# Calculating Light Output for an Organic Scintillator

July to
August 2016
Laboratory for Laser Energetics

A fusion reaction, shot 81371, in the OMEGA-60 at the Laboratory for Laser Energetics was observed using an organic liquid scintillator 13.4 m from the reaction. High energy neutrons were produced by the fusion of deuterons with tritons (DT) as well as deuterons with deuterons (DD). DT neutrons have a kinetic energy of 14.1 MeV and DD neutrons have a kinetic energy of 2.8 MeV. Neutrons from these reactions were directly incident on the scintillation detector.

Neutrons were also incident on a vessel of deuterated hexane, 9 cm from the fusion reaction. Some of these neutrons scattered from the vessel and were measured in the scintillator. Those that elastically scattered from deuterium (nD) had an energy of 1.55 MeV and those that elastically scattered from carbon (nC) had an energy of 10.14 MeV. Neutrons also inelastically scattered from carbon, exciting the nucleus, causing it to release neutrons (nC$$^{*}$$) at 8.11 MeV, a fraction of which were measured by the scintillation detector.

DT Neutrons Incident on Scintillator from Vessel
Of the neutrons produced by DT fusion, $$N_{total}$$, a fraction were incident on the scintillator, $$N_{incident}$$, after scattering from a vessel positioned opposite the scintillator. A simple calculation was done to determine $$N_{incident}$$. Neutrons from the reaction scattered off a vessel of deuterated hexane (D$$_6$$C$$_6$$), 9 cm from the fusion reaction to reach the scintillation detector 13.4 m from the reaction at 180$$^{\circ}$$.

Scintillator Incident Neutron Yield
The number of neutrons incident on the scintillator that scattered from the vessel was calculated as $$N_{incident} = \Omega_{vessel}*L*N_{DT},$$where $$\Omega$$ is the solid angle of the vessel from the fusion reaction and $$L$$ is the fraction of neutrons incident on the vessel that scattered 180$$^{\circ}$$ from their original trajectory.
Although the actual vessel was conical with a 0.5 cm radius face towards the fusion reaction and a depth of 8.25 cm, it was modeled as a cylinder at 12 cm from the reaction with a face of 1 cm in radius and the same depth as the conical vessel to produce a simplified approximation of the geometry. The solid angle of the vessel from the fusion reaction was calculated as$$\Omega_{vessel} = \frac{\pi r^2}{4\pi R^2} ,$$where r was the radius of the cylindrical vessel and R was the distance from the center of the fusion reaction to the face of the vessel. Both values were known from the experimental apparatus, $$r=1$$ cm and $$R=12$$ cm. As such, the solid angle was calculated to be $$\Omega = 0.0017$$.

The fraction of neutrons incident on the vessel that scattered 180$$^{\circ}$$, $$L$$, was calculated as $$L = 1-e^{-\mu x} ,$$where $$\mu$$ is the attenuation coefficient and $$x$$ is the height of the cylindrical vessel. The attenuation coefficient is $$\mu = \frac{N_A\sigma_{(nX)}\rho}{m_{X}} ,$$where $$N_A$$ is Avogadro's number, $$\sigma_{(nX)}$$ is the cross section of a neutron of 2.56 MeV on deuterium or carbon, $$\rho$$ is the density of the deuterated hexane, and $$m_{X}$$ is the molar mass of deuterium or carbon. The attenuation coefficient for neutrons from the DD reaction was calculated to be $$\mu_{deuterium}=\dfrac{6.022*10^{23}\textrm{ mol}^{-1}*0.6388\textrm{ barns}*0.95\frac{\textrm{g}}{\textrm{mL}}}{2.014\frac{\textrm{g}}{\textrm{mol}}}=0.0396 \textrm{ cm}^{-1}$$for neutrons incident on deuterium, and$$\mu_{carbon}=\dfrac{6.022*10^{23}\textrm{ mol}^{-1}*0.8314\textrm{ barns}*0.95\frac{\textrm{g}}{\textrm{mL}}}{12.01\frac{\textrm{g}}{\textrm{mol}}}=0.1815 \textrm{ cm}^{-1}$$for neutrons incident on carbon.

