Layered Shield Attenuation Calculator

Three materials in series: the total transmission, which layer is doing the work as a share of the attenuation, and the single slab that would do the same job — with the order of the layers making no difference, which is checked rather than asserted.

Last updated 7 Oct 2026

This is narrow-beam attenuation through uniform material: the physics, not a shielding design. Real geometry scatters, and the scattered photons a narrow-beam calculation ignores can add tens of per cent to the dose behind a thick shield — that is what a build-up factor is for, and where a page omits one it says so on the page. A shield arrived at from a figure here has not been designed: shielding is specified by a qualified expert against the regulation that applies, and then verified by measurement.

Layered shield attenuation

Three materials and thicknesses → the total, and the shares

Photon energySix sources, and two of the six are conventions rather than lines. Technetium-99m, iodine-131, fluorine-18 and caesium-137 each have one photon that decides the shielding, and the preset is that photon. Cobalt-60 has two lines and iridium-192 has nine, so their presets are single-energy stand-ins: the conventional 1.25 MeV for cobalt-60 is 0.2 per cent from the yield-weighted mean of its two lines and is an exact point on the NIST grid, which makes it the better input of the two; the conventional 380 keV for iridium-192 sits between its yield-weighted mean of 371.6 keV and its energy-weighted mean of 397.8. No single energy reproduces a nine-line spectrum through a thick shield, because the soft lines are stripped out first and what is left is harder than what went in.

…or the energy itselfIn MeV. Locked while a source is selected above; choose the last option and this becomes the input. The NIST grid runs from 1 keV to 20 MeV and this page returns nothing outside it rather than an extrapolation, which is deliberate: the coefficient curve is not a polynomial and there is nothing honest to extend it with.

DensitiesOne switch for all three layers. NIST’s coefficient is per gram, so each layer needs a density before it becomes a per-centimetre attenuation, and the conventional values are only a starting point: ordinary structural concrete runs 2.2 to 2.4 g/cm3 and far higher loaded with barytes or magnetite. In a layered shield the density error does not cancel — it shifts that layer’s share of the work as well as the total.

Layer 1 — innermost, nearest the sourceThe innermost layer, nearest the source. The ORDER of the three layers makes no difference whatever to the answer, because the exponents add and a sum commutes — the page proves it by giving the same total for any ordering. Order does matter for two things this page cannot compute: bremsstrahlung from a beta source, which wants a low-Z layer first, and characteristic X-rays from the shield itself, which want a lower-Z layer last.

…thicknessIn millimetres. Set it to zero to switch this layer off: a zero-thickness layer contributes exactly nothing, exactly, which is how a two-layer or single-layer stack is entered here.

…densityIn g/cm3. Locked to the conventional value for the material above unless the override at the top of the page is switched on. This is the input that most often makes a layered calculation wrong, because a shield is built out of what the builder had: a 2.2 g/cm3 pour and a 2.4 g/cm3 pour differ by nine per cent on this layer’s whole contribution.

Layer 2 — middleThe middle layer. The ORDER of the three layers makes no difference whatever to the answer, because the exponents add and a sum commutes — the page proves it by giving the same total for any ordering. Order does matter for two things this page cannot compute: bremsstrahlung from a beta source, which wants a low-Z layer first, and characteristic X-rays from the shield itself, which want a lower-Z layer last.

…thicknessIn millimetres. Set it to zero to switch this layer off: a zero-thickness layer contributes exactly nothing, exactly, which is how a two-layer or single-layer stack is entered here.

Layer 3 — outermost, nearest the personThe outermost layer, nearest the person. The ORDER of the three layers makes no difference whatever to the answer, because the exponents add and a sum commutes — the page proves it by giving the same total for any ordering. Order does matter for two things this page cannot compute: bremsstrahlung from a beta source, which wants a low-Z layer first, and characteristic X-rays from the shield itself, which want a lower-Z layer last.

…thicknessIn millimetres. Set it to zero to switch this layer off: a zero-thickness layer contributes exactly nothing, exactly, which is how a two-layer or single-layer stack is entered here.

