In hydrogen and hydrogen-like atoms,the ratio of the difference of energies $E_{4n}-E_{2n}$ and $E_{2n}-E_{n}$ varies with its atomic number $Z$ and $n$ as:

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

Inner shell electrons in atoms moving from one energy level to another lower energy level produce

$A$ 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:

$A$ hydrogen atom (ionisation potential $13.6 \ eV$) makes a transition from the third excited state to the first excited state. The energy of the photon emitted in the process is.......$eV$.

The energy of a hydrogen atom in the $n^{th}$ orbit is $E_n$. What will be the energy in the $n^{th}$ orbit of a singly ionized helium atom?