By incorporating the depth of the vessel as $$x=8.25$$ cm, the fraction of neutrons that scattered was evaluated to be $$L_{deuterium} = 0.2787$$and $$L_{carbon} = 0.7762,$$respectively. Thus, the number of neutrons incident on the scintillator that scattered from the vessel was calculated using Equation 1. With a DT neutron yield of $$N_{total}=6*10^{12}$$, the neutron yield incident on the scintillator after scattering from deuterium was evaluated to be $$N(nD)=0.0017*0.2787*6*10^{12} \textrm{ neutrons}=2.842*10^{9}\textrm{ neutrons}.$$For DT neutrons that scattered from carbon in the vessel, the yield incident on the scintillator was evaluated to be $$N(nC)=0.0017*0.7762*6*10^{12} \textrm{ neutrons}=7.917*10^{9}\textrm{ neutrons}$$

Energy of nD and nC
Neutrons produced by DT fusion that scattered elastically from the vessel, lost energy when they collided with the nuclei of the deuterium and carbon. The energy of the products of the elastic scattering were calculated non-relativistically by assuming that the neutrons collided head-on with stationary nuclei. The velocity of the 14.1 MeV neutron was calculated classically and found to be slow enough that classical calculations would suffice. $$v_{in}=\sqrt{\dfrac{2*14.1\textrm{ MeV}}{939.56\frac{\textrm{MeV}}{\textrm{c}^2}}}=0.17325\textrm{ c}.$$From the classical equations of momentum and energy conservation during elastic scattering, the following relationship between the initial velocity of the neutron and final velocity of the target, $$v_{fX}$$, was derived, $$v_{fX}=\dfrac{2v_{in}}{\frac{m_X}{m_n}+1},$$where $$m_X$$ is the mass of the target and $$m_n$$ is the mass of the neutron. The final velocity of the target was related back to the final velocity of the neutron using classical momentum, $$v_{fn}=v_{in}-\dfrac{m_X}{m_n}\dfrac{2v_{in}}{\frac{m_X}{m_n}+1}$$and subsequently to the final energy of the neutron, $$E_{n}=\dfrac{1}{2}m_nv_{in}^2(1-\dfrac{2m_X}{m_X+m_n})^2.$$The kinetic energy of the neutrons that scattered from deuterium was evaluated to be $$E_{nD}=1.5532$$ MeV; those that scattered from carbon were evaluated to be $$E_{nC}=10.1377$$ MeV.

Neutrons from Excited Carbon Nuclei
Neutrons from the DT reaction that were incident on carbon nuclei in the vessel were able to inelastically scatter from the carbon. This scattering raised the carbon nuclei to its first excited state. De-excitation of the carbon nuclei results in the emission of a neutron with a kinetic energy of 8 MeV. The yield of neutrons from these excited nuclei was calculated by $$N_{^*carbon} = \Omega_{vessel}*M*N_{DT} ,$$where $$M$$ is $$M = e^{-\mu x}.$$The attenuation coefficient, $$\mu$$, was determined using the cross section of 0.775 b. It was evaluated to be $$\mu_{^*carbon}=\dfrac{6.022*10^{23}\textrm{ mol}^{-1}*0.775\textrm{ barns}*0.95\frac{\textrm{g}}{\textrm{mL}}}{12.01\frac{\textrm{g}}{\textrm{mol}}}=0.03653 \textrm{ cm}^{-1}.$$As such, $$M_{^*carbon}$$ was evaluated as to be 0.7519 and $$N_{^*carbon} = 0.0017*0.7519*6*10^{12} \textrm{ neutrons} = 7.669*10^{9} \textrm{ neutrons}.$$The number of these neutrons, emitted by de-exciting carbon nuclei that were incident on the scintillation detector was calculated using the solid angle of the detector from the vessel, $$\Omega_{scint} = \dfrac{\pi (20 \textrm{ cm})^2}{4\pi (1349 \textrm{ cm})^2} = 5.495*10^{-5}.$$Using this, the number of neutrons from de-excitation of carbon that were incident on the scintillator was evaluated to be $$N_{nC^*} = 4.2212*10^5$$ neutrons.

