My research project intends to make a plasma generating chamber and due to the applied high voltage, there is a possibility of creating X-ray. The voltage I intend to apply is 20,000 volts and the lids of the chamber are made of Aluminium. To make sure the safety of personnel working I need to apply a lead sheet covering around the chamber. I need to know the thickness of the sheet required to stop the X-ray created at this voltage.
First, it depends on the anode material. If the anode material is Aluminium (as the chamber lids), the maximum obtainable X-ray energy is about 1.5 keV. At this energy value, the linear attenuation coefficient of Pb is about 27 cm-1, which means that a 1.7 mm thick Pb sheet is able to attenuate 99% of the radiation.
By the way, such a thick Pb sheet could be not necessary if the X-ray intensity is already below or near the radioprotection safety threshold. Note that the intensity of the produced X-ray not only depends on the accelerating voltage (20 kV), but also on the plasma electron current (i.e. on the plasma power).
I suggest to check with a Geiger counter the radioactivity just outside the chamber after plasma ignition, and then (if the detected value is over the safety threshold) you can choose the appropriate thickness of the Pb sheet which attenuates the X-ray intensity down to safe values.
Dear Indula,
I attached a table with half-value thicknesses. Take the value for 25 kV.
With Regard
R. Mitdank
Some Marco's statements are correct, but there are some very large serious errors.
- "If the anode material is Aluminium (as the chamber lids), the maximum obtainable X-ray energy is about 1.5 keV." This is false. You will get additional intensity at the energy of the anode's characteristic x-rays, but you will also get x-rays with energies up to your excitation voltage, 20 kV in your case. This is a very serious error because higher energy x-rays are typically much more penetrating that low energy x-rays.
- "At this energy value, the linear attenuation coefficient of Pb is about 27 cm-1, which means that a 1.7 mm thick Pb sheet is able to attenuate 99% of the radiation." These numbers are grossly in error. X-rays of energy this low are much easier to shield than this. Al characteristic x-rays do have 1.5 keV energy, but the rest is wrong. At this energy the mass linear attenuation coefficient is 27000 cm^-1. (If you do have a 1.7 mm of a material with a linear attenuation coefficient of 27 cm^-1, the transmission will be 1%, for 99% attenuation, but that is quite irrelevant). You actually need 1.7 microns of Pb for 1% transmission of 1.5 keV x-rays. X-rays of this energy will only penetrate a few inches of air and not much of your skin. They are unlikely to be detectable by your survey meter (e.g., Gieger counter). However the highest energy x-rays will penetrate more and you need to worry about what will penetrate. For lead, the linear attenuation coefficient at 20 keV is 980 cm^-1, so you would need 50 microns of 0.05 mm of lead for 1% transmission.
However
- You do not need something this thick for 1% transmmision because that is the thickness because only a very small fraction of the x-rays will have the maximum possible energy of 20 keV; all the rest will be stopped more easily or much more easily. You ask how many x-rays will have that energy? It depends! Very difficult to say exactly.
- You do not necessarily need to design for 1% transmission. Maybe you have low current plasma so 50% transmission is fine. Maybe you have a high current and 1% transmission would be instantly fatal, so you need much much better shielding than 99% attenuation. (I doubt it, though).
My recommendation for this low energy is to try 0.1 mm of steel; if you have any detectabl transmission through that, try 1 mm of steel. Of course use whatever thickness you have laying around or are cheapest. The transmission of 20 keV through 1 mm of steel will be
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