Question 2 of 7: Cyclotron Vault Mixed-Field Dose Rates — Sv to Gy

Question 2: Cyclotron Vault Mixed-Field Dose Rates — Sv to Gy (9 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Measured equivalent dose rates outside the vault shield: gamma 5 μSv/h, thermal neutrons 10 μSv/h, fast neutrons 12 μSv/h; positrons made via the $^{18}\text{O}(p,n)$ reaction on $^{18}$O-enriched water.

Find. (a)–(e) qualitative mechanisms; (f) each field's absorbed dose rate $\dot D=\dot H/w_R$ in Gy/h.

Approach. Balance the $(p,n)$ reaction to identify the product nuclide and its decay mode, trace how the primary fast-neutron and proton beams generate the secondary thermal-neutron and gamma fields, then divide each reported equivalent dose rate by the ICRP-60 radiation weighting factor appropriate to its energy to recover absorbed dose.

1. Part (a) — product isotope and positron emission. Conserving $Z$ and $A$ across $p+{}^{18}_{\ 8}\text{O}\rightarrow{}^{18}_{\ 9}\text{F}+n$ gives fluorine-18, $^{18}$F. Stable fluorine is $^{19}$F ($Z=9,N=10$); $^{18}$F has $Z=9,N=9$, one neutron short of stability, i.e. it is proton-rich. A proton-rich nuclide moves toward stability by converting a proton into a neutron, which is exactly $\beta^+$ (positron) decay ($p\rightarrow n+\beta^++\nu$) or the competing process electron capture — hence $^{18}$F is a positron emitter, and its annihilation-photon pair (511 keV, back-to-back) is what PET imaging detects.
2. Part (b) — how neutrons arise in a proton accelerator. Neutrons are produced whenever the accelerated protons strike a nucleus with enough energy to induce a nuclear reaction that emits a neutron — here directly via the target $(p,n)$ reaction itself, and more generally via $(p,xn)$ reactions and spallation on the target holder, beamline components, and shielding wherever the beam or its secondaries deposit energy above the relevant reaction thresholds.
3. Part (c) — thermal-neutron dose from a fast-neutron source. Fast neutrons lose energy by successive elastic scattering collisions with light nuclei (chiefly hydrogen) in the shield, target assembly and surrounding materials; each collision removes a substantial fraction of the neutron's kinetic energy. After enough scattering events the neutrons are moderated down to thermal energies ($\sim$0.025 eV) within and around the shield, so a population of thermalized neutrons — and hence a thermal-neutron dose — necessarily accompanies any sufficiently thick fast-neutron shield, even though the source itself emits only fast neutrons.
4. Part (d) — gamma-ray production mechanisms. Gammas arise from several concurrent processes: (i) neutron radiative capture, $(n,\gamma)$, when thermalized neutrons are absorbed by shield/structural nuclei; (ii) inelastic neutron scattering off shield nuclei, which de-excite by prompt gamma emission; (iii) 511 keV positron-annihilation photons from the $^{18}$F decay itself; and (iv) bremsstrahlung radiated by decelerating charged particles (protons, positrons, recoil electrons) in the target and shielding materials.
5. Part (e) — absence of a reported proton dose. Protons have a very short range in matter compared with neutrons or gammas of similar energy (they lose energy rapidly via ionization, following a Bragg-curve profile) and are essentially completely stopped within the target and the first layer of shielding. None of the primary proton beam escapes to the point outside the shield where the survey was performed, so there is no external proton dose to report there — the proton dose is entirely absorbed internally, at the target itself.
6. Part (f) — converting equivalent dose rate to absorbed dose rate. Equivalent dose and absorbed dose are related by the radiation weighting factor, $\dot H=w_R\dot D$, so $\dot D=\dot H/w_R$. Using the ICRP-60 step values — photons $w_R=1$; neutrons $E<10$ keV (thermal, $\sim$0.025 eV) $w_R=5$; neutrons 100 keV–2 MeV (the "fast" bracket, $w_R$ peaks here) $w_R=20$: $$\boxed{\dot D_\gamma = \frac{5\ \mu\text{Sv/h}}{1} = 5\ \mu\text{Gy/h} = 5\times10^{-6}\ \text{Gy/h}}$$ $$\boxed{\dot D_{\text{thermal}} = \frac{10\ \mu\text{Sv/h}}{5} = 2\ \mu\text{Gy/h} = 2\times10^{-6}\ \text{Gy/h}}$$ $$\boxed{\dot D_{\text{fast}} = \frac{12\ \mu\text{Sv/h}}{20} = 0.6\ \mu\text{Gy/h} = 6\times10^{-7}\ \text{Gy/h}}$$ The fast-neutron field, despite carrying the largest equivalent dose rate of the three (12 of 27 $\mu$Sv/h), corresponds to the smallest absorbed dose rate, because its large $w_R=20$ means a comparatively small physical energy deposition produces a disproportionately large biological equivalent dose.