# Answer to Question #9869 Submitted to "Ask the Experts"

*Category: Radiation Basics — Neutrons*

The following question was answered by an expert in the appropriate field:

We use a Pu-Be neutron source for irradiating thermoluminescent material.

The activity of the neutron source is 4.6 E6 neutron/s, the thickness of the surrounding water is 20 cm, and the energy is 12 MeV.

Please help me calculate the radiation dose received by the material.

While I can provide you with an estimative method for approximating the soft-tissue dose at the presumed location of the thermoluminescent material, which I assume is the intended target, the approximation should not be used for purposes of calibration or as a tool for delivering known neutron doses to the material. The more accurate calculations required to establish the doses to the level of accuracy required for calibration or dosimetry testing (e.g., for dosimeter accreditation purposes) would require much more information (e.g., source construction and dimensions, exact placement location of the source in the water, geometry of the water containment, and characteristics of the water-containment vessel). With the additional information, Monte Carlo simulations could be run to determine expected dose rates, but such are beyond our discussion here.

For estimative purposes, we may note that the neutrons produced by the (n,a) reaction on beryllium are fast neutrons with an average energy of about 4.5 MeV. At this energy, most of the interactions in water are by elastic scattering with hydrogen, and the mass interaction cross section for hydrogen in water, based on the hydrogen content of water (0.111 g H per g water) and the cross section for hydrogen (1.75 x 10^{-24} cm^{2} per atom) can be calculated as 0.117 cm^{2}g^{-1}, and for unit density the linear macroscopic cross section may be estimated as 0.117 cm^{-1}. Assuming a single interaction will remove a neutron from further consideration, we may then estimate the transmission factor of fast neutrons through 20 cm of water as

T = e^{-(0.117)(20)} = 0.0963.

The unattenuated fast neutron fluence rate at 20 cm from the source may be estimated, assuming the source may be treated as a point isotropic source, as

F_{unatt} = 4.6x10^{6} s^{-1}/(4p(20 cm)^{2}) = 9.15x10^{2} cm^{-2} s^{-1}.

The attenuated fluence rate of fast neutrons would then be

F_{att} = (0.0963)(9.15x10^{2} cm^{-2 }s^{-1}) = 88.1 cm^{-2 }s^{-1}.

If we use conversions consistent with current U.S. Nuclear Regulatory Commission (NRC) regulations the fluence rate to yield an equivalent dose rate of 10 µSv h^{-1} (1 mrem h^{-1}) is about 6.4 cm^{-2 }s^{-1}. The above attenuated fluence rate would then yield a fast neutron dose rate of

H = (88.1 cm^{-2 }s^{-1})(10 µSv h^{-1}/6.4 cm^{-2} s^{-1}) = 138 µSv h^{-1} (13.8 mrem h^{-1}).

If current recommendations of the International Commission on Radiological Protection (ICRP) (ICRP Publication 103) are used, the radiation weighting factor at 4.5 MeV would be somewhat larger than that used in NRC regulations (about 13 compared to 8), and this would lead to an equivalent dose rate of about 225 µSv h^{-1}.

The above calculations estimate the fast neutron dose rate, treating all neutrons as if they had the average neutron energy. Accounting for the actual spectrum of neutrons would produce some change in the result. Additionally, we assumed that a single interaction with hydrogen removed the neutron from further consideration. In actuality, some neutrons will interact with oxygen and some that scatter elastically with hydrogen may undergo additional interactions and some may escape to the dose point. There is also a small component of dose associated with thermal neutrons that could contribute a few tens of µSv h^{-1}. There is also a small gamma dose component associated particularly with the capture of thermal neutrons by hydrogen in which 2.2 MeV gamma rays are produced.

As has been stated, the approach used here is approximate and does not account for all that is transpiring to contribute to dose at the location of interest. If you require more accurate estimations you will have to use more sophisticated methods. I wish you well in your continuing work.

George Chabot, PhD, CHP