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 ten below that value to be the maximum allowable neutron irradiation.  
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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 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.   
  
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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 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).
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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]]
  
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*The formula used in the graph above is,
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*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

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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 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 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 here for discussion of source rates).


MaxHPGeCF252Time.png

  • 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]


References

<references />

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