Not a circuit: the same three-layer stack drawn twice, and the fact that the two bars are different shapes is the entire point of the page. The top bar divides the stack by THICKNESS — where the material physically is, with the densest hatch for layer 1. The bottom bar divides exactly the same stack by CONTRIBUTION to the attenuation, μᵢxᵢ as a share of Σμᵢxᵢ. For the stack this page opens with, 5 mm of lead is four per cent of the thickness and eighteen per cent of the work, while 100 mm of concrete is eighty per cent of the thickness and half of the work — so the lead is punching far above its weight and the concrete is still doing more in total, because there is twenty times as much of it. Read the two bars together and the design question answers itself: a layer that is wide on the top bar and narrow on the bottom one is taking up space and buying little, and a layer that is narrow on top and wide below is the one to make thicker. The total at the right of each bar is the other half of the story — the thickness in millimetres and the attenuation in half-value layers, which are the two currencies a shield is paid for in. Rearranging the layers reorders the segments in both bars and changes neither total, because the exponents add and a sum commutes.

2.733836%Example

5 mm of lead, then 100 mm of concrete, then 20 mm of steel, against caesium-137 — and the same three layers in any other order

Transmissions multiply, so the exponents add — and a sum commutes

T = ∏_i e^−μₕxₕ = e^{−Σμₕxₕ} · shareₕ = μₕxₕ / Σμ℔x℔ · x_equiv = Σμₕxₕ / μ_ref

Σμᵢxᵢ: the only quantity the answer depends on. Each layer contributes its own μ times its own thickness, the contributions add, and the total goes into one exponential. Because it is a SUM, the order of the layers is irrelevant: any arrangement of the same layers gives the same number to the last bit
shareᵢ: the fraction of the total attenuation that belongs to layer i, and the design insight on this page. A layer contributing two per cent of the attenuation is costing money and weight and buying nothing; the layer with the largest share is the one worth making thicker. Note that it is μx and not thickness: a thin high-Z layer can dominate a thick low-Z one, and 100 mm of concrete can dominate 5 mm of lead
x_equiv: the thickness of a single material that would do the same job, which is the right way to compare a stack with an alternative. It is Σμᵢxᵢ divided by the reference material’s own μ, and it is usually LIGHTER than the stack as well as thinner when the reference is lead
0 thickness: exactly a no-op, and that is how an unused layer is switched off here. A zero-thickness layer contributes zero to the sum, multiplies the transmission by exactly 1, and takes a zero share. Two-layer and single-layer stacks are entered that way
what is missing: build-up, as everywhere in narrow-beam attenuation, and here there is an extra difficulty: published build-up factors are tabulated for single homogeneous materials and the approximations for a layered shield disagree with each other. The honest statement is that Σμᵢxᵢ is a lower bound on what a real stack needs, by a fifth or so, and that the dominant layer’s own build-up factor is the best rough guide

Worked example

5 mm of lead, then 100 mm of concrete, then 20 mm of steel, against caesium-137 — and the same three layers in any other order
THE THREE COEFFICIENTS, at 661.7 keV from the NIST grid and each material's conventional density. Lead μ = 1.261213 per cm, concrete 0.181195, steel 0.578456. Note the span: lead is seven times concrete per centimetre at this energy and only 1.4 times it per gram
EACH LAYER'S CONTRIBUTION TO THE EXPONENT, which is μ times thickness in centimetres. Lead: 1.2612 × 0.5 = 0.6306. Concrete: 0.1812 × 10 = 1.8119. Steel: 0.5785 × 2 = 1.1569
THE TOTAL, AND THE WHOLE ANSWER. Σμx = 0.6306 + 1.8119 + 1.1569 = 3.5995, so T = e^−3.5995 = 2.7338 per cent, one part in 36.6. Check it the other way: the three layers transmit 0.5323, 0.1633 and 0.3145 on their own, and the product of those three is 0.027338 — the same number, because transmissions multiply
WHICH LAYER IS DOING THE WORK, and the answer is not the lead. The shares are 17.5 per cent for the lead, 50.3 per cent for the concrete and 32.1 per cent for the steel. The concrete has the smallest coefficient of the three and does half of the work, because there is 100 mm of it. If this stack had to be improved, the lead is the layer to thicken — it gives the most attenuation per millimetre — and the concrete is the layer to thin if space is wanted back, at a known cost
AND NOW THE ORDER. Rearrange the three layers however you like — steel first, concrete first, lead in the middle — and Σμx is still 3.5995, because addition commutes. All six orderings give the same transmission to one part in 10¹⁶, which is where a double runs out of bits, and the proof module behind this page checks all six rather than asserting it. That is a genuine property of narrow-beam photon attenuation and not an approximation
THE EQUIVALENT SINGLE SLAB. The same Σμx in lead alone would be 3.5995/1.2612 = 28.54 mm — one slab, 29 mm thick and 324 kg/m², against the stack's 125 mm and 444 kg/m². The lead version is both thinner and lighter, which is why lead is specified at all; it is also more expensive per kilogram and has to be hung on something
WHERE THIS STOPS BEING TRUE. Everything above is narrow-beam, so the real stack transmits MORE than 2.734 per cent — build-up adds scattered photons that the exponential cannot see, and for a layered shield there is no clean published correction, because build-up factors are tabulated for single materials. And if the source is a beta emitter rather than a gamma emitter the ORDER matters after all: low atomic number first, so the electrons stop somewhere that makes as little bremsstrahlung as possible

