Energy of Orbitals

Niels Bohr's 1913 model of the hydrogen atom was a landmark in physics and chemistry, introducing the revolutionary idea that electrons occupy discrete, quantized energy levels rather than any arbitrary orbit around the nucleus. By combining classical mechanics with a quantum condition, Bohr derived an explicit formula for the energy of each allowed orbit, showing that it depends on two quantities: the nuclear charge Z and the principal quantum number n. The resulting expression, E_n ∝ −Z²/n², elegantly explains why different elements emit light at different frequencies and why it takes a specific amount of energy to remove an electron from an atom. Central to Bohr's model is the distinction between the ground state, the lowest energy configuration an electron can occupy, and excited states, higher-energy arrangements that atoms temporarily adopt when they absorb energy. When an electron falls from a higher to a lower energy level it releases a photon whose energy exactly equals the difference between those levels, producing the characteristic spectral lines observed in emission spectra. Conversely, the minimum energy required to completely remove an electron from its ground state is called the ionization energy, a quantity that Bohr's formula predicts with remarkable accuracy for hydrogen-like species. This article derives the orbital energy formula from first principles using Coulomb's law and the Bohr quantization condition, explores how Z and n control the energy of each level, and applies the result to calculate ionization energies and photon frequencies for one-electron systems such as \ce{H}, \ce{He+}, and \ce{Li^{2+}}. Mastering these ideas provides a foundation for understanding modern quantum mechanical treatments of atomic structure, periodic trends, and chemical bonding.

Key takeaways
  • Coulomb Force Provides Centripetal Acceleration
  • Bohr's Quantization Condition
  • Total Mechanical Energy
  • Ground vs. Excited States
  • Ionization Energy

Coulomb Force Provides Centripetal Acceleration

For an electron (charge −e) orbiting a nucleus of charge +Ze at radius r, the electrostatic attraction provides exactly the centripetal force needed for circular motion: \frac{kZe^2}{r^2} = \frac{m_e v^2}{r}, where k = 8.99 \times 10^9 \text{ N m}^2\text{C}^{-2} is Coulomb's constant and m_e = 9.109 \times 10^{-31}\text{ kg} is the electron mass. Rearranging gives the kinetic energy: \frac{1}{2}m_e v^2 = \frac{kZe^2}{2r}.

Bohr's Quantization Condition

Bohr postulated that the angular momentum of the electron is quantized in integer multiples of \hbar = h/(2\pi): m_e v r = n\hbar, where n = 1, 2, 3, \ldots is the principal quantum number. Solving for v and substituting back into the centripetal force equation yields the allowed orbital radii: r_n = \frac{n^2 \hbar^2}{m_e k Z e^2} = \frac{n^2}{Z} a_0, where a_0 = 0.529 \text{ Å} is the Bohr radius for hydrogen.

Total Mechanical Energy

The total energy E_n is the sum of kinetic energy (KE = \frac{kZe^2}{2r}) and potential energy (PE = −\frac{kZe^2}{r}), giving E_n = KE + PE = −\frac{kZe^2}{2r_n}. Substituting r_n from Step 2: E_n = -\frac{m_e k^2 Z^2 e^4}{2\hbar^2} \cdot \frac{1}{n^2}. The factor \frac{m_e k^2 e^4}{2\hbar^2} = 13.6\text{ eV} is the Rydberg energy E_R, so the final result is E_n = -\frac{13.6 Z^2}{n^2}\text{ eV}.

Ground vs. Excited States

The ground state corresponds to n = 1, the most negative (lowest) energy level. For hydrogen (Z = 1) the ground-state energy is −13.6 eV. Excited states have n ≥ 2 and are less negative, meaning the electron is further from the nucleus and less tightly bound. The energy difference between levels n_i (initial) and n_f (final) determines the photon energy: \Delta E = E_{n_f} - E_{n_i} = 13.6 Z^2 \left(\frac{1}{n_i^2} - \frac{1}{n_f^2}\right)\text{ eV} for emission (n_i > n_f).

Ionization Energy

Ionization corresponds to removing the electron entirely, equivalent to raising it from the ground state (n = 1) to n = ∞, where E_\infty = 0. Therefore the ionization energy is IE = E_\infty - E_1 = 0 - \left(-\frac{13.6 Z^2}{1^2}\right) = 13.6 Z^2\text{ eV}. For hydrogen IE = 13.6 eV; for \ce{He+} (Z = 2) IE = 54.4 eV; for \ce{Li^{2+}} (Z = 3) IE = 122.4 eV. The Z² dependence reflects the increased nuclear attraction as atomic number grows.

The Key Equation

En=−13.6 Z2n2 eVE_n = -\frac{13.6\, Z^2}{n^2} \text{ eV}
  • E_n — Orbital energy (eV (electron-volts)): Total mechanical energy of the electron in the nth Bohr orbit; always negative, indicating a bound state.
  • 13.6 — Rydberg energy (E_R) (eV): The ground-state ionization energy of hydrogen, equal to m_e k^2 e^4 / (2ℏ²). Encapsulates all fundamental constants in the formula.
  • Z — Atomic (nuclear charge) number (dimensionless): Number of protons in the nucleus. Larger Z increases nuclear attraction, making all orbitals more negative in energy.
  • n — Principal quantum number (dimensionless (positive integer)): Labels the allowed orbits: n = 1 is the ground state, n = 2, 3, … are excited states. Increasing n raises energy toward zero.

Worked Example

Problem. A singly ionized helium atom \ce{He+} (Z = 2) has its electron in the n = 3 excited state. (a) Calculate the energy of this state in eV. (b) The electron drops to n = 1. Find the energy and wavelength of the emitted photon.

E3=−6.04 eV;λ=25.6 nmE_3 = -6.04\text{ eV};\quad \lambda = 25.6\text{ nm}

Answer. Energy of n = 3 state: −6.04 eV. The emitted photon has energy 48.36 eV and wavelength 25.6 nm (extreme UV).

Figure
Diagram
Bohr Energy Level Diagram — Hydrogen (Z = 1)Energy(eV)n=1−13.6n=2−3.4n=3−1.51n=4−0.85n=5−0.54n=∞0.00← Ionization limitEmissionΔE = 1.89 eVλ = 656 nm(Hα, red)IonizationIE = 13.6 eVE_n = −13.6/n² eV (for H)E_n = −13.6 / n² eV (Z = 1)
Origins · 1 minNiels Bohr and how this was discovered
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Niels Bohr
1885–1962 · Danish

Bohr proposed his quantized model of the hydrogen atom in 1913, deriving the orbital energy formula E_n = −13.6 Z²/n² eV and explaining atomic emission spectra for the first time. He received the Nobel Prize in Physics in 1922 for this work. He later made foundational contributions to quantum mechanics through his principle of complementarity and the Copenhagen interpretation. Bohr was an avid football (soccer) player and his brother Harald was an Olympic silver medallist in football for Denmark in 1908. Niels reportedly remained a respectable goalkeeper well into his academic career.