Formula used
- Macroscopic cross section:
Σ = N · σ, whereΣis in 1/length,Nis number density, andσis microscopic cross section. - Number density from material data:
N = ρ · NA / A, with densityρ, Avogadro constantNA, and molar massA. - Mean free path:
λ = 1/Σ, the average travel distance between interactions. - Uncollided transmission:
I/I0 = exp(−Σx)for thicknessx.
How to use this calculator
- Select Density + molar mass or Direct number density.
- Enter the microscopic cross section σ and its unit.
- Provide ρ and A, or enter N directly.
- Optionally add porosity or packing factor for effective density.
- Optionally enter thickness to estimate I/I0 transmission.
- Press Calculate to view results above the form.
Example data table
| Material | ρ (g/cm³) | A (g/mol) | σ (barn) | Σ (1/cm) | λ (cm) |
|---|---|---|---|---|---|
| Iron (Fe), example σ | 7.87 | 55.845 | 2.5 | 0.211 | 4.74 |
| Water (H₂O), example σ | 1.00 | 18.015 | 0.66 | 0.0221 | 45.2 |
| Concrete, example σ | 2.30 | 100.0 | 1.2 | 0.0166 | 60.2 |
Macroscopic Cross Section Guide
1) What the macroscopic cross section represents
The macroscopic cross section, Σ, expresses how strongly a bulk material interacts with a particle beam. It is an interaction probability per unit path length, typically reported in 1/cm or 1/m. Larger values imply more frequent interactions inside the material.
2) Core relation used by the calculator
The standard relation is Σ = N σ, where N is number density (atoms per cm³) and σ is microscopic cross section (cm², often given in barns). Remember 1 barn = 10-24 cm².
3) Number density from density and molar mass
If you know density ρ and molar mass M, number density is N = (ρ / M) NA, with NA as Avogadro’s constant. For typical solids, N is often on the order of 1022 atoms/cm³. For aluminum (ρ = 2.70 g/cm³, M = 26.98 g/mol), N ≈ 6.0×1022 atoms/cm³; with σ = 1 barn, Σ ≈ 0.060 1/cm.
4) Unit handling and conversions
Inputs can be entered in barns, cm², or m² for σ. Density can be g/cm³ or kg/m³, and molar mass can be g/mol or kg/mol. The calculator converts to consistent internal units and reports Σ in both 1/cm and 1/m.
5) Total, scattering, and absorption components
When separate microscopic values are available, compute Σs = Nσs and Σa = Nσa, then combine them as Σt = Σs + Σa. This separation is useful for shielding and detector design.
6) Mean free path and attenuation intuition
The mean free path is λ = 1/Σ. If Σ = 0.50 1/cm, then λ = 2.0 cm, so interactions occur within a few centimeters on average. A simple transmission model is e-Σx for thickness x.
7) Mixtures and compounds
For mixtures, a common approximation is a weighted sum: Σ = Σi Niσi. If you only have an effective microscopic value, estimate N from the bulk density and an effective molar mass, then apply Σ = Nσ.
8) Practical checks and typical ranges
Microscopic cross sections can range from <1 to hundreds of barns depending on energy and isotope. If your Σ looks unrealistic, re-check unit selections (barns vs m²) and confirm density and molar mass are for the same material composition.
FAQs
1) What is the difference between microscopic and macroscopic cross section?
Microscopic cross section (σ) describes a single nucleus interaction area, usually in barns. Macroscopic cross section (Σ) scales that by number density, giving an interaction probability per unit distance in bulk material.
2) Why does the calculator ask for density and molar mass?
Density and molar mass determine the number density N of atoms in the material. Once N is known, the macroscopic cross section follows directly from Σ = Nσ.
3) Which unit should I use for σ?
Use barns when you have nuclear data tables, or cm²/m² if your source provides SI units. The tool converts units internally, so consistent selection is more important than the unit choice itself.
4) What does a higher Σ mean physically?
A higher Σ means interactions occur more frequently as the particle travels through the material. This typically corresponds to stronger attenuation and a shorter mean free path λ = 1/Σ.
5) Can I compute scattering and absorption separately?
Yes. If you have σs and σa, compute Σs = Nσs and Σa = Nσa. Add them for the total Σt.
6) How do I estimate Σ for a compound?
Ideally, sum contributions from each element using their own number densities and microscopic values. If only an effective microscopic value is available, use the compound density and effective molar mass to estimate N, then apply Σ = Nσ.
7) Why do my results look too large or too small?
Most issues come from unit mismatch: barns vs m², or kg/m³ vs g/cm³. Also confirm molar mass units and that density corresponds to the same material state and composition as your cross section source.