Eight stacks at caesium-137, including the same two layers in both orders

Rows four and five are the same two layers in opposite orders and every figure in them is identical, to the last digit printed and far beyond it. That is not a coincidence or an approximation: the exponents add, a sum commutes, and narrow-beam photon attenuation is genuinely order-independent. The one pedantic qualification is that floating-point addition is commutative but not associative, so a three-term sum taken in a different order can differ by one unit in the last place of a double — about one part in 10¹⁶, fifteen decimal places below anything on this page. The share column is where the design lives. In the first row the 100 mm of concrete is doing 50 per cent of the work and the 5 mm of lead 18, which is not what a materials table would lead you to expect — thickness beats coefficient when there is enough of it. Compare rows four and six: taking the lead from 2 mm to 25 mm moves its share from 12.2 per cent to 63.5 and takes the transmission from 12.69 per cent to 0.70, which is the honest way to decide whether more lead is worth buying. And read the last two columns together: the lead-equivalent column says what single slab of lead would do the same job, and it is almost always lighter than the stack as well as thinner. This is NARROW-BEAM attenuation: one uniform slab, a parallel beam, and every scattered photon assumed lost. A real shield scatters photons back into the beam, and the published half-value layers in shielding tables are 11 to 33 per cent LARGER than ln2/μ for exactly that reason. A narrow-beam answer therefore UNDER-states the shield a real room needs, which is the unsafe direction, and closing that gap is what a build-up factor is for.
Stack	Σ μx	Half-value layers	Transmission (%)	One part in	Share per layer (%)	Lead equivalent (mm)	Mass (kg/m²)
5 mm lead + 100 mm concrete + 20 mm steel	3.5995	5.193	2.7338	1 in 36.6	17.5 / 50.3 / 32.1	28.54	444.2
2 mm lead only	0.2522	0.364	77.7056	1 in 1.3	100.0	2.00	22.7
100 mm concrete only	1.8119	2.614	16.3336	1 in 6.1	100.0	14.37	230.0
2 mm lead + 100 mm concrete	2.0642	2.978	12.6921	1 in 7.9	12.2 / 87.8	16.37	252.7
100 mm concrete + 2 mm lead, the SAME stack reversed	2.0642	2.978	12.6921	1 in 7.9	87.8 / 12.2	16.37	252.7
25 mm lead + 100 mm concrete	4.9650	7.163	0.6978	1 in 143.3	63.5 / 36.5	39.37	513.8
200 mm concrete + 10 mm steel	4.2023	6.063	1.4960	1 in 66.8	86.2 / 13.8	33.32	538.7
3 mm tungsten + 50 mm water	0.9992	1.442	36.8181	1 in 2.7	57.1 / 42.9	7.92	107.9

