top of page
Writer's picturePaul V

Shielding and Half value layers (HVL)

According to the U.S. Nuclear Regulatory Commission we control our personal exposure to Ionizing Radiation through a combination of Time, Distance, and Shielding.


Today I want to talk a little about shielding. Shielding is anything we can put between ourselves and the source of the radiation that has enough density to absorb the radiation or at least absorb as much as possible. The denser the material the better. The density of the material is basically how much does it weigh vs how big is it. So something with high density vs something with low density would be like a pound of feathers vs a pound of Lead (Pb). The lead will be much smaller physically than the pound of feathers, from this we get that the lead has a greater density than the feathers. What this means for us is that the atoms that make up the lead are first, more tightly packed together and second, is typically made of atoms with a higher Z number (Atomic number, from the periodic table of elements). The greater the density of a material, the more chance there is that the photons from the Ionizing radiation source will interact with one or more of those atoms and have their energy absorbed through the Photoelectric effect, and/or Compton scatter.


Lead is the most common shielding material used, it is low cost and very dense, easy to shape. Other common shielding materials would be Concrete, Tungsten and for exposure devices Depleted Uranium (DU). Steel works well but not as well as lead and is more costly.


A half value layer of any material is the thickness of that material that we need to place between us and the source to absorb half of the intensity (half the radiation). We use Absorb and attenuate interchangeably in this, when we absorb radiation we are attenuating it. With Ir-192 (Iridium 192), a HVL of Lead (Pb) is 0.19 inches (4.8mm), Steel is 0.5 inches (12.7mm). You can look up half value layer thicknesses for various materials in the NRC publication "Working Safely in Gamma Radiography: A training Manual for Industrial Radiographers", it is free to download from the NRC (www.nrc.gov/reading-rm/doc-collections/nuregs/brochures/br0024/index.html). it's an older book but still relevant.


Once we know the thickness of a material we can calculate how many HVL of it that we have.


For instance: 1.5 inches of steel would be 3 HVL (half value layers), we get there by simply dividing the thickness of material by it's HVL thickness. 1.5" of steel/0.5" (HVL) = 3


How can I apply this? well if I know that I'm receiving 100mR/hr at the location I'm standing, and I place 1.5 inches of steel between the source and myself I can easily calculate the new intensity by dividing 100 by 2, 3 times. 100/2=50, 50/2=25, 25/2=12.5mR/hr. The other way to do this, if you don't have a nice round number would be to use the following formula: 100 mR/hr/2^3=12.5 mR/hr. this symbol '^' means "to the power of", basically 2 multiplied by itself 3 times (2x2x2=8) 100/8=12.5


Now, what if I need to know how many half value layers I need to add in order to get to a specific intensity? Something like this, I am receiving a dose rate (intensity) of 100 mR/hr at my current location, how many inches of steel do I need to put between myself and the source to bring that down to 2 mR/hr?


I know from earlier that 0.5 inches of steel equals 1 HVL, one way to do this would be to divide 100 by 2 till I get below the 2 mR/hr threshold.

100/2=50

50/2=25

25/2=12.5

12.5/2=6.25

6.25/2=3.125

3.125/2=1.56


6 HVLs will bring me below 2 mR/hr, which would be 6 x 0.5=3 inches of steel


Another method for those who like math


Log(Io/Id)/Log(2)=# of HVLs needed

Io = original intensity

Id = Desired intensity


on my calculator (yours might work differently), The key strokes go like this:

[Log](100[/]2)[/][Log](2)=5.644 HVLs


to prove it we can run this formula from above 100/2^5.644=1.9998 mR/hr


we calculate the number of inches of steel like this:


0.5 x 5.644= 2.822 inches


All the math works the same for SI units as it does for the English Standard units used here in these examples.


As always very with a survey meter!!


Recent Posts

See All

Let’s talk about Penetrant testing.

Did you know that all penetrants are solvent removable? Water washable penetrants can be removed using water or solvent wipe methods...

Comentários


bottom of page