Half value layer and tenth value layer
Photons are removed from a narrow beam exponentially: I = I0 x e^(-mu x x), where mu is the linear attenuation coefficient of the shield at that photon energy. The thickness that halves the intensity is the half value layer, HVL = ln2 / mu. Because the law is exponential, each successive HVL halves what is left, so n half value layers transmit 2^-n of the incident beam.
The tenth value layer is the same idea to a factor of ten: TVL = ln10 / mu = HVL x log2(10) = 3.322 x HVL. Shielding designers work in TVLs because the attenuation factors needed on a radiography bay are usually several orders of magnitude, and counting in tens is quicker than counting in halves.
Attenuation depends steeply on both photon energy and the atomic number of the shield. At iridium energies lead has an HVL of about 4.8 mm while steel needs 12.7 mm and concrete 44.5 mm for the same effect. At cobalt energies the whole picture stiffens: lead needs 12.5 mm, and lead’s advantage over steel narrows because Compton scattering, which scales with electron density rather than Z to the fourth or fifth power, takes over from photoelectric absorption.
These values are narrow-beam figures. In a real broad, uncollimated field, photons that scatter within the shield still emerge and contribute to the dose behind it. That effect is the build-up factor, and it means a shield designed on HVL arithmetic alone always performs worse than the calculation says. See the shielding thickness calculator for how much margin to add.
Worked example
| Isotope | ir192 |
| Shield material | pb |
| Shield thickness | 24 mm |
| Unshielded dose rate | 1000 uSv/h |
| Build-up factor B (1 = narrow beam) | 1 |
| Half value layer | 4.8 mm |
| Tenth value layer | 15.95 mm |
| Half value layers in this shield | 5 |
| Transmission | 3.125 % |
| Dose rate behind shield | 31.25 uSv/h |
Ir-192 in lead has HVL 4.8 mm, so TVL = 4.8 x 3.3219 = 15.95 mm. A 24 mm shield is 24/4.8 = 5.00 half value layers, transmitting 2^-5 = 0.03125 = 3.125 percent. A 1000 uSv/h incident field therefore leaves 31.25 uSv/h behind the lead, before build-up.
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