What one half-value layer costs in each material, at caesium-137

The last two columns are the whole materials argument and they point in different directions. On THICKNESS lead wins easily: a half-value layer of concrete is seven times thicker, water fifteen times, air thirteen and a half THOUSAND times. On MASS the ranking nearly collapses: concrete needs only 1.41 times the kilograms of lead for the same attenuation at this energy, water 1.30 and iron 1.51, because at 662 keV most of the attenuation is Compton scattering, which counts electrons per gram and cares very little which element they belong to. So lead still wins on mass here, but by 30 to 50 per cent rather than by the factor of seven it wins on thickness — and above 1.01 MeV it loses to water per gram outright. Lead buys SPACE, not weight, and tungsten buys more space for more money. Note what the mass column does NOT explain: a spent-fuel pool is water because water is cheap, self-circulating and a coolant, not because it is light per half-value layer — it is the second heaviest in this table. Use this table to decide which layer to make thicker: the layer with the largest μ gives you the most attenuation per millimetre, and the layer with the smallest mass per half-value layer gives you the most per kilogram. This is NARROW-BEAM attenuation: one uniform slab, a parallel beam, and every scattered photon assumed lost. A real shield scatters photons back into the beam, and the published half-value layers in shielding tables are 11 to 33 per cent LARGER than ln2/μ for exactly that reason. A narrow-beam answer therefore UNDER-states the shield a real room needs, which is the unsafe direction, and closing that gap is what a build-up factor is for. The mass attenuation coefficient is per gram, so a density has to come from somewhere else to turn it into a per-centimetre attenuation, and the density used is not NIST’s. Ordinary structural concrete runs anywhere from 2.2 to 2.4 g/cm3 and much higher when loaded with barytes or magnetite; lead sheet is close to theoretical but lead glass and leaded acrylic are not. A density wrong by five per cent makes every thickness on this page wrong by five per cent.
Material	Density (g/cm³)	μ (per cm)	Half-value layer	…weighs (kg/m²)	Thickness relative to lead	Mass relative to lead
Lead (Pb)	11.3500	1.26121	5.496	62.38	1.0000	1.0000
Concrete, ordinary	2.3000	0.18119	38.254	87.98	6.9605	1.4105
Water	1.0000	0.08569	80.890	80.89	14.7183	1.2968
Iron / mild steel (Fe)	7.8740	0.57846	11.983	94.35	2.1803	1.5126
Tungsten (W)	19.3000	1.90243	3.643	70.32	0.6630	1.1273
Aluminium (Al)	2.6990	0.20137	34.421	92.90	6.2631	1.4894
Soft tissue (ICRU-44)	1.0600	0.08999	77.022	81.64	14.0145	1.3088
Air, dry (sea level)	0.0012	0.00009	7,464.87 cm	89.94	13,582.6763	1.4418

When the order DOES matter, and why this page cannot see it

The first row is what this page computes and the rest is what it cannot. The beta row is the one worth memorising, because the intuition runs the wrong way: lead is the better shield for almost everything and it is the WORSE first layer for a pure beta emitter, since it converts the beta energy into penetrating bremsstrahlung more efficiently than a plastic does. Yttrium-90 and strontium-90 are the two nuclides in this site’s tables with no useful gamma emission at all, and both are handled low-Z first. The fluorescence row is the other practical one: a lead shield exposed to photons above 88 keV emits its own 72 to 88 keV X-rays, which are soft enough that a millimetre or two of steel on the outside removes them and weak enough that most people never notice they exist until a spectrum is taken. This is a first-pass calculation on an idealised geometry — a point source, a uniform slab, no self-absorption in the source, no container, no floor and no walls. Real sources have extent and encapsulation, and real rooms scatter. A closed-form answer cannot see any of that.
Situation	Does the order matter?	What to do
Photons, narrow beam — what this page computes	No, not at all. The total is Σμᵢxᵢ and a sum commutes	Order the layers for buildability and cost. The answer is the same either way
A beta emitter — strontium-90, yttrium-90, phosphorus-32	YES, and getting it wrong makes the problem worse	LOW-Z FIRST. Bremsstrahlung yield rises roughly with atomic number, so electrons stopping in lead produce far more penetrating X-rays than electrons stopping in perspex. The standard arrangement is perspex or acrylic thick enough to stop the betas, then lead if anything is needed for the bremsstrahlung — never lead facing the source
Characteristic X-rays from the shield itself	Yes, at the far side	Lead fluoresces at 72 to 88 keV whenever it absorbs a photon above its K edge. A thin layer of a lower-Z material on the OUTSIDE — tin, copper, steel — absorbs those fluorescence X-rays, which is why graded shields exist and why a shield for a high-energy source often has a cheap steel skin
Neutrons alongside gammas	Yes, and the materials are different	Not this page’s subject at all. Neutrons want hydrogen to slow them and boron or cadmium to capture them, and capture produces gammas that then need a high-Z layer. The usual order is moderator, absorber, gamma shield
A broad beam with build-up	Slightly, and nobody should rely on it	Published build-up factors are tabulated for single homogeneous materials; for a stack there is no simple rule and the approximations in the literature disagree with each other. Use the dominant material’s factor as a rough guide and expect it to be wrong
The same two materials with an air gap between them	No, for photons	An air gap changes nothing worth computing — ten centimetres of air at caesium-137 transmits 99.9 per cent — but it changes the geometry, which changes the build-up, and it is also where ducts and cables go. Gaps leak more than they attenuate

