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 actually occurring in a given pulse is denoted by the random variable [math]n[/math], and is assumed to follow the Poissonian 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, since in both cases, a single neutron is emitted that is uncorrelated with all other neutrons.

Variable reference

variable	Description
[math]n[/math]	A random variable for the number of neutron producing reactions occurring in a single pulse.
[math]v[/math]	A random variable for the number of correlated neutrons produced by a single neutron-producing reaction in a given pulse.
[math]v_i[/math]	A random variable with the same distribution as [math]v[/math]. The index only distinguishes between distinct and independent instances of [math]v[/math].
[math]\lambda[/math]	Poissonian mean for the number of neutron-producing interactions per pulse.

Section Title

The rate of detected accidentals can be estimated from the two-neutron "coincidence" rate of different pulse (DP) events. Under suitable conditions, the rate of accidentals is very close to 1/2 of two-neutron rate in DP events.

During the derivation of this result, a pulse is considered in which there are [math]n[/math] neutron-producing interactions, where each individual reaction produces [math]v_i[/math] correlated neutrons, where [math]i[/math] ranges from 1 to [math]n[/math]. This event is denoted as, [math]v_1...v_n[/math], and its probability by the expression, [math]p(v_1...v_n)[/math].

Assumptions

There are two approximations which are required to make this result valid.

1. Poissonian statistics

The number of neutron-producing reactions occurring per pulse, [math]n[/math], follows the Poissonian distribution. This can be considered as an approximation to the binomial distribution, where the number of independent trials is much greater than the typical number of successes. Each electron can constitute an independent binomial trial, with success being that the electron produces a Bremsstrahlung photon, which also makes it's may to the target and induces a neutron-producing reaction. Out of trillions of electrons per pulse, there were on the order of one photo-nuclear reactions per pulse.

2. Small detection probability

The probability of detecting two specific uncorrelated neutrons in a given pulse, denoted by [math]p(d_{n_1}d_{n_2})[/math], is independent of how many other neutrons are emitted in that same pulse. These two neutrons could be among, say, 8 other neutrons which were emitted in the same pulse, and [math]d_{n_1}d_{n_2}[/math] would refer to the event that both (and only both) neutrons are detected.

This is only an approximation, however, since each detector can register at most one hit per pulse, and so the effective efficiency of the entire array drops as the number of particles emitted increases. In other words, it is being assumed that neutrons do not "compete" against one another for a chance to be detected. This approximation is justified because each detector covers only 0.5% out of [math]4\pi[/math] sr, and the rate of detected two-neutron coincidences per pulse is on the order of 10E-5, and no triple neutron events were recorded. I am making a point to address this because it accounts for the fact that the SP/DP accidental ratio is significantly greater than 1/2 for photons, all of which are accidentals.

This assumption can be expressed mathematically as:

[math]p(d_{n_1}d_{n_2}|v_1...v_n) = p(d_{n_1}d_{n_2})[/math]

As a result, to find the probability of detecting an accidental in a given pulse, simply count all possible accidental pairs in that pulse and multiply the result by [math]p(d_{n_1}d_{n_2})[/math]:

[math]p(SP_{acc}) = M_{SP}*p(d_{n_1}d_{n_2})[/math]

where,

[math]SP_{acc}[/math] is the event that an accidental is detected.

[math]M_{SP}[/math] is the number of possible distinct pairs of two-neutron accidentals in a given pulse.

Combinatorics

Same pulse

The total number of accidentals in a given single pulse can be calculated directly

[math]M_{SP} = \prod_{i\neq j}v_iv_j[/math]

For example, consider a pulse with three fission events producing sets of 2, 4, and 3 correlated neutrons. The number of all possible uncorrelated pairs is

[math]2*4 + 2*3 + 3*2 = 20[/math]

If there were 5 fissions producing 2, 5, 2, 1 and 2 neutrons, then we have:

[math] 2*5 + 2*2 + 2*1 + 5*2 + 5*1 + 2*1 = 33[/math]

The most important point to note here is that the number of terms in the sequence, [math]\prod_{i\neq j}v_iv_j[/math], depends only on the number of reactions, [math]n[/math]:

[math]T_{SP} = \frac{1}{2}n(n-1)[/math]

where [math]T_{SP}[/math] is the number of terms.

The importance of [math]T_{SP}[/math], is that when taking the expectation value of this series, the result can be expressed in terms of only [math]\bar{T_{SP}}[/math] and [math]\bar{v}[/math], where the bar denotes the expected value.

The expected number of accidentals per pulse, [math]M_{SP}[/math], is an expectation of a product of independent random variables, and by the linearity of expectation, this becomes the product of expectation values:

[math]E(M_{SP}|n) = E(\prod_{i\neq j}v_iv_j|n) = T_{SP}\bar{v}^2 = \frac{1}{2}n(n-1)\bar{v}^2[/math]

The expression [math]E(M_{SP}|n)[/math] is a conditional expectation, meaning that n is fixed for the calculation. Now taking another expectation, but this time w.r.t. n:

[math]E(M_{SP}) = E \left (\frac{1}{2}n(n-1)\bar{v}^2 \right ) = \frac{1}{2} \left (E(n^2)- E(n)\right ) *\bar{v}^2[/math]

[math]=\frac{1}{2} \lambda^2 *\bar{v}^2[/math]

Different pulses

When forming pairs of uncorrelated neutrons across separate pulses, the total number of pairs is given by:

[math]M_{DP} = (\sum_{i=1}^{n_1}v_i)(\sum_{j=1}^{n_2}v_j)[/math]

Here, [math] T_{DP}[/math], or the total number of terms of the form [math]v_iv_j[/math] after expanding, is the product of the number of reactions in each pulse:

[math] T_{DP} = n_1 * n_2 [/math]

[math] E(M_{DP}|n_1,n_2) = n_1*n_2 *\bar{v}^2\lt math\gt Since the n's of two separate pulse are independent, the final expected number of two-neutron pair available for detection when looking at DP events is: \lt math\gt E(M_{DP}) = \lambda^2 \bar{v}^2[/math]