Bohr Model

12.6 Bohr's Bold Synthesis (1913)

By 1913 physics had two rival pieces of evidence that could not be reconciled within classical theory:

• Rutherford's nuclear atom (correct in structure but unstable in classical electrodynamics).
• Planck's quantum hypothesis (1900) and Einstein's photon (1905), showing that energy comes in discrete packets.

The young Danish physicist Niels Bohr (then 28) bravely combined the two. He kept Rutherford's nuclear atom but added three radically new quantum postulates. The result was a model that, for the first time, gave the precise wavelengths of the hydrogen spectrum.

12.7 Bohr's Three Postulates

12.8 Deriving the Bohr Radius and Energy Levels

Combine Newton's second law for circular motion with Coulomb's law for the H-atom electron (charge -e orbiting nucleus of charge +e):

\[\frac{m_e v^2}{r} = \frac{1}{4\pi\varepsilon_0}\frac{e^2}{r^2} \;\Longrightarrow\; m_e v^2 = \frac{e^2}{4\pi\varepsilon_0 r}\quad\text{...(i)}\]

Bohr's quantisation condition (Postulate 2):

\[v = \frac{nh}{2\pi m_e r}\quad\text{...(ii)}\]

Substitute (ii) into (i) and solve for r:

For \(n = 1\) we obtain the Bohr radius:

\[a_0 = \frac{h^2 \varepsilon_0}{\pi m_e e^2} = 5.29\times10^{-11}\,\text{m} \approx 0.529\,\text{Å}\]

So orbital radii are \(r_n = n^2 a_0\) — they grow rapidly with n: 0.529 Å, 2.12 Å, 4.76 Å, …

Speed of the orbiting electron

From (ii) at n = 1:

\[v_1 = \frac{h}{2\pi m_e a_0} = \frac{e^2}{2\varepsilon_0 h} \approx 2.19\times10^{6}\,\text{m/s} \approx \frac{c}{137}\]

This famous ratio \(\alpha = e^2/(2\varepsilon_0 hc) \approx 1/137\) is the fine-structure constant — a dimensionless measure of the strength of electromagnetic interactions.

Energy of the n-th level

From the orbit equation (i), the kinetic energy is:

\[K_n = \frac{1}{2}m_e v^2 = \frac{e^2}{8\pi\varepsilon_0 r_n}\]

The Coulomb potential energy is:

\[U_n = -\frac{e^2}{4\pi\varepsilon_0 r_n} = -2K_n\]

Total energy:

\[E_n = K_n + U_n = -K_n = -\frac{e^2}{8\pi\varepsilon_0 r_n} = -\frac{m_e e^4}{8\varepsilon_0^{2} h^{2}}\cdot\frac{1}{n^{2}}\]

n	rn (Å)	En (eV)	Name
1	0.529	−13.60	Ground state
2	2.116	−3.40	1st excited
3	4.761	−1.51	2nd excited
4	8.464	−0.85	3rd excited
5	13.225	−0.54	4th excited
∞	∞	0	Free electron (ionised)

12.9 Ionisation, Excitation and Binding Energies

Three energies are most often asked about in problems on the Bohr atom:

• Ionisation energy — energy to take the electron from the ground state (n=1) to free space (n=∞): \(I = E_\infty - E_1 = 0 - (-13.6) = 13.6\) eV.
• Excitation energy — energy required to lift the electron from n=1 to a higher bound state. n=1 → n=2: \(\Delta E = -3.4 - (-13.6) = 10.2\) eV. n=1 → n=3: 12.1 eV. n=1 → n=4: 12.75 eV.
• Binding energy at level n — same as |E_n| = 13.6/n² eV. The least bound state is n=∞ (binding 0); the most bound is n=1 (binding 13.6 eV).

Hydrogen-like ions

For one-electron systems with nuclear charge Ze (e.g. He⁺ has Z=2, Li²⁺ has Z=3), the radii and energies scale as:

\[r_n = \frac{n^2 a_0}{Z}, \qquad E_n = -\frac{13.6\,Z^2}{n^2}\,\text{eV}\]

So He⁺ has E₁ = −54.4 eV (much more tightly bound) and r₁ = 0.265 Å (smaller).

Worked Examples