Why lead can be used for protection ?

[lionelwang]

Hi all,
anybody knows why x-ray cannot penetrate into a sheet of lead paper deeply? Do only the big amount of electrons in lead atoms diffract the x-ray, or the protons in the atoms also contribute the diffraction ?

Thank you for your time!

 

[Drakkith]

I believe it is a combination of the large amount of electrons present in Lead as well as the very high density of Lead. The density will pack more of those large amounts of electrons into a smaller space. The nucleus should be very good at shielding from Neutrons however.

 

[Bob S]

Here are thumbnail plots comparing the x-ray attenuation coefficients μ of aluminum and lead, plotted against x-ray energy. The attenuation is of the form e^-μx, where x is the thickness in grams per cm². X-rays are attenuated entirely by the electrons in the material, not the nuclei. Aluminum has more electrons per gram than lead (why?). Yet the x-ray attenuation coefficient in lead is nearly ten times larger than in aluminum between 10 and 100 KeV. Why?

Bob S

Ref: http://physics.nist.gov/PhysRefData/XrayMassCoef/tab3.html

 

[mathman]

From my recollection, there is something called the K edge (in lead) which is an absorption resonance ~ 10 kev.

 

[brainstorm]

I believe it is a combination of the large amount of electrons present in Lead as well as the very high density of Lead. The density will pack more of those large amounts of electrons into a smaller space. The nucleus should be very good at shielding from Neutrons however.

Does this mean that if a molecule would have a similar electron-density as lead and be cooled to a temperature where the density of the molecules was equivalent to room-temp lead, that the x-ray resistance would be the same as well?

 

[lionelwang]

It seems that the discussion has not yet ended...

 

[Drakkith]

Does this mean that if a molecule would have a similar electron-density as lead and be cooled to a temperature where the density of the molecules was equivalent to room-temp lead, that the x-ray resistance would be the same as well?

It would make sense to me based on my understanding, but don't take my word for it and stand in front of a source of x-rays with your custom shielding. ;)

 

[lionelwang]

I believe it is a combination of the large amount of electrons present in Lead as well as the very high density of Lead. The density will pack more of those large amounts of electrons into a smaller space. The nucleus should be very good at shielding from Neutrons however.

You are right, finally I found the reason from a textbook: some of the electrons the ground state in lead will absorb x-ray photons and then excited to a higher energy level. Thanks!

 

[Cleonis]

finally I found the reason from a textbook: some of the electrons the ground state in lead will absorb x-ray photons and then excited to a higher energy level.

If I remember correctly an alternative to lead to shield off X-rays has been developed. The ions of a certain salt (don't remember which one) had a similar (or even higher) absorption as lead. So medical workers who need to protect themselves could put on clothing made of fabric that had been immersed in a solution of that salt. This clothing weighed just a fraction of the protection with slabs of lead, but gave the same shielding.

 

[Cleonis]

If I remember correctly an alternative to lead to shield off X-rays has been developed.

http://www.radshield.com/pdf/Scientic%20American%20May%202003.pdf" (The link downloads a PDF)

It's not less heavy, but it folds and bends much more easily, and it allows body heat to escape.

 

[Bob S]

The effectiveness of x-ray (photon) shielding can be divided into three regions; below ~200 KeV, 0.5 to ~2 MeV, and above ~2 MeV.

Below ~200 KeV (all-x-ray machines)
The shielding mechanism below ~200 KeV is the deep-core photoejection (photoelectron) total absorption of incident x-rays, very dependent on the binding energy of inner atomic electrons in the shielding material. The K-edge, L-edge, and M-edge peaks in the lead attenuation coefficient (shown in the post #3 thumbnail above) represent the binding energies of K, L, and M-shell electrons. Very roughly, the K-shell binding energy is ~13.6(Z-1)² eV, where Z is the atomic number of the material, and 13.6 eV is the binding energy of the electron in the hydrogen atom. For lead, Z=82, and the K edge is ~ 88 KeV, while for aluminum (Z=13), the K edge is only ~1.7 KeV. Compare the plots for aluminum and lead in thumbnail above. Bismuth (Z=83), used in the Demron radiation protection clothing (see previous post), is slightly better (K edge ~ 90.5 KeV) than lead.

~0.5 to 2 MeV (some nuclear gammas)
The shielding mechanism is Compton scattering of x-rays (photons), which is proportional to the number of electrons per gram (aluminum is better than lead; see thumbnail). So aluminum per gram is slightly better than lead for the primary 1.1 and 1.3 MeV gammas from Cobalt-60.

Above ~2 MeV (mainly bremsstrahlung from electron accelerators)
Here, the shielding (photon absorption) mechanism is pair production by high-energy photons in the vicinity of high-Z nuclei (like lead and bismuth). The pair production cross-section is proportional to ~Z². Personnel protection clothing is no substitute for area perimeter protection (microswitches on access points to radiation areas).

