Difference between revisions of "Fast neutron damage to HPGe Detector"
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| − | 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>. I choose a factor 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>. I choose a factor of 50 below that value to be the maximum allowable neutron irradiation. |
The maximum neutron flux from a point source will occur exactly at the center of the detector face, where the expression for integral flux over a period <math>\Delta t</math> is simply: <math>\Delta t\times n_{rate}\times \frac{1}{4\pi d^2}</math>, where <math>n_{rate}</math> is the neutron rate of the source. | The maximum neutron flux from a point source will occur exactly at the center of the detector face, where the expression for integral flux over a period <math>\Delta t</math> is simply: <math>\Delta t\times n_{rate}\times \frac{1}{4\pi d^2}</math>, where <math>n_{rate}</math> is the neutron rate of the source. | ||
| − | The number of days it would take to reach an integral flux of | + | The number of days it would take to reach an integral flux of 10^6 n/cm^2, as a function of the distance from source to HPGe face is shown below. The calculation uses a neutron rate of 19,066 <math>\pm</math> 300 n/s, which was the neutron rate of the CF-252 source on 01/2017 (see [[Cf-252 FTC-CFZ-431| here]] for discussion of source rates). |
[[File:MaxHPGeCF252Time.png|700px]] | [[File:MaxHPGeCF252Time.png|700px]] | ||
| − | *The formula used in the graph above is, | + | *The formula used in the graph above is (black solid line), |
<math>y=\frac{5\times 10^6}{(19066 * 60^2 * 24 * \frac{1}{(4*\pi * x^2)})}</math> | <math>y=\frac{5\times 10^6}{(19066 * 60^2 * 24 * \frac{1}{(4*\pi * x^2)})}</math> | ||
Latest revision as of 22:32, 30 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>. I choose a factor of 50 below that value to be the maximum allowable neutron irradiation.
The maximum neutron flux from a point source will occur exactly at the center of the detector face, where the expression for integral flux over a period is simply: , where is the neutron rate of the source.
The number of days it would take to reach an integral flux of 10^6 n/cm^2, as a function of the distance from source to HPGe face is shown below. The calculation uses a neutron rate of 19,066 300 n/s, which was the neutron rate of the CF-252 source on 01/2017 (see here for discussion of source rates).
- The formula used in the graph above is (black solid line),
References
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