Q.Calculate the ionisation energy of the He$^+$ ion ($Z=2$) in its ground state ($n=1$).

Combining the orbit-radius and orbital-speed formulas with the definition of total energy (kinetic plus electrostatic potential) gives the central result of Bohr's model: the energy of an electron in the nth orbit of a hydrogen-like atom is $E_n=-\frac{m_eZ^2e^4}{8\epsilon_0^2h^2n^2}=-\frac{13.6Z^2}{n^2}$ eV. The negative sign signals a BOUND electron -- energy must be supplied to bring it up to zero (free) -- and $E_n$ becomes less negative (higher) as n increases, approaching but never reaching zero as $n\rightarrow\infty$.

The lowest-energy orbit (n=1) is the ground state, the atom's natural, most stable configuration; any n > 1 is an excited state. The excitation energy of a given excited state n is the energy needed to move an electron there FROM the ground state, $E_n-E_1$; the ionization energy is the fixed energy needed to remove the ground-state electron completely from the atom (13.6 eV for hydrogen), numerically identical to that electron's binding energy, since freeing it and binding it are exactly reverse processes. …