Photocathode Charge Measurement
Neutrons were incident on a liquid organic scintillator. The energy of the photons produced from scintillations with the energy of the incident neutrons. These photons were measured by four photomultiplier tubes (PMTs) - labeled A, B, C, and D - which each produced a total electric charge that was proportional to the yield of neutrons.

The PMTs were all set at different levels of amplification. PMT A was used to measure the DT peak while PMT D was used to measure the DD and nD peaks. PMT A was normalized to PMT D by comparing DT peaks measured by each during shot 82071. The peak of neutrons that elastically scattered from carbon, nC, was too low to be observed by this method.

Fitting and Integrating Charge for DT Neutrons
The peak of neutrons measured by the DT reaction had no significant background noise because of its high yield signal and the low amplification of PMT A. It was fit using an Exponentially Modified Gaussian (EMG) in time-space, $$EMG(t) = \dfrac{A\lambda}{2}*\textrm{exp}(\dfrac{\lambda}{2}(2\mu+\lambda\sigma^2 -2t))*\textrm{erfc}(\dfrac{\mu+\lambda\sigma^2 -t}{\sigma\sqrt{2}})$$where  $$A$$ was a fitting constant, $$\lambda$$ was the inverse relaxation time of the exponential, $$\mu$$ was the mean of the Gaussian, and $$\sigma$$ was the standard deviation of the Gaussian. The equation was fit to data using a method of least squares. Both the peak of DT neutrons and the fit are plotted in Figure 1.

Figure 1. Voltage of the DT peak with respect to time. The blue points are data from PMT A, normalized to PMT D. The green curve is the EMG fit to this data.

Removing Background for nD and DD peaks
Before the nD and DD peaks were integrated to produce a measurement of charge, the background was removed by fitting the peak and background using two Exponentially Modified Gaussians (EMG) plus an exponential decay in time space, $$EMG_i(t) = \dfrac{A\lambda_i}{2}*\textrm{exp}(\dfrac{\lambda_i}{2}(2\mu_i +\lambda_i\sigma_i^2 -2t))*\textrm{erfc}(\dfrac{\mu_i +\lambda_i\sigma_i^2 -t}{\sigma_i\sqrt{2}})$$
$$Y(t) = EMG_1(t) +EMG_2(t) +B\textrm{exp}(-Ct)$$where $$A$$, $$B$$, and $$C$$ were fitting constants, $$\lambda_i$$ was the inverse relaxation time of the exponential, $$\mu$$ was the mean of the Gaussian, and $$\sigma_i$$ was the standard deviation of the Gaussian. The exponential term of Equation 10 was then subtracted from the data, removing the background from the peaks. The nD and DD peaks, before and after background removal, are shown in Figure 2.

Figure 2. A plot of voltage imparted to PMT D with respect to time. The dark blue points are raw data from the PMT, the red curve is a plot of Equation 10 fit to the data, the green curve is the exponential term of Equation 10 that was subtracted to give the light blue points which are data with the background removed.

The background for the nC$$^*$$ and DD Peaks was removed using the same method. Equation 10 was fit to the data and the exponential term was subtracted from the peaks. This is illustrated in Figure 3.

Figure 3. A plot of voltage imparted to PMT D with respect to time. The dark blue points are raw data from the PMT, the green curve is a plot of Equation 10 fit to the data, the orange curve is the exponential term of Equation 10 that was subtracted to give the light blue points which are data with the background removed.

