Difference between revisions of "Theoretical analysis of 2n accidentals rates"

Revision as of 03:06, 20 January 2018

Introduction

A given photon pulse may cause multiple neutron-producing reactions, ranging from zero to "infinity" reactions. The number of neutron-producing reactions in a pulse is hereafter denoted by [math]N[/math]. Being the number of neutron-producing reactions actually occurring per pulse, [math]N[/math] is assumed to follow the Poissonian distribution as a limiting case of the binomial distribution. Each neutron-producing interaction is said to produce [math]V_{i}[/math] correlated neutrons, where the random variable [math]V_{i}[/math] is the distribution of the number of neutrons produced in a single neutron-producing reaction. Each of the [math]V_i\text{'}s[/math] are independent and identically distributed random variables, so the purpose of the subscript is to distinguish between several distinct neutron-producing interactions which may occur in a single pulse.

The beam has a Bremsstrahlung end point of 10.5 MeV, which energetically allows for only two possible neutron-producing interactions, 1n-knochout and photofission. Thus, [math]V_{i}[/math] is equal to the photofission neutron multiplicity plus a contribution at [math]V_{i}=1[/math] from 1n-knockout events. The analysis that follows does not need to distinguish between 1n-knockout events and photofission events that emit a single neutron. In both cases, a single neutron is emitted and is uncorrelated with all and any other neutrons.

@@ Line 6: / Line 6: @@
 ==Probability of detecting a given pair of neutrons in a single pulse==
-=Old, needed to be simplified=
-Consider a pulse that causes three neutron producing reactions, two 1n-knockouts and a photofission event with multiplicity of 4. In terms of the notation, <math>N=3</math>, <math>V_{1}=1</math>, <math>V_{2}=1</math>, and <math>V_{3}=4</math>. Now let's find the probability, <math>p_{acc}</math>, of detecting only and exactly the following pair of neutrons:
-# the neutron from the 1n-knockout, <math>V_{2}</math>, and
-# A given neutron from the photofission event, <math>V_{3}</math>, of which there are four neutrons to choose from, but here we consider the detection of a single particular neutron out of the four.
-This example is an accidental, since each neutron is created in a separate reaction. Let <math>D1</math> represent the event where the neutron from the 1n-knockout in <math>V_{2}</math> is detected, and <math>D2</math> for the detection of the photofission neutron. Define <math>\overline{D_{others}}</math> as the event that all other neutrons are ''not'' detected. Using standard notation, the probability can be written:
-::<math>p_{acc} = P\left(D_1,D_2,\overline{D_{others}},N=3,V_1=1,V_2=1,V_3=4\right)</math>
-,which is interpreted as the Boolean product (AND) between all events separated by the commas, which are used to enhance readability.
-By an application of the general product rule of probability (also called the chain rule), <math>p_{acc}</math> can be factored as follows:
-::<math>p_{acc} = P\left(D_1,D_2,\overline{D_{others}}\big| N=3,V_1=1,V_2=1,V_3=4\right)\times P(N=3,V_1=1,V_2=1,V_3=4)</math>
-where the factor on the left is a conditional probability which may be read:
-:"The probability of the detection of both and only the two neutrons under consideration, given that two 1n-knochouts and a single photofission event, with a multiplicity of four, occurred during this pulse."
-The factor on the right can be simplified into four factors by recognizing that the events are all mutually independent, since each photo-nuclear interaction is assumed to be uncorrelated. This gives
-::<math>p_{acc} = P\left(D_1,D_2,\overline{D_{others}}\big| N=3,V_1=1,V_2=1,V_3=4\right)\times P(N=3)P(V_1=1)P(V_2=1)P(V_3=4)</math>
-In order to save space, the event <math>(N=3,V_1=1,V_2=1,V_3=4)</math> will hereafter be denoted as <math>E^{3}_{1,1,4}</math>.
-Now to tackle the factor on the left, <math>P(D_1,D_2,\overline{D_{others}}|E^{3}_{1,1,4})</math>. Application of the product rule gives:
-::<math>P(D_1,D_2,\overline{D_{others}}\big|E^{3}_{1,1,4}) =P(D_1,D_2 \big|,\overline{D_{others}},E^{3}_{1,1,4})P(\overline{D_{others}}) </math>
-With ten detectors in total, each covering a solid angle of 0.5% of <math>4\pi</math>, and having an intrinsic efficiency of less than 25%, <math>P(\overline{D_{others}})</math> may be approximated by 1, or quantitatively:
-::<math>P(\overline{D_{others}})\approx (1-8*0.005*0.20)^{n_{total}}>0.96 \qquad  \text{for } n_{total}<7 </math>
-::thus,
-::<math>P(\overline{D_{others}})\approx 1</math>
-:where,
-:<math>8</math> is the number of detectors available.
-:<math>n_{total}</math> is the total number neutrons emitted.
-Seven neutrons emitted in a single pulse would be an extreme case under the conditions of the experiment, and so  <math>P(\overline{D_{others}})</math> may be set to unity as a approximation.
-The expression for the probability of detecting our two neutrons becomes:
-::<math>p_{acc} = P(D_1,D_2,|E^{3}_{1,1,4})\times P(E^{3}_{1,1,4})</math>
-Since we are looking at accidentals only, <math>D_1</math> and <math>D_2</math> are uncorrelated, and thus their only dependence on <math>E^{3}_{1,1,4}</math> is the existence of the neutrons which <math>D_1</math> and <math>D_2</math> are referring to. There's no use in picking either <math>D_1</math> or <math>D_2</math> which refer to neutrons that were never created, so the product, <math>D_1D_2</math>, may be considered independent of <math>E^{3}_{1,1,4}</math>.
-This brings us to an important point in the analysis, where <math>p_{acc}</math> is evidently a product of mutually independent variables:
-::<math>p_{acc} = P(E^{3}_{1,1,4})\times P(D_1D_2) = P(N=3)P(V_1=1)P(V_2=1)P(V_3=4)P(D_1D_2)</math>
-==probability of detecting any pair of accidentals in a given pulse==
-The above result for <math>p_{acc}</math> would be the same had <math>D_1</math> and <math>D_2</math> had been chosen to refer to different neutrons. In the specific example above, there are 9 possible pairs of accidentals, the four fission neutrons paired with each 1n-knockout neutron, plus the case of the two knockout neutrons pairs with each other. Thus, the probability of detecting ''any'' accidentall:
-::<math>P(A) = 9*P(N=3)P(V_1=1)P(V_2=1)P(V_3=4)P(D_1D_2)</math>
-The number of pairs of accidentals is, in general:
-::<math>\sum V_iV_j \qquad \text{for every} \{V_i,V_j\} \text{that is a subset of } \{V_1,V_2,...V_N\}</math>

Difference between revisions of "Theoretical analysis of 2n accidentals rates"

Revision as of 03:06, 20 January 2018

Introduction

Probability of detecting a given pair of neutrons in a single pulse

Navigation menu

Search