I received detailed plans for the liquid helium Dewar that's going to be used in the h-ray experiment yesterday and spent some time calculating the amount of attenuation we can expect to the x-rays theoretically produced by our quenched Pb superconductor. I'm making the assumption that most of the attenuation of x-ray flux will come from the aluminum walls and aluminum backed insulation, not from the plastic walls. A cross-section of the walls is shown to the left, (picture 1). The wall on the left of the cross-section is made of .125 inch thick aluminum. The five sheets of insulation between the inner and outer walls are Mylar coated with .003 inch thick aluminum.The inner wall on the right hand side of the cross section is constructed of .1 inch of 'low thermal conductivity plastic'. This gives me a grand total of 0.35 cm of aluminum between the source and the detector. The graph shown below (picture 2) is the amount of flux transmitted through the aluminum with respect to energy. The attenuation values were taken from an online attenuation calculator[1]. The attenuation value without coherent scattering was used and was multiplied by the density of aluminum to arrive at a linear attenuation number mu. The percent of transmitted flux was then calculated as
exp(-mu*thickness of material)
In addition to the attenuation due to the Dewar walls, I did two more calculations to get a ball park figure for how much attenuation I should expect from the superconducting solenoid magnet that will surround the sample and that is made from copper clad niobium alloy wire that is .045 cm thick. The first graph is for copper of this thickness and the second graph is for Niobium, (pictures 3 and 4).
The theory predicts x-rays of about 382 keV energy. If x-rays are produced at a much lower energy, say around 50 keV, it's conceivable that they could be missed with the detector outside the liquid helium Dewar.
References:
1. Aluminum attenuation
http://atom.kaeri.re.kr/cgi-bin/w3xcom?m=13
2. Copper attenuation
http://atom.kaeri.re.kr/cgi-bin/w3xcom?m=29
3. Niobium attenuation
http://atom.kaeri.re.kr/cgi-bin/w3xcom?m=41
4. x-ray attenuation documentation
http://en.wikibooks.org/wiki/Basic_Physics_of_Nuclear_Medicine/Attenuation_of_Gamma-Rays
exp(-mu*thickness of material)
In addition to the attenuation due to the Dewar walls, I did two more calculations to get a ball park figure for how much attenuation I should expect from the superconducting solenoid magnet that will surround the sample and that is made from copper clad niobium alloy wire that is .045 cm thick. The first graph is for copper of this thickness and the second graph is for Niobium, (pictures 3 and 4).
The theory predicts x-rays of about 382 keV energy. If x-rays are produced at a much lower energy, say around 50 keV, it's conceivable that they could be missed with the detector outside the liquid helium Dewar.
Any thoughts, pointers, or questions are always welcome!
References:
1. Aluminum attenuation
http://atom.kaeri.re.kr/cgi-bin/w3xcom?m=13
2. Copper attenuation
http://atom.kaeri.re.kr/cgi-bin/w3xcom?m=29
3. Niobium attenuation
http://atom.kaeri.re.kr/cgi-bin/w3xcom?m=41
4. x-ray attenuation documentation
http://en.wikibooks.org/wiki/Basic_Physics_of_Nuclear_Medicine/Attenuation_of_Gamma-Rays
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