Bohr Radius — Physica

Equivalent forms

a_0 = \hbar / (m_e c \alpha)

Four constants combine to give the size of every atom — a number engraved into the fabric of chemistry.

Symbol	Name	SI unit	Dimension	Range
$a_0$	Bohr radiusoutput Radius of ground-state hydrogen orbit	m	L	0 – 1e-10
$ε_0$	Vacuum permittivity Electric constant of vacuum	F/m	M⁻¹·L⁻³·T⁴·I²	8.8541878128e-12 – 8.8541878128e-12
$ℏ$	Reduced Planck constant Planck constant divided by 2π	J·s	M·L²·T⁻¹	1.054571817e-34 – 1.054571817e-34
$m_e$	Electron mass Electron rest mass	kg	M	9.1093837015e-31 – 9.1093837015e-31
$e$	Elementary charge Magnitude of electron charge	C	I·T	1.602176634e-19 – 1.602176634e-19

Unit systems

Symbol	SI	CGS	Natural
$a_0$	m	cm	1/MeV
$ε_0$	F/m	—	—
$ℏ$	J·s	erg·s	1
$m_e$	kg	g	MeV
$e$	C	statC	√α

Where it holds

Strictly valid for non-relativistic hydrogen-like atoms in the Bohr model. Modern quantum mechanics gives the same value as the expectation ⟨r⟩ for 1s, up to a factor.

Dimensional analysis

a_

0 \to [L]

\varepsilon _0 \hbar ^{2} / (m_e e^{2}) \to [M^{-1}L^{-3}T^{4}I^{2}][M^{2}L^{4}T^{-2}] / ([M][I^{2}T^{2}]) = [L]

Both sides are length.

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Discovery

Niels Bohr · 1913

Bohr postulated quantized angular momentum L = nℏ to explain hydrogen spectra. The resulting ground-state radius a_0 ≈ 0.529 Å became a fundamental atomic length scale.

Try this

How big is a hydrogen atom?

Compute the radius of the ground-state orbit of an electron in hydrogen using only fundamental constants.

Research status: stable

Real-world applications

Sets the length scale for all atomic physics calculations
Defines the Hartree unit of length in quantum chemistry
Used in muonic and exotic atom experiments to test QED

Common misconceptions

Electrons do not actually orbit like planets; a_0 is the expectation value of radius in the 1s state, not a classical orbit.
a_0 applies to hydrogen-like (one-electron) systems; multi-electron atoms require corrections.
Bohr's model is wrong about angular momentum of the ground state (the correct value is $L = 0$ , $not \hbar )$ .

Experimental verification

Spectroscopic measurements of hydrogen's Rydberg constant give a_

0 = 0.529177210903(80) \times 10^{-10} m

, agreeing with theory to better than

10^{-9}

.

Derivation

Equate centripetal force with Coulomb force:

m_e v^{2}/r = e^{2}/(4\pi \varepsilon _0\,\mathrm{r}^{2})

.

Combine with quantized angular momentum

L = m_e v r = n\hbar

.

Solve for

r_n = (4\pi \varepsilon _0 \hbar ^{2} n^{2})/(m_e e^{2})

.

Setting

n=1

gives a_0.

Limiting cases

m_e \to \infty

⟶ a_

0 \to 0

Heavier orbiting particle (e.g., muon) gives a smaller orbit — basis of muonic atoms.

e \to 0

⟶ a_

0 \to \infty

No Coulomb attraction means the electron escapes — no bound atom.

\hbar \to 0

⟶ a_

0 \to 0

Classical limit: the electron would spiral into the nucleus (the original atomic collapse paradox).

What if…

What if the electron were replaced by a muon?

a_ $0 \to a_0 \times (m_e/m_\mu ) \approx a_0 / 207$ — muonic hydrogen is $207\times$ smaller than ordinary hydrogen.

What

if \alpha

were doubled?

Atoms would be half the size; chemistry would change dramatically.

1

Compute the Bohr radius from constants

Given ·

ε 0:: 8.8541878128e-12
ℏ:: 1.054571817e-34
m e:: 9.1093837015e-31
e:: 1.602176634e-19

Find · a_0

Steps

Step 1: Numerator $= 4\pi \times \varepsilon _0 \times \hbar ^{2} = 4\pi \times 8.854e-12 \times 1.113e-68 = 1.238e-78$ .
Step 2: Denominator $= m_e \times e^{2} = 9.109e-31 \times 2.567e-38 = 2.338e-68$ .
Step 3: a_ $0 = 1.238e-78 / 2.338e-68 = 5.29 \times 10^{-11} m = 0.529 \text{Å}$ .

Result · a_

0 = 4\pi \times 8.854e-12 \times (1.055e-34)^{2} / (9.109e-31 \times (1.602e-19)^{2}) = 5.29 \times 10^{-11} m

Rydberg Formula→fine structure constant