Exponents add, so the order cannot matter — and the useful question is which layer is doing the work

Transmissions multiply, so the exponents add, and that single fact is the whole page. A photon that gets through three layers has got through all three, so the probability is the product of the three probabilities — and because each is an exponential, the product is one exponential of the SUM Σμₕxₕ. There is nothing else to compute. The practical consequence is the one everybody wants and nobody quite believes: the ORDER of the layers makes no difference whatever. Lead then concrete transmits exactly what concrete then lead transmits, because addition commutes. The table on this page prints the same two-layer stack both ways round so the identical figures can be seen rather than taken on trust, and the verification behind the page checks all six orderings of a three-layer stack rather than asserting it.

The useful question is which layer is doing the work, and the answer is often not the one you would guess. Layer i’s share of the attenuation is μₕxₕ divided by the total, and it depends on the thickness as much as on the coefficient. In the stack this page opens with — 5 mm of lead, 100 mm of concrete, 20 mm of steel against caesium-137 — the CONCRETE is doing 50 per cent of the work, the steel 32 and the lead 18, even though lead’s coefficient is seven times the concrete’s, because there is twenty times as much concrete. That is the number that makes a design decision: a layer contributing five per cent of the attenuation is costing money and weight and buying almost nothing, and a layer contributing half of it is where any extra thickness should go. The logarithmic chart shows the same thing geometrically — each layer is a straight segment whose slope is its own μ and whose drop is its contribution.

A zero-thickness layer is exactly a no-op, and that is how a two-layer shield is entered. There is no mode selector and nothing to switch off. Set a thickness to zero and that layer contributes exactly zero to the sum, multiplies the transmission by exactly one, and takes a zero share. Set all three to zero and the stack transmits exactly 100 per cent. Those are the degenerate cases, they are checked explicitly rather than assumed, and the reason to be careful about them is that a degenerate case is where every wrong version of a formula agrees with the right one — so the verification behind this page also checks that pairing each coefficient with the wrong thickness gives a visibly different answer, which is what makes the order-independence test worth running.

The equivalent single slab is the right way to compare a stack with an alternative. Σμₕxₕ divided by any one material’s μ gives the thickness of that material alone which would do the same job. For the default stack it is 28.5 mm of lead against 125 mm and 444 kg/m² of mixed materials — the lead version is both thinner and lighter, which is the honest reason lead gets specified. It is also more expensive per kilogram, it has to be hung on something, and it cannot be the structure. A concrete wall is seven times thicker and 40 per cent heavier per unit of attenuation at this energy, and it holds the building up, which is why most shielding in the world is concrete.

Where the order does matter, this page is the wrong tool and says so. Three cases. A pure beta emitter — yttrium-90, strontium-90, phosphorus-32 — must be shielded LOW-Z FIRST, because bremsstrahlung yield rises roughly with atomic number and stopping electrons in lead makes far more penetrating radiation than stopping them in perspex. A lead shield exposed to photons above its K edge at 88.0 keV fluoresces, emitting 72 to 88 keV characteristic X-rays from its outward face, which is why a graded shield puts a thin lower-Z layer on the OUTSIDE. And neutrons are a different problem with different materials in a required order — moderator, absorber, then a gamma shield for the capture gammas. None of those three is computed here. This page attenuates the photons it is given, in a narrow beam, and creates nothing.

And the correction that is missing is harder to supply here than anywhere else in this batch. Every number on this page is narrow-beam, so a real stack transmits MORE than it says, by of the order of a fifth for a single material. For a LAYERED shield there is no clean published correction at all: build-up factors are tabulated for single homogeneous materials, and the approximations in the literature for stacks disagree with one another, which is itself the honest finding. The best rough guide is the dominant layer’s own build-up factor, applied to the whole, with the expectation that it is wrong. Treat Σμₕxₕ as a lower bound on what the stack needs, use the share column for the design decision it is actually good at, and verify the result with an instrument.

