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

From New IAC Wiki
Jump to navigation Jump to search
 
(24 intermediate revisions by the same user not shown)
Line 1: Line 1:
−
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.
+
[[2nCor 44 | Go back]]
  
−
The maximum flux occurs right at the center of the detector, where the expression for flux simplifies to <math>n_{rate}\frac{1}{4\pi d^2}</math>.  The number of days it would take to reach an integral flux of <math>5\times 10^6 n/cm^2 </math> as a function of the CF-252 source distance HPGe face is shown below. The graph assumes the activity of the Cf252 source on 01/2017, which is 19066 n/s.
 
  
 +
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 [[Cf-252 FTC-CFZ-431| here]] for discussion of source rates).
 +
 +
 +
[[File:MaxHPGeCF252Time.png|700px]]
 +
 +
*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 ==
 
<references />
 
<references />

Latest revision as of 22:32, 30 December 2016

Go back


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 />

Retrieved from "https://wiki.iac.isu.edu/index.php?title=Fast_neutron_damage_to_HPGe_Detector&oldid=111027"