# To investigate the attenuation with thickness of material of a beam of x-ray photons

1. Feb 6, 2012

### 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 knowledgable 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