Fast neutron damage to HPGe Detector - Revision history

Burgjeff at 22:32, 30 December 2016

2016-12-30T22:32:59Z

Burgjeff at 22:11, 30 December 2016

2016-12-30T22:11:13Z

Burgjeff at 21:55, 29 December 2016

2016-12-29T21:55:14Z

Burgjeff at 21:54, 29 December 2016

2016-12-29T21:54:51Z

Burgjeff at 09:10, 29 December 2016

2016-12-29T09:10:40Z

Burgjeff at 09:10, 29 December 2016

2016-12-29T09:10:27Z

Burgjeff at 09:10, 29 December 2016

2016-12-29T09:10:06Z

Burgjeff at 09:09, 29 December 2016

2016-12-29T09:09:27Z

Burgjeff at 09:07, 29 December 2016

2016-12-29T09:07:14Z

Burgjeff at 07:00, 29 December 2016

2016-12-29T07:00:11Z

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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.
+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 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).
+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,
+*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>

← Older revision		Revision as of 22:11, 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 ten below that value to be the maximum allowable neutron irradiation.		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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← Older revision		Revision as of 21:54, 29 December 2016
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		+	[[12-23-2016]]
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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 ten below that value to be the maximum allowable neutron irradiation.		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 ten 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}\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 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).

← Older revision		Revision as of 09:10, 29 December 2016
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	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}\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}\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 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, or the neutron rate of the CF-252 source on 01/2017 (see [[Cf-252 FTC-CFZ-431\| here]] for discussion of source rates).	+	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).

← Older revision		Revision as of 09:09, 29 December 2016
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	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}\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}\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 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 was~~ 19,066 <math>\pm</math> 300 n/s (see [[Cf-252 FTC-CFZ-431\| here]] for discussion of source rates).	+	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, or 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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-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.
+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.
-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 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}\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 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 was 19,066 <math>\pm</math> 300 n/s (see [[Cf-252 FTC-CFZ-431| here]] for discussion of source rates).
 [[File:MaxHPGeCF252Time.png|700px]]

← Older revision		Revision as of 07:00, 29 December 2016
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−	An 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 I consider a factor of ten below this to be a good conservative amount to make the maximum.	+	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>.