Osłony RTG / Knowledge / Shielding thickness
How much lead does an X-ray room need?
Shielding thickness cannot be given "for 200 kV" — it follows from the attenuation factor, that is from how many times the dose behind the barrier has to be reduced. Below is the full calculation to PN-86/J-80001 on real numbers, step by step.
The formula
For primary radiation (clause 2.5.1.2 of the standard):
k = (D' · I · t · y) / (D · l²)
- D' — tube dose rate at 1 m per 1 mA [mSv/(min·mA) at 1 m],
- I — anode current [mA],
- t — irradiation time per week [min] = single exposure time × number of exposures,
- y — attenuation of the beam in the object under test,
- D — weekly dose limit behind the barrier [mSv],
- l — distance from the focus to the barrier [m].
There is no single dose limit: 20 mSv per year for a worker and 1 mSv per year for a member of the public (Polish Atomic Law, art. 14, following Directive 2013/59/Euratom). On top of that comes the occupancy factor T — the fraction of time someone is actually behind the barrier.
Worked example
200 kV tube, 3.7 mA (the maximum for this voltage), 20 s exposure, 20 exposures per week, barrier 2 m from the focus, T = 1 for workers and T = 0.25 for the public:
| Quantity | Workers | Public |
|---|---|---|
| Weekly dose limit [mSv] | 0.12 | 0.006 |
| Exposure time per week [min] | 6.67 | 1.67 |
| Attenuation factor k | 1 151 | 5 756 |
| log₁₀(k) ≈ number of TVLs | 3.06 | 3.76 |
| Pb from primary radiation [mm] | 2.75 | 3.75 |
| Pb from scattered radiation [mm] | 0.50 | 1.25 |
| Required thickness [mm Pb] | 2.75 | 3.75 |
Design thickness with a margin: 4.00 mm Pb. At 350 exposures per week instead of 20 the same room needs 4.75 mm for workers and 6.50 mm for the public — the number of exposures enters the calculation linearly, which is why it usually decides the thickness.
Why every wall comes out different
Each barrier carries its own distance and its own use factor U: 1 for a barrier hit by the primary beam, 0.25 for the remaining walls, 0.05 for the ceiling (clause 2.3 of the standard). For a 5 × 4 × 3 m room with the tube in the middle and the beam pointing down:
| Barrier | Distance [m] | Required [mm Pb] |
|---|---|---|
| Walls 1 and 3 (nearer) | 2.00 | 1.25 |
| Walls 2 and 4 (farther) | 2.50 | 1.00 |
| Ceiling (U = 0.05) | 2.00 | 0.75 |
| Floor (beam down, U = 1) | 1.00 | 4.75 |
The floor comes out thickest because it takes the primary beam from the shortest distance. That is the typical result when radiographing downwards, and the most common reason for changing the design: raise the focus, or aim the beam at a wall that is easier to shield.
What this calculation does not cover
- Doors and windows have to meet the requirement of the wall they sit in — frame and threshold included.
- For a layered barrier (steel + lead + concrete, say) build-up has to be taken into account and the layers computed in the order in which the beam meets them.
- The curves of the standard cover a limited range; above it the result is an extrapolation of the curve and the design has to say so.
Calculate your own room in the calculator See the demo on prepared rooms
These calculations support the shielding design. The final assessment, the classification of workers and the approval of the design are made by the head of the organisation in consultation with the radiation protection officer.
Frequently asked questions
How much lead is needed at 200 kV?
It depends on the tube current, the exposure time and their number, the distance and on who stays behind the barrier. For a 200 kV / 3.7 mA tube, 20 s per exposure, 20 exposures per week and a barrier 2 m from the focus the result is 2.75 mm Pb for workers and 3.75 mm Pb for the public; with a margin 4.00 mm Pb is adopted. At 350 exposures per week the same room already needs 4.75 and 6.50 mm Pb.
What formula does PN-86/J-80001 use?
The attenuation factor k = (D' · I · t · y) / (D · l²), where D' is the tube dose rate at 1 m per 1 mA, I the anode current, t the irradiation time per week, y the attenuation in the object, D the weekly dose limit behind the barrier and l the distance from the focus to the barrier. The lead thickness is read from the attenuation curve (Fig. 1 of the standard) for that k and the tube voltage.
Are primary and scattered radiation calculated separately?
Yes. Primary radiation comes from the attenuation factor k (Fig. 1), scattered radiation from the reduced dose rates C₁ and C₂ (Figs. 3 and 4). For every barrier the larger of the two required thicknesses is taken.
Is the thickness the same for every wall?
No. Each barrier has its own distance from the focus, its own use factor (1 for the wall hit by the primary beam and for the floor, 0.25 for the other walls, 0.05 for the ceiling) and a different use of the room behind it. In a 5 × 4 m room with the beam pointing down the result is 1.25 mm Pb on the near walls, 1.00 mm on the far ones, 0.75 mm on the ceiling and 4.75 mm on the floor.