X-ray shielding, why is lead used to shield us when taking X-ray images?

As far as I remember (but can't find it on wikipedia ... ), the deflection on (high energy) photons increases the more heavier the nuclei are. (Don't remember and don't find if it's really the mass or rather the proton number.)

In either case, there are heavier, more dense materials with higher proton numbers.

The material is not consumed nor altered by exposure to X-rays. So why don't we use gold or depleted uranium (just to name some alternatives)?

(Not sure about tags, if anyone knows better, please feel free to suggest/add some others.)

Edit: as the answers and comments here helped me to clear my mind to change the question, but the new question is sufficiently different, I've asked a follow-up here: Formula for scattering and energy change of photons on (naked) nuclei

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    $\begingroup$ Near duplicate: physics.stackexchange.com/questions/11506/… $\endgroup$ Feb 23 '16 at 21:28
  • $\begingroup$ @dmckee thanks for quoting it as a near duplicate. The mentioning over there of concrete indicates the question is more likely to ask for protection against atomic explosions, as concrete with its water contents is a good shielding against neutrons but not against gamma- or X-rays. $\endgroup$ Feb 23 '16 at 21:35
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    $\begingroup$ That question asks explicitly about gammas. He might or might not have known what he wanted, but that's what he asked for. $\endgroup$ Feb 23 '16 at 21:37
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    $\begingroup$ Why don't we use [...] depleted uranium? What makes you think "we" don't? I have done work related to medical linear accelerators (used in radiation oncology) that employed uranium shielding. $\endgroup$ Feb 26 '16 at 20:35

Lead : 1.65 €/kg, Gold : 35,000€/kg, Uranium (even depleted) : cost unknown, with possible radioactive contamination.

So I think it's just a question of cost and practical use. Lead is cheap and available.

  • $\begingroup$ There are a lot of investors, right now, in view of Mario Draghi willing to print any amount of money backed with nothing, who are willing to buy gold at nearly any price, doing no job and gaining no interests, wouldn't that be a market to lend the gold away for X-ray protection? OK, so far it's not about physics. But what about other elements? $\endgroup$ Feb 23 '16 at 21:27
  • $\begingroup$ I upvote it, because it's true and reasonable, but I won't accept it as an answer because I'd like to have an answer based on physics (density, laws of deflection, proton number, nuclear number or whatever plays the crucial role). $\endgroup$ Feb 23 '16 at 21:45
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    $\begingroup$ @GyroGearloose There isn't a pure physics reason. We use lead for economic reasons, because this stuff is done in the real world. $\endgroup$ Feb 23 '16 at 21:47
  • $\begingroup$ @dmckee right you are, but this is physics.stackexchange, so my question could only be about the physic aspects. Putting all thoughts about costs and availability aside, what would be the best shielding material? $\endgroup$ Feb 23 '16 at 21:53
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    $\begingroup$ What you asked is "why is lead used to shield us when taking X-ray images?". If you meant what material provides the best gamma shielding you should have asked that in the first place. The best I know is heavymet, but uranium might be better aside from it's intrinsic tendency to add radiation to the environment where it is used. $\endgroup$ Feb 23 '16 at 22:01

I pulled out my notes from a shielding class and found that the absorption cross section per atom follows a rule: $$\sigma_a\sim\frac{Z^p}{E^3},$$ where $z$ is the atomic number of the absorber atom, $E$ is the energy of the photon, and $p$ is an energy dependent value between 3 and 5. For most x-rays, $p\simeq 4$.

While the cross-section per atom does indeed get larger for increasing $Z$, the density of the material is important, too. The density peaks at osmium ($Z=76$), then drops off, then climbs again in the actinides, but never reaches densities near osmium and iridium ($Z=77$).

When considering the effectiveness of an shield/absorber, one must consider the combined effects of cross-section per atom and density. The result of this is a quantity known as the linear attenuation coefficient, $\mu$, which is typically quoted in $\mathrm{cm}^{-1}$. This is used to calculated the intensity of radiation after travelling through a thickness, $x$ of a material: $$I(x)=I_0 e^{-\mu x}.$$

A quick search of the internet turned up a compilation of x-ray absorption coefficients called the McMaster Tables.

Medical and dental x-rays fall in the energy realm of $5-150\ \mathrm{keV}$. Mammography x-rays use a filtered $20\ \mathrm{keV}$ discrete x-ray from a Mo anode, so I chose to compare linear absorption coefficients at $20\ \mathrm{keV}$ for several elements:

Element  Z   linear atten. (1/cm) @ 20 keV  density (g/cm3)
   W    74     1293                19.3
   Re   75     1446                21.0
   Os   76     1585                22.5
   Ir   77     1648                22.4
   Pt   78     1629                21.4
   Au   79     1515                19.4
   Pb   82      975                11.3
   Th   90     1152                11.7
   U    92     1307                19.1

All these will, in general, smoothly scale up for lower energy and scale down for higher energy except when a binding energy edge of a K, L, or M shell electron falls at that energy. Then there will be a sharp spike upward.

One must also keep in mind that absorption is followed by a follow-up (lower energy) x-ray because the absorption displaces an electron from an atom, and that hole must be filled. In designing a shield you need to know the incoming energies and the consequential energies that follow the absorption. That's why low-background shields for low-energy gamma and x-ray counting systems use lead, lined with cadmium, then lined with copper.


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