lucas

Against Electromagnetic Radiation

As per the reference post, when discussing electromagnetic radiation as a source of ionizing radiation, we are discussing high energy photons, which is to say, photons with short wavelengths. These are known as x-rays or gamma-rays, depending on their energy level.

The energy of those rays is also the condition that make electromagnetic radiation ionizing (harmful) and non-ionizing (less harmful**). The ionizing effect start when we enter UV level of energy (just above the visible light spectrum)

**Please note that “Less harmful” refers to the type of physical damage expected upon exposure. For example, visible light is a form of non-ionizing radiation. However, looking directly into a strong light source may damage someone’s retinas! Your DNA is not expected to be affected, however 👍

The goal then becomes to lower the energy level of this radiation until it becomes non-ionising or null. This will happen naturally when photons interact with matter by the following phenomenon:

Photoelectric absorption (low energy interaction) ­­→ The energy is absorbed by the matter, which results in the emission of photoelectrons from this interaction.

Scattering (mid energy interaction, X-rays) ­­→ The photon is reflected by the matter, which can result in a change of direction (think of a bouncing ball). Scattering can lead to a reduction in energy (Crompton Scattering), which is an ionizing interaction, or not (Rayleigh Scattering), which is non-ionizing.

Pair Production (high energy interaction, happens at Gamma-rays level) ­­→ The photon “explodes” upon interacting with matter and creates a new photon with new energy and direction.

As we are talking about sub-atomic and atomic size interactions, it is important to remember that matter is actually mostly vacuum. All atoms composing a material are actually quite far from each other in comparison to their size and the physical effect of the existence of the material they create comes from the strength of the bond between these atoms.

Therefore, when discussing interaction between photons and matter, we must think in terms of probability, which is why everything has the potential to interact with this type of radiation, even air.

This also informs us on how to recognize an appropriate shield. Heavy or dense material, like lead (Pb), will absorb much better than light material like aluminum (Al). In the same way, composite material (which combines different elements) can offer better protection than a shield composed of each of their separate elements.

The probability of a material to interact with photons, and the type of interaction, is called the mass attenuation coefficient and characterizes how easy it is for energy to penetrate a material for a given mass of this material. This probability is affected by the type of material but also the energy of the photon.

See below a chart showing the mass attenuation of different interactions for different materials (note, x-rays are within 1.00E-03 (or 10keV) and 1.00E+00 MeV and gamma-rays are ranged above). These charts are called Photon Cross Sections.

eV stands for electron-Volt, a unit of energy (Such as Joules or Calories)

Example:

These are known by experimentation. The ray is shot at the material and a geigercounter (measure instrument of radiation) is positioned behind the shield to measure how much radiation passes through. The test is repeated for different energy level.

And here is a comparison of the total mass attenuation of lead, blood and stainless steel.

The best summary for this is the following: The higher the Mass attenuation coefficient, the more effective the material will act as a shield. As we can see, generally, the more energetic photons will interact less than lower energy photons.

Finally, we can know how much radiation will pass through a material by using the following formula:

Where I is the resultant intensity, I0 is the initial intensity, 𝜇/⍴ is the Mass attenuation coefficient, ⍴ is the material’s density and x is the thickness of the material

(as a reminder e is the symbol for Euler's number, a mathematic constant, like pi)

Let’s take stainless steel as an example. We want to shield against 99% of the radiation of a CT Scan, which operates at roughly 100keV, or 0.1 MeV.

And I/I0 = 0.01 (remaining 1% of the initial intensity value)

Conclusion, 0.62 in of 304 Stainless Steel would attenuate 99% of the radiation emitted by a CT scan.

In the world of Iron Lung, we can see that blood is a decent radiation "absorber". Meaning that the camera would most likely not have a very great range, 2m at most, to capture what's in the depth of AT-5.

To be noted as well that pair production would render any camera useless as the new photons created would create a massive amount of artifacts and debris in a picture. Therefore, despite the huge amount of radiation emitted by the camera, the energy put to the photons would probably be lower, which, in turns, translates to more photons shot than high energy photons.

It also means that, for the most part, very little radiation would make its way back to the Iron Lung — or its pilot.

Other Factors - Inverse Square Law

The other consideration to be taken when analyzing how much radiation would affect someone is the Inverse Square Law, which stipulate that the intensity of an electromagnetic radiation is inversely proportional to the distance from its source

Where ID is the intensity at a calculated distance.

For example, if I measure 1 mSv at 1m from the source of the radiation, I would measure ¼ of this intensity at 2m from the source, or 0.25 mSv

You can imagine this to be similar to how it is with a flashlight, the further you are from the flashlight, the less it illuminates your surroundings.

The following illustrate this law:

S being the Source, A being the subject and r being the distance, you can see that the further A is from the source, the less concentrated the rays are at its position.

If I was to put this in the context of Iron Lung.

This is the shot Simon takes in the hangar. Jack, the welder, is standing more or less in front of the camera. Ava and the Pilot are further in the background.

I have terrible depth perception and cannot estimate distances for the life of me so... I'll illustrate how much radiation Ava would have gotten based on possible distances.

Let's say Jack got the 7400mSv at 0.5m from the camera.

Conclusion, it is fair to suppose that Ava was mostly joking about Simon expected apologies to her tumors...

100mSv is roughly the yearly dose needed to see a significant increase in cancer risk.

Can't save Jack with science, however. R.I.P Jack.

References

https://www.osha.gov/ionizing-radiation/introduction/ionizing-radiation-equations

https://www.nist.gov/pml/x-ray-mass-attenuation-coefficients

https://www.makeitfrom.com/material-properties/AISI-304-S30400-Stainless-Steel

https://radetco.com/your-complete-guide-materials-that-block-radiation/