Frequently asked questions

Does the order of the layers matter?

Not for photon attenuation. The transmission is e^(−Σμx), the total is a sum, and a sum commutes exactly, so any arrangement of the same layers gives the same answer to within one unit in the last place of a double — one part in 10¹⁶, which is where the arithmetic runs out of bits and fifteen decimal places below anything printed. The table on this page prints a two-layer stack in both orders to show the identical figures, and the verification behind the page checks all six orderings of three layers. It DOES matter for three things the page cannot compute: a beta source wants a low-Z layer first, because lead turns electrons into bremsstrahlung far more efficiently than perspex does; a lead layer wants a lower-Z layer outside it to absorb its own fluorescence X-rays; and neutrons want moderator, then absorber, then gamma shield.

How do I work out the transmission through two different materials?

Multiply their individual transmissions, or equivalently add their μx values and take one exponential. For 2 mm of lead and 100 mm of concrete at caesium-137: lead gives μx = 1.261 × 0.2 = 0.2522 and concrete 0.1812 × 10 = 1.8119, so Σμx = 2.064 and the transmission is e^(−2.064) = 12.7 per cent. The same answer comes from 0.777 × 0.163. The thing NOT to do is add the thicknesses, or average the coefficients, or add the half-value layers in any way that is not a weighted sum — it is the products μx that add, and nothing else.

Which layer should I make thicker?

Not necessarily the one with the biggest share. Three different questions have three different answers and this page prints all three. For the most attenuation per MILLIMETRE, thicken the layer with the largest μ — at caesium-137 that is tungsten, then lead, then iron. For the most per KILOGRAM, use the mass per half-value layer in the materials table, where water wins and iron loses. For the most per POUND STERLING, concrete wins almost always and is also the structure. The share column answers a different and equally useful question: which layer could you REMOVE without much loss, which is the one with the smallest share.

Is a lead-lined concrete wall better than either alone?

Better than the same thickness of concrete, and better than the same MASS of lead only in the sense that it holds the building up. The arithmetic itself has no synergy: Σμx simply adds, and a combination is worth exactly the sum of its parts. What makes the combination attractive is practical rather than physical. Concrete is cheap, structural and thick; lead is expensive, non-structural and thin. But be careful how much you expect of a lead LINING at gamma-source energies. At caesium-137, a 200 mm concrete wall with a 2 mm lead sheet in it attenuates like 214 mm of plain concrete: 2 mm of lead is worth only 13.9 mm of concrete there, because it is 0.36 of a half-value layer. At technetium-99m the same 2 mm is 7.77 half-value layers and is worth 15.8 cm of concrete, which is why lead-lined plasterboard is a diagnostic-energy product. The real synergy in layered shields is in the two effects this page cannot see: a low-Z layer stopping betas before a high-Z layer can make bremsstrahlung from them, and a low-Z outer layer catching the high-Z layer’s fluorescence.

Why does 100 mm of concrete beat 5 mm of lead?

Because there is twenty times as much of it and lead is only seven times better per centimetre at this energy. μ for lead at caesium-137 is 1.261 per cm against concrete’s 0.1812, so 5 mm of lead is μx = 0.631 and 100 mm of concrete is 1.812 — nearly three times as much attenuation. This is the most common misreading of a materials table: a large coefficient tells you what a millimetre is worth, not what a layer is worth, and in a real shield the thicknesses differ by far more than the coefficients do. The share column on this page exists to stop the mistake.

Can I use this for a lead apron over a concrete wall?

For the attenuation arithmetic, yes — it is two layers and the order is irrelevant. For the answer to be useful, three cautions. An apron’s lead EQUIVALENCE is quoted at a specific kV and is not a thickness of metallic lead; 0.25 mm lead equivalent means it attenuates like 0.25 mm of lead at the stated beam quality and may be a loaded rubber or a lead-free composite that behaves differently elsewhere in the spectrum. The density override exists for exactly that case, but if the composition is not lead then the mass attenuation coefficient is not lead’s either and scaling the density is not enough. And an apron at diagnostic energies is a real shield while the same apron at caesium-137 is almost nothing: lead’s half-value layer is 0.26 mm at technetium-99m and 5.50 mm at caesium-137, a factor of twenty-one.

What about an air gap between the layers?

For the photon arithmetic it changes nothing worth computing. Air’s half-value layer at caesium-137 is 74.6 metres, so ten centimetres of it transmits 99.91 per cent, and you can enter it as a layer and watch it take a share of a few hundredths of a per cent. What a gap does change is everything the page cannot see. It changes the geometry, and therefore the build-up. It is where cables, ducts and conduits go, and a penetration leaks far more than the gap attenuates. And it means the two layers are not in contact, so a shield relying on one layer to absorb the other’s fluorescence may not be doing it. Gaps are a leakage question, not an attenuation one.

How much does build-up add for a layered shield?

Nobody can tell you accurately, and that is the honest answer rather than a hedge. Published build-up factors are tabulated for single HOMOGENEOUS materials; for a stack there is no standard formula, and the approximations in the literature disagree with each other. What can be said is the size and the direction. For single materials the published broad-beam half-value layers are 11 to 33 per cent above ln2/μ, with the larger excesses in the lower-Z materials, so a layered answer from this page is a LOWER bound on the transmission problem by something of that order. A reasonable working practice is to apply the dominant layer’s build-up factor to the whole stack, expect it to be wrong, add margin, and measure.

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References

J. H. Hubbell and S. M. Seltzer, Tables of X-Ray Mass Attenuation Coefficients and Mass Energy-Absorption Coefficients, NIST Standard Reference Database 126, physics.nist.gov/PhysRefData/XrayMassCoef/ (read 7 October 2026). Every coefficient on these pages comes from here: eight materials, 1 keV to 20 MeV, 369 tabulated rows. A work of the United States Government and therefore free of domestic copyright, which is the reason this vertical exists in the form it does — the alternative sources for the same numbers are copyrighted standards. Three things were checked against it before anything was written: every absorption edge sits at its published energy (lead K at 88.0045 keV, tungsten K at 69.525, iron K at 7.112); μ/ρ at 1 MeV reproduces the published spot value for all eight materials to the last printed digit; and μ_en/ρ is at or below μ/ρ in every one of the 369 rows, as it must be.
Table 1: Values of <Z/A>, I and Densities for Elemental Media and Table 2: … for Compounds and Mixtures, NIST X-Ray Mass Attenuation Coefficients (read 7 October 2026). Two things are taken from these. The <Z/A> column, which is what turns a Klein–Nishina cross-section per electron into a coefficient per gram: lead 0.39575, tungsten 0.40250, iron 0.46556, aluminium 0.48181, ordinary concrete 0.50932, dry air 0.49919, ICRU-44 soft tissue 0.54996, water 0.55508. And the density column, which matters because a mass attenuation coefficient is per gram and cannot become a thickness without one: 11.35, 19.30, 7.874, 2.699, 2.300, 1.205×10⁻³, 1.060 and 1.000 g/cm³. Note what the <Z/A> column shows on its own: lead carries 0.396 electrons per nucleon against water’s 0.555, so a gram of lead holds 29 per cent FEWER electrons than a gram of water. Where attenuation is Compton scattering and nothing else, which for these two materials is 1.01 MeV to 2.47 MeV, that makes lead the worse material per gram, and only its density rescues it.
Ionizing Radiation — Shielding Layer Examples, Occupational Safety and Health Administration, osha.gov/ionizing-radiation/introduction/shielding-layer-examples (read 7 October 2026). A US Government work and reproduced here. Two rows, six materials-worth of numbers, all in centimetres: caesium-137 at 0.66 MeV, HVL 4.8 concrete / 1.6 steel / 0.7 lead and TVL 15.7 / 5.3 / 2.1; cobalt-60 at 1.17 and 1.33 MeV, HVL 6.6 / 2.1 / 1.2 and TVL 20.8 / 6.9 / 4.0. These are BROAD-BEAM figures and every one of them is larger than ln2/μ from the NIST coefficient. At caesium-137 the excess is 25 per cent for concrete, 33 for steel and 27 for lead; at cobalt-60 it is 27, 28 and 15. The ratio of TVL to HVL in the table is also worth reading: narrow-beam it would be exactly ln10/ln2 = 3.322 for every entry, and here it runs 3.27, 3.31, 3.00, 3.15, 3.29 and 3.33, which is the rounding of one-significant-figure entries rather than physics.
V. Steiner, A. Malki, T. Ben-Yehuda and M. Moinester, Concrete and Lead Shielding Requirements for PET Facilities, arXiv:2407.12991 (read 7 October 2026). The one source found that MEASURES the narrow-beam and wide-beam cases side by side on the same wall, which is what makes it worth citing above any table. Through 20 cm of Israeli B30 concrete (density 2.30 g/cm³) at 511 keV, the narrow-beam transmission was “T = (3.0±1.0)%, corresponding to a B = 1.4±0.5 buildup factor” and the wide-beam transmission “T = (8.8±1.8)%, corresponding to a buildup factor B = 4.0±0.8, consistent with Monte Carlo (MC) simulations B = 3.9±0.6”. It uses μ_m = 0.0833 cm²/g, μ = 0.196 cm⁻¹ and a mean free path of 5.10 cm, where NIST’s ordinary concrete interpolates to 0.08831 cm²/g — 6.0 per cent apart, which is two different concretes rather than an error. It also states that “a 3.3 cm lead wall results in <0.5% transmission”, which the NIST coefficient reproduces at 0.29 per cent.
J. Eakins, An MCNP-4C2 Determination of Gamma Source Shielding, HPA-RPD-030, Health Protection Agency, Centre for Radiation, Chemical and Environmental Hazards, Radiation Protection Division (September 2007; read 7 October 2026). Crown copyright, quoted briefly. It is a Monte Carlo calculation done explicitly for BROAD-beam geometry — its stated aim is transmission “corresponding to a source that is considered plane parallel and effectively infinite in extent” — which makes it the cleanest available statement of how much a narrow-beam answer under-states. For 511 keV photons it gives lead a half-value thickness and tenth-value thickness of “approximately 5 mm and 17 mm”, against ln2/μ = 3.90 mm and ln10/μ = 12.96 mm from the NIST coefficient: ratios of 1.28 and 1.31 — larger than any of the shielding-table ratios, which is the point. A build-up factor belongs to a GEOMETRY and not to a material, and an infinite plane source is a harsher geometry than the collimated room a shielding table has in mind. The report’s iron figures are read off a plotted curve rather than tabulated and are used here only for that qualitative statement: “approximately 21 mm and 55 mm” against ln2/μ = 10.57 and ln10/μ = 35.12 give ratios of 1.99 and 1.57, and a pair that inconsistent cannot both be right.
Lead shielding thickness — what are the lead codes?, IonActive Consulting radiation protection resource hub (read 7 October 2026). The source for the six BS EN 12588 rolled lead sheet codes and for the practical advice around them: Code 3 is 1.32 mm, Code 4 1.80, Code 5 2.24, Code 6 2.65, Code 7 3.15 and Code 8 3.55, with “a ± 5% tolerance”. Three statements are taken from it directly. Sheet lead codes “tend to be used for relatively low energy applications (i.e. up to about 150 kVp)”. “Where significantly thick lead is required (e.g. 10’s – 100’s mm), lead bricks are often the better option”. And “since lead codes relate to specific thickness dimensions there is often a compromise required, so you may need to use a lead code above what you actually require” — which is the reason the shield-thickness page prints the code at or above the answer rather than the nearest one. It also gives a useful anchor at the diagnostic end: “at 150 kV it is often reported that the TVT is 0.95 mm lead (so practically 1 mm lead)”.
Gamma ray, Wikipedia (read 7 October 2026). Read for the shielding section and quoted for one comparison that is worth having because it is in the units people actually think in: “gamma rays that require 1 cm (0.4 inch) of lead to reduce their intensity by 50% will also have their intensity reduced in half by 4.1 cm of granite rock, 6 cm (2.5 inches) of concrete, or 9 cm (3.5 inches) of packed soil”, and for the statement that “a lead (high Z) shield is 20–30% better as a gamma shield than an equal mass of a low-Z shielding material”. Both are consistent with the NIST coefficients at around 1 MeV, with one correction. The energy at which lead’s narrow-beam half-value layer is exactly 1.000 cm is 1.194 MeV, and concrete’s there is 5.07 cm, not 6 — so the quoted pairing is about 18 per cent generous to concrete, which is roughly the size of a build-up correction and may be where it came from. The second statement needs more care still: per unit MASS lead beats water below 1.012 MeV, LOSES to it from 1.012 to 2.471 MeV, and wins again above that as pair production takes over. “20–30% better” is true at diagnostic energies and false at cobalt-60.