In the hydrogen atom spectrum,let $E_1$ and $E_2$ be the energies for the transitions $n=2 \rightarrow n=1$ and $n=3 \rightarrow n=2$ respectively. The ratio $E_2 / E_1$ is

If the ionization energy of a hydrogen atom in the ground state is $13.6 \ eV$,find the ionization potential of the third orbit in $eV$.

$A$ hydrogen atom makes a transition from $n = 2$ to $n = 1$ and emits a photon. This photon strikes a doubly ionized lithium atom $(Z = 3)$ in an excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for this process is:

In an atom, the difference between two energy levels is $3.31 \ eV$. The wavelength of the radiation emitted when the transition takes place between these levels is nearly: (in $Å$)

$A$ hydrogen atom absorbs radiation of wavelength $975 \, \text{\AA}$ to transition from the ground state to an excited state. To which energy level will the atom transition?

The ground state energy of the hydrogen atom is $-13.6 \ eV$. The kinetic and potential energy of the electron in the second excited state are,respectively: