Gamma dose rate and shielding
Point-source dose rate from the emission spectrum, with attenuation and buildup, and the shield thickness solved backwards.
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How the dose rate is obtained
Most calculators of this kind look up a gamma constant in a table. This one computes it. For each photon line of the nuclide, the energy and the emission probability come from the evaluated nuclear data, and the mass energy-absorption coefficient of air at that energy comes from the NIST tables. Summing E·y·(µen/ρ) over the lines and dividing by 4π gives the air kerma rate constant directly. The result is not set against published gamma constants, which describe a different quantity and often do not state the cutoff they apply. The validation report checks the parts instead: the unit chain by hand, the interpolation of the NIST tables, and a sample of emission lines against an independent evaluation.
Computing rather than copying has a practical benefit beyond provenance: the low-energy cutoff becomes a control you can move. Published constants apply a cutoff, usually 20 keV, to exclude photons that the source encapsulation absorbs before they reach anyone. That choice is invisible in a table and decisive in the answer. For Am-241 a 10 keV cutoff instead of 20 keV raises the constant by roughly a factor of eight, because the 17 keV neptunium L X-rays are absorbed strongly in air and dominate the kerma integral while contributing nothing at a metre through any real capsule.
Attenuation, and why it is not enough
A shield removes photons following exp(−µx), where µ is the linear attenuation coefficient — the mass attenuation coefficient at that energy multiplied by the density. The calculator applies this line by line, so a nuclide with a broad spectrum hardens correctly as it passes through: the low-energy lines are removed preferentially and the average energy behind the shield is higher than in front of it.
Exponential attenuation alone describes a narrow beam, where any photon that interacts is gone. A real shield is a broad beam. Photons Compton-scatter inside the material, change direction, and a substantial fraction still emerge toward the receptor. Ignoring them can underestimate the dose rate behind a thick shield by a factor of several. The half-value layers this tool reports are therefore narrow-beam values, noticeably thinner than the broad-beam figures in shielding handbooks — 5.5 mm rather than about 6.5 mm of lead at 662 keV. The note on half-value and tenth-value layers tabulates these layers for six common emitters in lead, iron and concrete.
Buildup factors
The correction is the buildup factor B, the ratio of the true dose rate to the attenuated-only dose rate. It grows with the number of mean free paths and depends on material, energy and geometry. This calculator implements the two-parameter Berger form B = 1 + a·µx·eb·µx but ships no coefficients. The most widely used tabulation, the geometric-progression fit, is the body of a paid consensus standard, and reproducing it here would not be legitimate. Enter a and b from the reference you are working to, and the tool will show what difference they make.
With attenuation only, the shield thickness this calculator returns is a lower bound. Treat it as the starting point for a design, never as the design.
Geometry assumptions
A point source in air, radiating isotropically, with a receptor at a stated distance and the shield between them. No room scatter, no skyshine, no source self-absorption, no capsule, and no correction from air kerma to a dose quantity in tissue. A result shown in µSv/h or mSv/h is that air kerma with 1 µGy counted as 1 µSv, not an ambient dose equivalent (how far apart the two are). Real sources are extended, and close to an extended source the inverse-square falloff does not hold — near a large plane the dose rate barely falls with distance at all.
Worked examples
A caesium-137 source on an open bench
A 37 GBq (1 Ci) Cs-137 calibration source is lifted out of its storage pot for a check and stands on a bench. How high is the air kerma rate where people stand, and what would a 5 cm lead block in front of it change?
| Input | Air kerma rate |
|---|---|
| Cs-137, 37 GBq, 1 m, no shield | 2.85 mGy/h |
| Same source at 2 m | 0.712 mGy/h |
| 1 m behind 5 cm lead, attenuation only | 5.11 µGy/h |
Doubling the distance divides the rate by four, as the inverse-square law says it must for a point source. The lead lets through 0.18% of the photons that travel straight through it, but that figure leaves out the photons scattered inside the block. Before the number is used to plan work, switch to “With buildup” and enter coefficients for lead at 662 keV; the shielded rate rises, and the honest answer is the higher one. The block also has to be wide enough to cover the whole line of sight from the source to the people it protects; placed close to the source, a small block does that job.
Where to rope off an iridium-192 exposure
Site radiography uses a 1.5 TBq Ir-192 source in a panoramic exposure with no collimator. Local rules put the barrier where the air kerma rate falls to 20 µGy/h. Enter 1500 GBq and 20 m with no shield.
| Quantity | Result |
|---|---|
| Air kerma rate at 20 m | 409 µGy/h |
| Distance for 20 µGy/h | 90.5 m |
The second line is one inverse-square step from the first: d = 20 m × √(409/20). A barrier ninety metres out is rarely practical, which is why the usual answer is a collimator, not more rope. The calculation also leaves out scatter from the ground and the object being radiographed, so the boundary is confirmed with a survey meter during the first exposure.
FAQ
What is the gamma constant of Cs-137?
Computed from the 662 keV line and the NIST energy-absorption coefficients of air, with a 20 keV cutoff, the air kerma rate constant of Cs-137 is 0.0769 mGy·m²/(GBq·h). The same method gives 0.306 for Co-60 and 0.109 for Ir-192. Older tables quoted in R·m²/(Ci·h) or as dose-equivalent constants look different because they describe a different quantity, not because the physics differs.
How do I calculate the dose rate from a point source?
Multiply the gamma constant by the activity and divide by the square of the distance: dose rate = Γ·A/d². The formula assumes a source small compared with the distance; at a distance of only a few source lengths it no longer holds and a line or area source model is needed.
What is the half-value layer of lead for Cs-137?
At 662 keV the NIST attenuation coefficient and a lead density of 11.35 g/cm³ give a half-value layer of about 0.55 cm, from HVL = ln 2/µ. That is a narrow-beam figure. Behind a real slab, scattered photons add to the transmitted beam, so the effective half-value layer of a broad beam is larger and the first few centimetres remove less than this suggests.
Why does my survey meter read more than the calculator?
The calculator gives air kerma rate in gray, while most survey meters read ambient dose equivalent H*(10) in sievert, which is about 20% higher for Cs-137 photons according to the conversion coefficients in ICRP Publication 74. The calculator's µSv/h and mSv/h count 1 µGy of air kerma as 1 µSv, so they do not include that difference. Scatter from walls, floor and the source holder adds more, and meters also have their own energy response.
Can the attenuation-only result be used for shielding design?
No. Attenuation alone counts only photons that pass through without interacting, so it understates the dose behind any thick shield. For design, use the buildup option, B = 1 + a·µx·exp(b·µx), with coefficients from a published source for the material and energy, and confirm the finished shield with a survey.