7}{3}E$ level to the $E$ level is ${\lambda _2}$. The ratio $\frac{{{\lambda _1}}}{{{\lambda _2}}}$ will be:

The energy levels for an electron in a certain atom are shown in the figure. Which of the following transitions will emit a photon with the highest energy?

The ratio of energies of photons produced due to the transition of an electron in a hydrogen atom from its $(i)$ $3^{\text{rd}}$ to $2^{\text{nd}}$ energy level and (ii) highest energy level to $3^{\text{rd}}$ level 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.

In a hydrogen-like atom,an electron makes a transition from an energy level with principal quantum number $n$ to another with principal quantum number $(n - 1)$. If $n >> 1$,the frequency of the emitted radiation is proportional to: