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 produces [math]V_{i}[/math] neutrons, where [math]V_{i}[/math] is the distribution of the number of neutrons produced from an individual neutron-producing reaction. 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.

Probability of detecting pairs of neutrons in a single pulse

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 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}})\approx (1-8*0.005*0.25)^n_{total}\gt 0.95 \qquad \text{for } n_{total}\lt 7 [/math].

This is done by adding up the probabilities of every possible way for two neutrons to be detected by two distinct detectors out of the ten in the array, where each term is also multiplied by the probability that the rest of the neutrons in the pulse are not detected.