Difference between revisions of "Fast neutron damage to HPGe Detector"

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A observable decrease in the energy resolution of a large HPGe 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 5*10^6 n/cm^2 is a conservative number to stay under.  
 
A observable decrease in the energy resolution of a large HPGe 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 5*10^6 n/cm^2 is a conservative number to stay under.  
  
The maximum neutron flux occurs right at the center of the detector, where the expression for integral flux is simply: <math>\Delta t n_{rate}\frac{1}{4\pi d^2}</math>.   
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The maximum neutron flux occurs right 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 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 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.  

Revision as of 06:41, 29 December 2016

A observable decrease in the energy resolution of a large HPGe 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 5*10^6 n/cm^2 is a conservative number to stay under.

The maximum neutron flux occurs right 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 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.

MaxHPGeCF252Time.png

Formula used in the graph above :

[math]\frac{5\times 10^6}{(19066 * 60^2 * 24 * \frac{1}{(4*\pi * d^2)})}[/math]


References

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