Yukawa Potential

Equivalent forms

V(r) = -g² e^(-r/λ) / r

An exponentially suppressed 1/r potential — the simplest model of any short-range force, used everywhere from nuclear physics to plasmas.

Symbol	Name	SI unit	Dimension	Range
$V$	Yukawa potentialoutput Potential energy between two particles	J	M·L²·T⁻²	-1e-10 – 0
$r$	Separation Distance between particles	m	L	1e-16 – 1e-14
$m$	Mediator mass Mass of the exchanged particle (e.g., pion)	kg	M	1e-29 – 1e-26
$g$	Coupling constant Strength of the interaction (units of √(J·m) in SI)	√(J·m)	M^(1/2)·L^(3/2)·T⁻¹	1e-26 – 1e-22
$ℏ$	Reduced Planck constant Planck constant over 2π	J·s	M·L²·T⁻¹	1.054571817e-34 – 1.054571817e-34
$c$	Speed of light Speed of light in vacuum	m/s	L·T⁻¹	299792458 – 299792458

Unit systems

Symbol	SI	CGS	Natural
$V$	J	erg	MeV
$r$	m	cm	1/MeV
$m$	kg	g	MeV
$g$	√(J·m)	—	—
$ℏ$	J·s	—	—
$c$	m/s	—	—

Where it holds

Phenomenological model of nuclear interactions at low energies; modern nuclear physics uses meson-exchange (OBE) or chiral effective field theory for higher precision.

Dimensional analysis

V \to [M\cdot L^{2}\cdot T^{-2}]

g^{2}/r \to [M\cdot L^{3}\cdot T^{-2}]/[L] = [M\cdot L^{2}\cdot T^{-2}]

Energy on both sides; the exponential is dimensionless since

mc\cdot r/\hbar has

units

[M][LT^{-1}][L]/[ML^{2}T^{-1}] = 1

.

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Discovery

Hideki Yukawa · 1935

Yukawa proposed a meson-mediated nuclear force with range ℏ/(mc) ~ 1 fm, predicting the existence of a particle with mass ~100 MeV/c². The pion (m_π ≈ 140 MeV/c²) was discovered in 1947, earning Yukawa the 1949 Nobel Prize.

Try this

What holds the nucleus together against electrostatic repulsion?

Compute the Yukawa potential between two nucleons at separation r = 1 fm, given pion mass ≈ 140 MeV/c².

Research status: stable

Real-world applications

Nuclear physics models (Bonn, Nijmegen, AV18 potentials)
Plasma screening (Debye-Hückel theory analogous)
Dark matter models with massive mediators
Condensed matter (screened impurities in metals)

Common misconceptions

Yukawa is not the full nuclear force — heavier mesons $(\rho$ , $\omega$ , $\sigma )$ contribute at shorter ranges.
QCD shows nucleons are made of quarks and the residual force is more subtle than simple meson exchange.
The range $is \hbar /(mc)$ , not 1/m — units matter.

Experimental verification

The pion mass predicted by Yukawa was confirmed in 1947 with cloud-chamber observations of cosmic-ray pions. Nucleon-nucleon scattering data are well-fit by one-pion-exchange Yukawa-like terms at long distances.

Derivation

Solve the static Klein-Gordon equation

(\nabla ^{2} - (mc/\hbar )^{2})\varphi = -g\rho for

a point source.

The Green's function is

e^(-mcr/\hbar )/(4\pi r)

.

The resulting potential between two coupled sources is

V(r) = -g^{2} e^(-r/\lambda )/r

with

\lambda = \hbar /(mc)

.

Limiting cases

m \to 0

⟶

V \to -g^{2}/r

Massless mediator gives Coulomb-like 1/r potential (e.g., photon → electromagnetism).

r ≫

\hbar /(mc)

⟶

V \to 0

(exponentially)Force is screened at distances larger than the Compton wavelength of the mediator.

r \to 0

⟶

V \to -\infty

Like Coulomb, the potential diverges at the origin in this simple form.

What if…

What if the pion were

10\times

heavier?

Nuclear range would shrink $to \sim 0.14\,\mathrm{fm}$ ; nuclei would barely bind.

What if

m = 0

?

We'd recover an infinite-range force, like Coulomb electromagnetism.

1

Yukawa range for pion exchange

Given ·

m:: 2.488e-28
ℏ:: 1.054571817e-34
c:: 299792458

Find ·

\lambda = \hbar /(mc)

Steps

Step 1: Compute $mc = 2.488e-28 \times 2.998e_{8} = 7.46 \times 10^{-20} kg\cdot m/s$ .
Step 2: $\lambda = \hbar /(mc) = 1.055e-34 / 7.46e-20 = 1.414 \times 10^{-15} m$ .
Step 3: This $\sim 1.4\,\mathrm{fm}$ range matches the observed range of the nuclear force.

Result ·

\lambda = 1.055e-34 / (2.488e-28 \times 3e_{8}) = 1.41 \times 10^{-15} m = 1.41\,\mathrm{fm}

coulomb potentialklein gordon equation