Bohr Radius — Physica

Equivalent forms

a_0 = \frac{\hbar}{m_e c \alpha}

a_0 \approx 5.29177 \times 10^{-11}\,\text{m}

Four fundamental constants combine to fix the size of every atom in the universe.

Symbol	Name	SI unit	Dimension	Range
$a_0$	Bohr radiusoutput Most probable electron-proton distance in hydrogen ground state	m	L	1e-12 – 1e-9
$\varepsilon_0$	Vacuum permittivity Electric constant	F/m	M⁻¹·L⁻³·T⁴·I²	8.8541878128e-12 – 8.8541878128e-12
$\hbar$	Reduced Planck constant h/(2π)	J·s	M·L²·T⁻¹	1.054571817e-34 – 1.054571817e-34
$m_e$	Electron mass Rest mass of the electron	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	Imperial
$a_0$	m	cm	in
$\varepsilon_0$	F/m	—	—
$\hbar$	J·s	erg·s	ft·lbf·s
$m_e$	kg	g	lb
$e$	C	statC	C

Where it holds

Strictly valid as the natural unit of length in the non-relativistic hydrogen problem. The ground-state expectation ⟨r⟩ in full quantum mechanics is

1.5\cdot a_{0}

, but

a_{0} is

the most probable radius and remains the standard atomic length scale.

Dimensional analysis

$[4\pi \varepsilon _{0}\hbar ^{2}/(m_e e^{2})] = (F/m)(J\cdot s)^{2}/((kg)(C^{2})) = L \checkmark$

Discovery

Niels Bohr · 1913

Bohr postulated that the electron's angular momentum is quantized in units of ℏ. Combined with classical Coulomb attraction, this fixed the orbit radius of the ground state and predicted hydrogen's emission spectrum with stunning accuracy. The model is wrong in detail but a₀ remains the natural atomic length scale.

Try this

Why does a hydrogen atom have exactly the size it does?

An electron orbits a proton in hydrogen. What balance of forces and quantum constraints sets the atom's characteristic size — and why is it about half an angstrom?

Research status: stable

Real-world applications

Atomic clocks — transitions in hydrogen-like atoms set the natural frequency scale.
Quantum chemistry — basis sets scale with $a_{0}$ ; bond lengths are typically 1– $4\cdot a_{0}$ .
Plasma physics — Debye length and ionization cross-sections reference $a_{0}$ .
Defining the SI second via cesium hyperfine transitions traces back to atomic scales set by $a_{0}$ .

Common misconceptions

The electron doesn't orbit at radius $a_{0}$ — it has a probability cloud peaked there.
$a_{0} is$ not the atom's 'edge' — the wavefunction extends to infinity, decaying as $\exp (-r/a_{0})$ .
Bohr's planetary picture is wrong, but $a_{0} is$ still the correct length scale.

Experimental verification

Hydrogen Lyman/Balmer series wavelengths match Bohr's prediction

to \sim 10^{-4}

. Modern measurements via Rydberg constant determine

a_{0} to 10^{-12}

precision. X-ray diffraction confirms atomic sizes

\sim a_{0}

across the periodic table.

Derivation

Bohr's quantization:

m_e v r = n\hbar

.

Coulomb force = centripetal:

e^{2}/(4\pi \varepsilon _{0}r^{2}) = m_e v^{2}/r

.

Solve simultaneously for

n=1

:

r = 4\pi \varepsilon _{0}\hbar ^{2}/(m_e e^{2}) = a_{0}

.

Limiting cases

m \to m_p

(positronium-like with proton orbiting)⟶

a \to a_{0}(m_e/m)

Heavier orbiter → smaller atom. Muonic hydrogen

(m_\mu \approx 207\cdot m_e)

has radius

\sim 256\,\mathrm{fm}

.

Z > 1 (hydrogenic ion)⟶

a \to a_{0}/Z

Higher nuclear charge pulls the electron closer.

He^{+} is

half the size of H.

\hbar \to 0

(classical limit)⟶

a \to 0

Without quantum mechanics, the electron would spiral into the nucleus — atoms wouldn't exist.

What if…

What if the electron were as heavy as a proton?

$a_{0}$ would shrink $by \sim 1836\times$ , $to \sim 29\,\mathrm{fm}$ — comparable to a nuclear radius. Atoms would dissolve into a quark–lepton soup.

What

if \alpha

(fine-structure constant) were

10\times

larger?

$a_{0} = \hbar /(m_e c \alpha )$ would shrink $10\times$ , atoms would $be \sim 5\,\mathrm{pm}$ , chemistry timescales drop, and the universe would look entirely different.

1

Compute a₀ from constants

Given ·

\varepsilon 0:: 8.8541878128e-12
\hbar:: 1.054571817e-34
m e:: 9.1093837015e-31
e:: 1.602176634e-19

Find · a_0

Steps

Numerator: $4\pi \cdot \varepsilon _{0}\cdot \hbar ^{2} = 4\pi \times 8.854e-12 \times (1.055e-34)^{2} \approx 1.241e-77$
Denominator: $m_e\cdot e^{2} = 9.109e-31 \times (1.602e-19)^{2} \approx 2.338e-68$
$a_{0} = 1.241e-77 / 2.338e-68 \approx 5.292e-11\,\mathrm{m} = 52.92\,\mathrm{pm}$

Result ·

a_{0} \approx 5.292 \times 10^{-11} m

2

Muonic hydrogen radius

Given ·

m e:: 1.8835e-28
\varepsilon 0:: 8.8541878128e-12
\hbar:: 1.054571817e-34
e:: 1.602176634e-19

Find · a (using

m_\mu in

place of m_e)

Steps

$m_\mu \approx 207\,\mathrm{m}_e$
a scales as 1/m: $a = 52.92\,\mathrm{pm} / 207 \approx 256\,\mathrm{fm}$
The muon orbits well inside the proton's charge radius — the basis of the proton-radius puzzle.

Result ·

a \approx a_{0} \times (m_e/m_\mu ) \approx 256\,\mathrm{fm}

Bohr Energy Levels (Hydrogen)→Rydberg Formula→De Broglie Wavelength→