Difference between revisions of "Fast neutron damage to HPGe Detector"
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| − | An observable decrease in the energy resolution of | + | An observable decrease in the energy resolution of large HPGe detectors was first seen after the irradiation of 5*10^7 n/cm^2<ref>P. H. Stelson, J. K. Dickens, S. Raman, and R. C. Trammell, “Deterioration of Large Ge(Li) Diodes Caused by Fast Neutrons,” Nuclear Instruments and Methods 98,481 (1972).</ref>, so I consider a factor of ten below this to be a good conservative amount to make the maximum. |
The maximum neutron flux occurs exactly at the center of the detector, where the expression for integral flux is simply: <math>\Delta t\times n_{rate}\frac{1}{4\pi d^2}</math>. | The maximum neutron flux occurs exactly at the center of the detector, where the expression for integral flux is simply: <math>\Delta t\times n_{rate}\frac{1}{4\pi d^2}</math>. | ||
Revision as of 07:00, 29 December 2016
An observable decrease in the energy resolution of large HPGe detectors was first seen after the irradiation of 5*10^7 n/cm^2<ref>P. H. Stelson, J. K. Dickens, S. Raman, and R. C. Trammell, “Deterioration of Large Ge(Li) Diodes Caused by Fast Neutrons,” Nuclear Instruments and Methods 98,481 (1972).</ref>, so I consider a factor of ten below this to be a good conservative amount to make the maximum.
The maximum neutron flux occurs exactly at the center of the detector, where the expression for integral flux is simply: .
The number of days it would take to reach an integral flux of 5*10^6 n/cm^2, as a function of the distance from source to HPGe face is shown below. The graph assumes the activity of the Cf-252 source as of 01/2017, which is 19,066 n/s.
- The formula used in the graph above is,
References
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