Mean Free Path — Physica

Equivalent forms

I(x) = I_0 e^{-x/\lambda}

A single fraction relates microscopic cross sections to macroscopic shielding thickness.

Symbol	Name	SI unit	Dimension	Range
$λ$	Mean free pathoutput Average distance between collisions	m	L	1e-10 – 10
$n$	Number density Targets per unit volume	m⁻³	L⁻³	100000000000000000000 – 1e+30
$σ$	Cross section Effective interaction area per target	m²	L²	1e-30 – 1e-24

Unit systems

Symbol	SI	CGS	Natural
$λ$	m	cm	(MeV)⁻¹·ℏc
$n$	m⁻³	cm⁻³	(MeV)³/(ℏc)³
$σ$	m²	cm²	(MeV)⁻²·(ℏc)²

Where it holds

Diffuse, weakly-absorbing media; breaks down when correlations between scatterers matter (dense plasmas, coherent scattering, channeling in crystals).

Dimensional analysis

\lambda \to [L]

1/(n\sigma ) \to 1/([L^{-3}][L^{2}]) = [L]

Both sides are length.

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Discovery

Rudolf Clausius · 1858

Clausius introduced the concept while developing the kinetic theory of gases to explain how diffusion can be slow despite high molecular speeds.

Try this

How far does a neutron travel through water before colliding with a nucleus?

Compute λ for a thermal neutron in water given n = 3.34e28 m⁻³ (H atoms) and σ = 20 barns.

Research status: stable

Real-world applications

Nuclear reactor shielding design (lead, water, concrete thickness)
PET scanner detector design (LSO/BGO crystal length)
Cosmic-ray propagation in interstellar medium
Gas discharge tubes and plasma diagnostics

Common misconceptions

$\lambda is$ an average, not a fixed distance; individual path lengths follow an exponential distribution.
Total cross section must be used; if multiple processes (absorption, scattering) contribute, $sum the \sigma 's$ .
Mean free path depends on the relative velocity for moving targets, not just their density.

Experimental verification

Neutron transmission through known thicknesses of moderator material (Fermi's Chicago Pile, 1942) directly measured macroscopic absorption coefficients consistent with

n\sigma

.

Derivation

Probability of interaction in dx is

dP = n\sigma dx

.

The survival probability after distance x satisfies

dN/dx = -n\sigma N

, giving

N(x) = N_{0} \exp (-n\sigma x)

.

The expected value of

x is \int x \cdot n\sigma \exp (-n\sigma x) dx = 1/(n\sigma ) = \lambda

.

Limiting cases

\sigma \to 0

⟶

\lambda \to \infty

No interactions — particle travels freely (neutrinos in matter).

n \to \infty

⟶

\lambda \to 0

Dense targets interact immediately (electrons in metals).

n = 3.34e28

,

\sigma = 2e-27

⟶

\lambda \approx 1.5\,\mathrm{cm}

Thermal-neutron mean free path in water, sets reactor moderator scale.

What if…

What

if \sigma

doubled?

$\lambda$ halves — twice as much shielding effectiveness per unit thickness.

What if the material were

10\times

denser?

$\lambda$ shrinks by $10\times$ ; thinner shielding suffices.

What if multiple isotopes contribute?

Replace $n\sigma$ with $\Sigma _i n_i \sigma _i$ — total macroscopic cross section.

1

Neutron in water

Given ·

n:: 3.34e+28
σ:: 2e-27

Find ·

\lambda

Steps

Step 1: $n\sigma = 3.34e28 \times 2e-27 = 66.8\,\mathrm{m}^{-1}$ .
Step 2: $\lambda = 1/66.8 \approx 0.015\,\mathrm{m}$ .
Step 3: Interpretation: a 5 cm slab attenuates flux by $\exp (-5/1.5) \approx 3.5$ %.

Result ·

\lambda = 1/(3.34e28 \times 2e-27) = 1/66.8 \approx 0.015\,\mathrm{m} = 1.5\,\mathrm{cm}

2

Solar neutrinos in Earth

Given ·

n:: 6e+29
σ:: 1e-44

Find ·

\lambda

Steps

Step 1: $n\sigma = 6e29 \times 1e-44 = 6e-15\,\mathrm{m}^{-1}$ .
Step 2: $\lambda \approx 1.7e14\,\mathrm{m} \approx 1000\,\mathrm{AU}$ .
Step 3: Conclusion: Earth (1.3e7 m) is essentially transparent to solar neutrinos.

Result ·

\lambda = 1/(6e29 \times 1e-44) = 1/6e-15 \approx 1.7e14\,\mathrm{m}

rutherford scattering cross sectionklein nishina cross sectionThomson Scattering Cross Section→