Bob S

 

[brainstorm]

Does lead absorb or reflect the x-rays, actually? Does it absorb and/or reflect them at a different wavelength?

 

[Bob S]

Does lead absorb or reflect the x-rays, actually? Does it absorb and/or reflect them at a different wavelength?

Very good question.

Below ~ 200 KeV (deep core photoejection), the incident x-ray energy is completely absorbed by the recoil electron, and most of the recoil electron energy is dissipated in the absorber as heat.

For the intermediate range, ~500 KeV to 2 MeV, the incident x-ray (photon) is Compton-scattered off the electrons in the absorber, and a lower energy secondary photon is emitted, usually in the forward direction. See illustration of Compton scattering and an on-line energy-angle calculator at http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/compton.html

Above ~ 2 MeV in lead, the incident photon creates positron-electron pairs, and when the positron stops, it annihilates with the emission of two 0.511 MeV back-to-back photons distributed isotropically. The positron rarely annihilates in flight.

[added] In post #3, the x-ray attenuation curves each have two lines. The higher solid line is for attenuation of the incident primary x-ray (photon), and the lower dotted line is for total energy attenuation (including secondary photons and electrons).

Bob S

 

[brainstorm]

Below ~ 200 KeV (deep core photoejection), the incident x-ray energy is completely absorbed by the recoil electron, and most of the recoil electron energy is dissipated in the absorber as heat.

That is crystal clear.

For the intermediate range, ~500 KeV to 2 MeV, the incident x-ray (photon) is Compton-scattered off the electrons in the absorber, and a lower energy secondary photon is emitted, usually in the forward direction. See illustration of Compton scattering and an on-line energy-angle calculator at http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/compton.html

I know I should be able to play with that calculator and interpret the output but I just don't understand the values and units at this scale. When you say "emitted in the forward direction," do you mean the x-ray photons get re-emitted as if they penetrated the lead, only in a lower frequency wavelength, e.g. UV, infrared, microwaves?

Above ~ 2 MeV in lead, the incident photon creates positron-electron pairs, and when the positron stops, it annihilates with the emission of two 0.511 MeV back-to-back photons distributed isotropically. The positron rarely annihilates in flight.

How far does it go and what happens to the atom with the electron it annihilates with? It ionizes?

[added] In post #3, the x-ray attenuation curves each have two lines. The higher solid line is for attenuation of the incident primary x-ray (photon), and the lower dotted line is for total energy attenuation (including secondary photons and electrons).

I looked and tried, really, and thanks for posting. I couldn't decode the labels, though. It looks very concise and transparent - just not for lay people, I think - but I can't speak for all lay people, of course, just the megalomaniac geniuses with math illiteracy issues.

 

[Bob S]

[Re Compton scattering]I know I should be able to play with that calculator and interpret the output but I just don't understand the values and units at this scale. When you say "emitted in the forward direction," do you mean the x-ray photons get re-emitted as if they penetrated the lead, only in a lower frequency wavelength, e.g. UV, infrared, microwaves?

By forward, I mean at laboratory angles less than 90 degrees. Backward would mean more than 90 degrees. The re-emitted Compton-scattered photon energy E' minimum is usually ~>=10% of the incident photon energy, or ~255 KeV, whichever is less (There is a formula, which I have not included).

[RE pair production] How far does [the positron] go and what happens to the atom with the electron it annihilates with? It ionizes?

The positron is always created with an electron, so this balances the electron lost when the positron annihilates. In a metal, the created electron finds the vacancy. In a dielectric, it doesn't.

I looked and tried, really, and thanks for posting. I couldn't decode the labels, though. It looks very concise and transparent - just not for lay people, I think - but I can't speak for all lay people, of course, just the megalomaniac geniuses with math illiteracy issues.

Definitions for the terms in the NIST tables can be found in
http://physics.nist.gov/PhysRefData/XrayMassCoef/intro.html and
http://physics.nist.gov/PhysRefData/XrayMassCoef/chap2.html
Basically, in Compton scattering for example, the higher attenuation coefficient curve in the plot applies to the attenuation of the incident photon only. while the lower curve applies to any (all) photons, including the incident photon and all Compton scattered photons. For pair production, the lower curve includes the incident photon, the ionizing pairs, and the annihilation photons.

Bob S

 

[Bob S]

Here in the thumbnail is a good plot of the mass attenuation coefficient for several elements, ranging from aluminum to uranium, and including lead. At energies below 500 KeV, the deep core photoejection attenuation dominates for high-Z materials, and above ~5 MeV, pair production dominates. Between 1 and 2 MeV, Compton scattering dominates for all Z. Lead is a good, inexpensive, and effective high-Z radiation protection material, especially below ~ 500 KeV and above ~ 10 MeV.

For further discussion of the mass attenuation coefficient, see

http://mightylib.mit.edu/Course%20Materials/22.01/Fall%202001/photons%20part%202.pdf
and

http://en.wikipedia.org/wiki/Mass_attenuation_coefficient

Bob S