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

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 of detecting a particular pair of neutrons: the neutron from the 1n-knockout, [math]V_{2}[/math], and a given neutron from the photofission, [math]V_{3}[/math]. 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{\Sigma_D}[/math] as the event that all other neutrons are not detected. Using standard notation, the probability can be written:

[math]P(D_1,D_2,\overline{\Sigma_D},N=3,V_1=1,V_2=1,V_3=4)[/math]

,which is interpreted as the AND between all events, each of which are separated by a comma.