Half Value Layer Calculator

For a narrow, monoenergetic beam in an absorber, intensity follows exponential attenuation:

I = I₀ · e^−μx

• I₀ is initial intensity, I is transmitted intensity.
• μ is the linear attenuation coefficient (per length).
• x is absorber thickness (length).

Half-value layer (HVL) is the thickness that halves the intensity:

HVL = ln(2) / μ

Tenth-value layer (TVL) and mean free path (MFP) are:

TVL = ln(10) / μ    |    MFP = 1 / μ

If you know the mass attenuation coefficient and density:

μ = (μ/ρ) · ρ

1. Select a calculation type that matches your data.
2. Choose thickness and output units to keep results consistent.
3. Enter your values (μ, μ/ρ and ρ, or I₀, I, and x).
4. Press Calculate to show results above the form.
5. Use Download CSV or Download PDF to export.

These examples are mathematically consistent and suitable for testing.

1) Why HVL matters in radiation work

Half-value layer (HVL) is the absorber thickness that reduces an unscattered beam to 50% of its original intensity. It is widely used for shielding comparisons, beam quality checks, and imaging quality control, because it converts complex attenuation behavior into a single, easy-to-compare length.

2) Exponential attenuation and what μ represents

The underlying model is I = I₀ e^−μx. The linear attenuation coefficient μ has units of inverse length, such as 1/cm or 1/m, and captures removal of primary photons by absorption and scattering out of the beam.

3) Converting μ into HVL, TVL, and MFP

Because halving is a constant ratio, HVL is computed with the constant ln(2) ≈ 0.6931, giving HVL = ln(2)/μ. For 10% transmission, TVL uses ln(10) ≈ 2.3026 as TVL = ln(10)/μ. Mean free path is 1/μ.

4) Using measurements to estimate μ

If you can measure initial and transmitted intensity, you can estimate attenuation by rearranging the model: μ = (1/x) ln(I₀/I). For example, if I₀ = 1000, I = 250, and x = 2 cm, then I/I₀ = 0.25 and the computed HVL is 1 cm.

5) Mass attenuation option for material workflows

Many tables provide mass attenuation coefficients μ/ρ (often cm²/g). When you also know the material density ρ (g/cm³), the linear coefficient follows μ = (μ/ρ)·ρ. This is useful when comparing materials with densities like lead (~11.34 g/cm³) and concrete (~2.3 g/cm³).

6) Thickness planning for a transmission target

Engineers often start with a permitted transmission, such as 10% or 1%. This calculator can solve thickness directly with x = −ln(T)/μ, where T is the fraction (e.g., 10% → 0.10). Lower transmission targets require thickness that grows logarithmically, not linearly.

7) How many HVLs is your shield?

A convenient interpretation is “number of HVLs”: N = x/HVL. Each additional HVL multiplies transmission by 0.5. For example, 3 HVLs gives 0.5³ = 12.5%, and 6 HVLs gives about 1.56%. This is a quick mental check for your design.

8) Practical cautions and beam geometry

The simple model best matches narrow-beam, monoenergetic conditions. In broad-beam setups, scatter can increase measured transmission, creating a larger “effective HVL.” Always match the same geometry and energy when comparing HVL values, and validate critical shielding with appropriate instrumentation and safety guidance. Record energy, distance, and detector setup so results remain comparable across repeated measurements.