Investigating Attenuation of X-Ray Photons with Al Thickness

[Jameshope]

Hi

Would anyone help/advise me? I've done 2 experiments as part of my radiography degree. One was to investigate the effect of the ISL on a beam of x-ray photons. I could understand this, produced 2 graphs (Mean reading against distance (uGy) and Mean reading against 1/d squared) and wrote a 1,000 page report with a knowledgeable discussion and good conclusion.

However, my other experiment is confusing me. Its about HVL and I'm getting totally confused with the limear attentuation co-efficient. The title is to investigate the attenuation with thickness of material of a beam of x-ray photons

We then set the x-ray tube at 100 cm above the dosemeter on the x-ray table and collimate to the sensitive plate. This distance and collimation was constant. We then made an exposure of 80 kV and 100 mAs and recorded the reading repeating twice more.

Then we attached 1mm sheet of aluminium beneath the light beam diaphragm and used the same exposure factors and record the dosemeter reading and repeated for 2, 3 and 4 mm thicknesses of aluminium.

I then plotted 2 graphs- one was HVL transmission % against Al)mm) thickness and the other was calculating Loge so it was HVL Loge Transmission.

The problem is I don't know wahat to put in my discussion? I can obviously see that with each additional mm of Al then the % of x-ray transmission falls and that the HVL at 50% was about 3.4 looking at the graph. But how do I prove this with a formula? How could I have improved the experient and what use will this mean in everyday practise?

Foe the other graph I simply calculated loge by the the % transmission. ie.. Ln on my calculator by 100% at 0mm, 79 at 1mm etc... and I came out with

0mm 4.6 loge
1 4.4
2 4.2
3 4
4 3.8

I have put both sets of data into an exponential graph but I'm struggling to understand everything. Help!

Regards

James

 

[mark726]

Dear James,

Thank you for reaching out for help with your experiment. I can see that you have put a lot of effort and thought into your work, and I am happy to assist you in any way I can.

Firstly, it is great that you were able to successfully conduct the experiment and obtain data for your graphs. From your description, it seems like you have a good understanding of the concept of HVL and how it affects the transmission of x-rays. The formula for HVL is HVL = ln(2)/µ, where µ is the linear attenuation coefficient. This formula can help you to calculate the HVL at different thicknesses of material.

In your discussion, you can mention the relationship between HVL and the thickness of the material. As the thickness increases, the HVL also increases, meaning that more material is required to reduce the transmission of x-rays by 50%. This is because the linear attenuation coefficient, which represents how much the material absorbs or scatters the x-rays, also increases with thickness.

To improve your experiment, you can try using different materials of varying thicknesses to see how the HVL changes. You can also try using different exposure factors, such as different kV and mAs, to see how they affect the HVL. This will help you to understand how to optimize exposure settings for different thicknesses of material.

In terms of practical applications, understanding the HVL is important in radiography as it helps to determine the appropriate thickness of shielding material needed to protect patients and staff from excessive radiation exposure. It also helps in selecting the appropriate exposure factors for different imaging scenarios.

For your second graph, it seems like you have correctly calculated the loge values for the HVL transmission percentages. You can mention in your discussion that taking the logarithm of the transmission percentages helps to determine the exponential relationship between the thickness of the material and the HVL.

I hope this helps to clarify some of your doubts. If you have any further questions, please do not hesitate to ask. Good luck with your studies!