Interactive calculator for Rad Techs for x-ray stopping power of different materials (water, lead, air, etc).

An interactive calculator for answering the question: what is the thickness of material A that we would need to match the x-ray transmission through a given thickness of material B. For instance how much water is equivalent to 1mm of lead in stopping an 50 keV x-rays? If you have been loosing sleep over such a question, this is the rad calculator for you.

material thickness header

Introduction

In this post we give information on this online rad calculator for comparing the x-ray attenuation of different materials. The motivation for this tool is to get a feeling for the relative difference between different materials in stopping x-rays, and to see how it changes with different energies.

This calculator is analogous to an online calculator which will let you know what your salary would need to be if you move from one city to another city to take into account the cost of living difference. The idea here is the same except we are comparing material instead of comparing cities, and the property we are interested in is the thickness of material needed to stop the same fraction of x-rays rather than the cost of living.

Different types of materials have different coefficients of attenuation which means that we need different thickness of each materials to get the same amount of x-ray transmission.

We will first give some examples of the calculator in action and then show the equation behind the calculator which is based off of Beer’s law.

This is a simplified calculator and just assumes x-rays traveling parallel so it does not take into account the 1/R^2 factor which also occurs with x-rays for an x-ray tube.

Beer’s Law

If we have the same incident flux of parallel x-rays coming in two different materials, we want to solve for the thickness of that materials in order to have the same output flux (see Figure).

material thickness header

Intensity of output x-rays is always less than that of input photons. We assume that material itself isn’t making any radiation. So, I is always less than I0. Attenuation is the amount that stopped in the material and transmission is the amount that goes through the material. So if we have the same attenuation, we have the same transmission.

Here is the relationship to solve for the new thickness given the reference thickness and a few quantities that can be looked up on the NIST website. If you want to see the details of solving for this from Beer’s law please keep reading after the examples of the calculator use.

pic1

Material Thickness Rad Calculator

We’re going to play around on the calculator now. Again, the idea is that we have a calculator and the purpose of that is to figure out if we start with one material and then we want to look at another material and ask, “How much of that material do I need to stop x-rays to the same degree?” So if x-rays are coming through to get the same level of x-ray transmission or we could say, the same level of x-ray attenuation in that single material.

We have what we’re calling the reference material. That’s the first material and then the new material will be compared to the reference material.

When we switch the energy, it is going to set us back to a reference scenario. Therefore, switch energy level first before you switch other parameters.

On to some examples. What if we take – water as the reference material and the new material is bone? Do we need more or less bone in order to stop the same fraction of x-rays?

pic2

The answer is we’re going to need less bone. You can see in this figure we need less than 1/3 the thickness of bone. This is because bone is significantly more attenuating as it has that high Z material in it (if this is new to you see our post on the photoelectric effect).

Now let’s look at our second example. Let’s say water is our reference again, and set the reference distance to be 100mm. We want to start with 100mm of water as we are next going to change to a material that is much more attenuating (i.e. lead).

How much lead do you think is going to be equivalent to 100 millimeters of water? Is it going to be 10 millimeters, or 1 millimeter? In fact, it is going to be even less than 1 millimeter.

As you can see from the next figure only 0.25 mm is required in order to get the equivalent transmission of 100mm of water. That is why we don’t use water to shield for x-rays in our walls, or in our doors, and luckily we’re not wearing water aprons that are 100 millimeters thick (if you think you back hurts after wearing a lead apron all day imagine 100mm of water).

This behavior is due to the fact that lead has a high Z and it is dense. Both of those things are working in its favor for it having really good stopping power for diagnostic x-rays.

Then what’s on the opposite end of the spectrum from lead? It’s dry air. We’re not talking about a vacuum, so there will be x-rays that are stopped within dry air. The x-rays will ionize the dry air (this is the mechanism that is used to measure radiation dose). But we know that the attenuation in dry air is going to be significantly lower attenuating compared with water.

Let’s start with a reference thickness of 1 millimeter of water, how much dry air do you think it will take to have the same x-ray transmission?

pic3

Is it going to be 10? Is it going to be 100 millimeters that we’re going to need for dry air? It’s going to be more than that. It’s going to be 872 millimeters of dry air to get the equivalent attenuation of 1 millimeter of water.

pic4

Let’s just do one more just for fun. In general, I want you to play around with the tool and really learn about the relative differences between these material. Let’s see if we go to fat. And if we go to fat, like we talked about, fat is less dense (e.g. oil floats in water). Therefore, we’re going to need a little bit more fat than we would water. For these 50 keV x-rays we need 1.16 millimeters of fat is equivalent to one millimeter of water in terms of x-ray transmission.

Summary of Example Comparisons

Here is a summary of what we learned from using the calculator in these examples. For x-rays with energy of 50 keV we have these relationships. In the examples above the input wasn’t always 100mm but we use that same input here so that we can easily compare.

Ref MaterialRef Thickness (mm)New MaterialNew Thickness (mm)
Water100Dry Air87269
Water100Fat116
Water100Bone28.9
Water100Lead0.25
svg_icon

Rad Take-home Point

  • There are many orders of magnitude difference between the extreme materials of dry air and lead in x-ray attenuation.
  • Fat is less attenuating than water so we need more of it to have the same x-ray transmission.
  • Bone attenuation is more different than fat in comparison to water which is why is has very good contrast on x-ray images.

Beer’s Law to Material Thickness

We have gone through all the fun examples and now we will go through quickly how to solve for the equation we presented above that the calculator uses.

pic5

In this figure we will have the same incident intensity so we want to set the intensity after passing though the material to be the same.

pic6

Like we said, in Beer’s Law we’re going to have I0 is that incident flux, and then we have an exponential term multiplying that which is less than one on each side.

pic7

The exponential term on each side is due to the reduction of intensity for that specific material. One the left is the reduction in intensity of the new material and on the right is the reduction in flux of the reference material.

The first term in the exponent is (μ/ρ) which is pronounced  mu over rho. This is the mass attenuation coefficient and this is dependent on the material and dependent on the x-ray energy.

The next term in the exponent is ρ. This is just the density of the material. The density is not dependent on the x-ray energy.

Then the final term in Beer’s law on both sides is the material thickness. This is just a distance measure (in mm for instance) of the thickness of material.

We can look up on a NIST website the values of (μ/ρ) and ρ for the different materials. In our calculator, we’ve done that for you just to make it easier.

The next thing we can do with our equality is to cancel out the I0 on each side so that we get this:

pic8

We see that we have the same exponential form on both sides, i.e. these terms look similar. What we can do is we can take the natural logarithm of both sides.

If you remember from math class, if you take the ln of an exponential, those two operations cancel out leaving us with.

pic9

Now we have just a linear equation where three terms on each side are being multiplied together.

We want to solve for Xnew the distance of the new material so we need to divide both sides by (μ/ρ)new and ρnew. This leaves us with the equation used in this Rad Calculator that we introduced above.

pic10
Scroll to Top