As was done to the DT peak, an EMG was fit to the nC$$^*$$, DD, and nD peaks using the method of least squares. These fits are shown in Figure 4.

Figure 4. Plots of voltage with respect to time. Left, the nC$^*$ peak where blue points are data with the background removed and the green curve is the EMG fit to data. Right, the DD and nD peaks where blue and green points are data with the background removed. The red curve is the EMG fit to the DD peak and the green curve is the EMG fit to the nD peak.

Measuring Charge from Neutron Peaks
The EMG for each peak, once fit to data, was integrated and divided by the resistance of the circuit to give the total charge imparted to PMT A, normalized to PMT D, by photons released in response to incident neutrons. The charges for each peak are listed in Table 1.
\begin{center}
Table 1. Charge calculated for each peak of neutrons.
$$\begin{tabular}{l l} \\Peak & Charge (pC) \\ DT 14.1 MeV & 804006.50 \\ nC^* 8 MeV & 187.65 \\ DD 2.8 MeV & 465.75 \\ nD 1.56 Mev & 15.77 \end{tabular}$$

Light Output
The amount of light produced by scintillation, $$H$$, is related to the energy of the incident neutron, $$E_n$$, nonlinearly as $$H = kE^\alpha,$$with $$k$$ and $$\alpha$$ being positive constants [1]. This function was fit to data by relating the charges listed in Table 1 to the light output by the scintillator as $$H = \dfrac{C_{tot}}{N(E)*\textrm{Yield}},$$where $$C_{tot}$$ is the total charge imparted to the PMTs, normalized to PMT D, Yield is the number of neutrons at that energy that are incident on the scintillator, and $$N(E)$$ is the amount of energy deposited in the scintillator per neutron as a function of energy. The function $$N(E)$$ was calculated by C. Forrest using MCNP to be $$N(E) = 1.36*E^{-0.96} +0.24$$ in units of MeV per neutron [2]. The light output for each neutron peak, found using Equation 10 is listed in Table 2.
$$Table 2. Experimental values for variables in Equation 10 which give the total light output for the peak. \\ \begin{tabular}{l l l l l} \\Peak & C_{tot} (pC) & N(E) (\frac{\textrm{MeV}}{\textrm{n}}) & Yield (n) & H (\frac{\textrm{pC}}{\textrm{MeV}})\\ DT 14.1 MeV & 804006.50 & 0.3472 & 5.43*10^{13} & 4.264*10^{-8} \\ nC^* 8 MeV & 187.65 & 0.4247 & 4.22*10^5 & 0.0010 \\ DD 2.8 MeV & 465.75 & 0.7461 & 1.37*10^{11} & 4.56*10^{-9} \\ nD 1.56 MeV & 15.77 & 1.1274 & 7.917*10^{9} & 1.77*10^{-9} \end{tabular}$$

Equation 9 was fit to the data, excluding the nC$$^*$$ peak using the method of least squares, giving $$k = 0.1051$$ and $$\alpha = 1.3993$$. The data and the fit are plotted in Figure 3.

Figure 4. Light output of the scintillation fluid as a function of the energy of incident neutrons.

As the light output is disproportionately high for the nC$^*$ peak when compared to the other peaks, it is thought that the yield of neutron from the excited carbon was not calculated correctly. Assuming that the integrated charge for the nC$$^*$$ is correct, for the light output to correspond with other measurements, the yield of neutron from the excited carbon must be $$2.354*10^{10}$$ neutrons. Figure 5 is a plot of light output with the neutron yield adjusted in this way.

Figure 4. Light output of the scintillation fluid as a function of the energy of incident neutrons.

References
1. Glenn F. Knoll, Radiation Detection and Measurement, 3$$^{\textrm{rd}}$$ ed. (Wiley, New York, 2000) p. 560.
2. C. Forrest, Measurements of the Fuel Distribution in Cryogenic D-T Direct-Drive Implosions, Thesis, University of Rochester, 2015.

Presentation