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Gamma dose rate, distance and shielding

How the point-source formula, the inverse-square law and half-value layers of lead, steel and concrete set the dose rate from a gamma source, worked out.

Practical guide

By Steven Michael Barber · Published October 10, 2026 · Updated November 3, 2026 · 6 min read

Three things set the gamma dose rate you get from a small radioactive source in this model: how much activity it has, how far away you are, and what stands between you and it. Time does not change the rate; it sets how much dose the rate adds up to, which is why health physics teaches the protective controls as time, distance and shielding. This guide works through the arithmetic for the last two, using the same formula and reference data as our free gamma dose rate calculator, so you can follow each step by hand and then check it.

The point-source formula

For a small source in open air, the exposure rate at a distance is:

Ẋ = Γ × A × T ÷ d²

• Ẋ is the exposure rate in roentgen per hour (R/h).
• Γ (gamma) is the specific gamma-ray constant: the exposure rate at 1 meter from a 1 curie source, in R·m²/(Ci·h). Each isotope has its own value. For cesium-137 the handbook figure is 0.33; for cobalt-60 it is 1.32; for iridium-192, 0.48.
• A is the activity in curies (Ci). One curie is exactly 3.7 × 10¹⁰ becquerels.
• d is the distance from the source to the point of interest, in meters.
• T is the fraction of radiation that gets through a shield. With no shield, T = 1.

The cesium-137 value of 0.33 includes the 662 keV gamma ray from its short-lived daughter, barium-137m, which is always present in an older sealed source. Published gamma constants for the same isotope differ by 5 to 20 percent, depending on whether weak x-rays are counted and how the source is encapsulated. That spread is one reason any result from this formula is an estimate, not a measurement.

Distance: the inverse-square law

The d² in the denominator is the inverse-square law. Radiation from a point spreads over the surface of a sphere, and the area of a sphere grows with the square of its radius. Double the distance and the same radiation covers four times the area, so the rate drops to one quarter. Triple it and the rate falls to one ninth.

That makes distance the cheapest protection there is. Stepping from half a meter back to two meters cuts the rate by a factor of 16 without any shielding at all.

The law has limits. It assumes the source is small compared with the distance, roughly when you are at least three times the largest dimension of the source away. Closer than that, real sources stop behaving like points, and the formula does not give a valid figure. It also ignores scatter from nearby walls and floors, and attenuation in the air itself. Over a few meters the air can usually be neglected; over tens to hundreds of meters a substantial share of the photons interact on the way (over 110 meters, about half of the 1.25 MeV photons and two thirds of the 600 keV photons), so the formula is a model there, not a measured boundary.

Worked example: a 100 mCi cesium-137 source

Take a 100 millicurie (0.1 Ci, or 3.7 GBq) cesium-137 source with no shield.

At half a meter: Ẋ = 0.33 × 0.1 ÷ 0.5² = 0.033 ÷ 0.25 = 0.132 R/h, or 132 mR/h.

Moving back changes only the denominator:

You can also run the formula backward. How far away do you need to be for the rate to fall to 2 mR/h (0.002 R/h)? Rearranged, d = √(Γ × A ÷ Ẋ) = √(0.33 × 0.1 ÷ 0.002) = √16.5 ≈ 4.06 meters.

Shielding: half-value and tenth-value layers

A shield removes a fixed fraction of the photons for each thickness of material, so attenuation is exponential rather than linear. Two thicknesses are convenient to remember:

• The half-value layer (HVL) is the thickness that cuts the dose rate in half. T = 2^−x/HVL, where x is the shield thickness.
• The tenth-value layer (TVL) is the thickness that cuts it to one tenth. T = 10^−x/TVL.

The calculator uses broad-beam values from NCRP Report No. 49, as reproduced in health physics texts. "Broad beam" means the numbers include photons that scatter inside a wide shield and still come out the far side, for the geometry and photon energies of those tables. They are a reasonable starting point for a wall or shield between a source and a person, not a solution for every geometry. For cesium-137 they are:

Density does most of the work. Lead stops cesium-137 photons in about a seventh of the thickness that concrete needs, but a thick concrete wall can match a thin lead sheet.

The two models give slightly different answers. If attenuation were perfectly exponential, one tenth-value layer would equal exactly log₂10 ≈ 3.32 half-value layers. The tabulated values come close but do not match: for lead, 3.32 × 0.65 cm = 2.16 cm, against a tabulated TVL of 2.1 cm. Buildup, the photon spectrum, the geometry and rounding in the table all play a part, and the first layers of a shield can differ from later ones. The calculator defaults to the tenth-value model. Both models apply one constant layer to the whole thickness, so read either result as an illustrative estimate, not a figure with a known accuracy. For a source with several photon lines or a thin shield, such as iridium-192, the spectrum hardens as it passes through, and a constant layer can misstate the attenuation.

Worked example: adding a shield

Put 2 cm of lead between you and the same 100 mCi source, with you at 2 meters, where the unshielded rate was 8.25 mR/h.

Tenth-value model: T = 10^−2/2.1 = 10^−0.952 ≈ 0.112, so the rate is 8.25 × 0.112 ≈ 0.92 mR/h.

Half-value model: T = 2^−2/0.65 = 2^−3.08 ≈ 0.119, so the rate is about 0.98 mR/h.

Now solve for thickness. Suppose the goal is 0.5 mR/h at 2 meters. The shield has to pass 0.5 ÷ 8.25 ≈ 0.0606 of the unshielded rate, or about 6.1 percent. With the tenth-value model, x = −TVL × log₁₀(T) = −2.1 × log₁₀(0.0606) ≈ 2.1 × 1.218 ≈ 2.56 cm of lead. The same target needs:

The models agree within a few percent here. That shows they are consistent with each other, not that either is accurate for a particular shield.

Turning roentgen into gray and sievert

The formula gives exposure, which measures ionization in air. The example's 0.92 mR/h corresponds to about 8.07 µGy/h of air kerma under the electronic-equilibrium assumption, a quantity in air rather than dose to tissue. The photon convention 1 R ≈ 1 rem instead gives roughly 9.2 µSv/h; that is not a calibrated measurement of ambient dose equivalent. Our guide to radiation units derives both conversions and explains their limits.

What the estimate leaves out

Real sources sit in capsules, holders and shielded containers that absorb some of their own radiation. Floors, walls and equipment scatter photons back toward you. Survey instruments respond differently at different energies. And the gamma constant itself carries several percent of uncertainty. All of this means the point-source formula is good for understanding how distance and shielding scale, for checking whether a survey reading is in the expected range, and for learning. It is not a substitute for a calibrated survey meter or a qualified shielding design.