Q.Calculate the energy of the electron in the $n=2$ state of the hydrogen atom, given that its ground-state energy is $-13.6$ eV.

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.

Applying Bohr's third postulate to this energy-level structure -- a photon of energy $E_m-E_n$ emitted on a transition from orbit m down to orbit n -- and converting to wavelength gives precisely the Rydberg formula $\frac{1}{\lambda}=R_HZ^2\left(\frac{1}{n^2}-\frac{1}{m^2}\right)$, with the Rydberg constant $R_H=1.097\times10^7$ m$^{-1}$ now DERIVED from first principles rather than simply fitted to data -- the single clearest triumph of Bohr's model, and the direct link between the energy-level picture and the observed line spectra of section 15.5.