Theoretical analysis of 2n accidentals rates

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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] neutrons, where the random variable [math]V_{i}[/math] is the distribution of the number of neutrons produced from a single neutron-producing reaction. Each of the [math]V_i\text{'}s[/math] are independent and identically distributed random variables, the purpose of the subscript is to distinguish between the several neutron-producing interactions in a single pulse.

The beam has a Bremsstrahlung end point of 10.5 MeV, energetically allowing only two possible neutron-producing interactions, 1n-knochout and photofission. Thus, [math]V_{i}[/math] is the photofission neutron multiplicity, but with a larger [math]P(V_{i}=1)[/math] from 1n-knockout events. In other words, a 1n-knockout event and a photo-fission event emitting exactly one neutron are considered identically in the analysis. In viewing it this way, the analysis is simplified, but the end result is not changed since the distinction is not needed, in both cases a single neutron is emitted that is uncorrelated with all other neutrons.

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:

1. the neutron from the 1n-knockout, [math]V_{2}[/math], and
2. 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}}\gt 0.96 \qquad \text{for } n_{total}\lt 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]