The energy $E$ of a hydrogen atom with principal quantum number $n$ is given by $E = \frac{-13.6}{n^2} \; eV$. The energy of a photon emitted when the electron jumps from the $n = 3$ state to the $n = 2$ state of hydrogen is approximately......$eV$.

Explain the formulas for the energy of an electron in an atom revolving around the nucleus in different orbits.

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:

Energy of an electron in the second orbit of a hydrogen atom is $E_{2}$. The energy of an electron in the third orbit of $He^{+}$ will be

As an electron makes a transition from an excited state to the ground state of a hydrogen-like atom/ion:

The energy levels with transitions for an atom are shown in the figure. The transitions corresponding to the emission of radiation of maximum and minimum wavelength are respectively: