Question 3 of 7: Gamma-Ray Shielding and Attenuation vs. an X-ray Source

Question 3: Gamma-Ray Shielding and Attenuation vs. an X-ray Source (17 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. ${}^{241}$Am gammas at 60 keV; mass attenuation coefficient $\mu/\rho=0.17\text{ cm}^2/\text{g}$ at 60 keV; concrete density $\rho=2{,}300\text{ kg/m}^3 =2.30\text{ g/cm}^3$; 60 kV X-ray TVL in concrete $=21.5$ mm.

Find. (a) the X-ray spectrum shape; (b) the ${}^{241}$Am TVL; (c)–(f) comparative attenuation, tabulated transmission at three thicknesses, dose-buildup reasoning, and an adequacy assessment of ${}^{241}$Am as a shielding-check surrogate for the real X-ray beam.

Approach. Compute the monoenergetic ${}^{241}$Am linear attenuation coefficient from the given mass coefficient and density, convert both TVLs to effective linear attenuation coefficients, apply simple exponential attenuation $I/I_0=e^{-\mu x}$ at each thickness for both sources, then reason physically about spectral hardness, buildup, and what a monoenergetic proxy source can and cannot tell you about a polychromatic X-ray beam's shielding adequacy.

1. Part (a) — X-ray spectrum sketch. The X-ray tube's unfiltered bremsstrahlung output is continuous: zero intensity at $E=0$, rising to a peak at roughly one-third to one-half of the tube's maximum (accelerating) energy, then falling to zero at the sharp high-energy cut-off set by the tube voltage (60 keV here, since no photon can carry more energy than the accelerated electron that produced it). No discrete characteristic lines are superimposed unless a specific high-$Z$ target line falls within this range.
   Figure 1 — unfiltered bremsstrahlung spectrum of the 60 kV X-ray machine: continuous from 0, peaking near 20 keV, cut off sharply at 60 keV.
2. Part (b) — TVL for ${}^{241}$Am 60 keV photons. Convert the mass attenuation coefficient to a linear coefficient with the given density, then to a TVL: $$\mu = (\mu/\rho)\,\rho = 0.17\times2.30 = 0.391\text{ cm}^{-1}$$ $$\boxed{\text{TVL}_{\gamma} = \frac{\ln10}{\mu} = \frac{2.3026}{0.391} = 5.89\text{ cm} = 58.9\text{ mm}}$$
3. Part (c) — comparison and physical basis. The ${}^{241}$Am TVL (58.9 mm) is nearly three times the 60 kV X-ray TVL (21.5 mm), even though both are nominally "60 keV" radiation: $$\text{TVL}_{\gamma}/\text{TVL}_{x} = 58.9/21.5 \approx 2.7$$ The physical basis is that the two sources are not equally energetic in practice. ${}^{241}$Am gammas are monoenergetic at 60 keV, so every photon is as penetrating as the number suggests. A 60 kV X-ray tube instead produces a continuous bremsstrahlung spectrum whose mean photon energy is well below its 60 keV maximum (roughly one-third of it, by the usual rule of thumb, i.e. $\approx20$ keV) — most of the beam's photons are much softer and far more attenuable than a true 60 keV photon, so the beam as a whole is attenuated much faster (smaller TVL) than the monoenergetic gamma source of the same peak energy.
4. Part (d) — tabulated transmission at 25/50/100 mm. Convert the X-ray TVL to an effective linear coefficient the same way, then apply $I/I_0=e^{-\mu x}$ with each source's own $\mu$: $$\mu_x = \frac{\ln10}{2.15\text{ cm}} = 1.071\text{ cm}^{-1}$$

   Transmitted fraction $I/I_0$ through concrete

   Thickness	${}^{241}$Am gamma (60 keV mono)	60 kV X-ray beam
   25 mm	0.376 (62.4% reduction)	0.0687 (93.1% reduction)
   50 mm	0.142 (85.8% reduction)	0.00473 (99.5% reduction)
   100 mm	0.0200 (98.0% reduction)	0.0000223 (>99.99% reduction)

   $$\boxed{I/I_0(50\text{ mm}):\ \ 0.142\text{ (gamma)}\quad\text{vs.}\quad 0.00473\text{ (X-ray)}}$$ Assumption stated (per the exam's own invitation): both beams are treated with a single, thickness-independent effective linear attenuation coefficient (narrow-beam, no buildup, and — for the X-ray beam — no spectral hardening with depth). A real polychromatic X-ray beam actually hardens as it penetrates (its softer photons are preferentially removed first, so its effective $\mu$ decreases with depth), which means the tabulated X-ray transmission values above are a slight underestimate of the true transmitted fraction at the larger thicknesses.
5. Part (e) — which source/thickness sees the most buildup. Dose buildup grows with (i) how many mean free paths ($\mu x$) the beam has traversed — more scattering interactions accumulate more scattered photons still contributing to dose — and (ii) how strongly Compton scattering (rather than photoelectric absorption, which removes a photon completely with no scattered survivor) dominates the interaction. At 60 keV in concrete (moderate effective $Z\approx11$), Compton scattering is already significant relative to the photoelectric effect, so the monoenergetic ${}^{241}$Am gammas generate a genuine buildup of scattered photons as they traverse the slab. The X-ray beam's transmitted photons, by contrast, are progressively dominated by whatever soft tail survives, and photoelectric absorption (which contributes no buildup) removes photons outright rather than scattering them. Combining both effects: the ${}^{241}$Am 60 keV gamma rays, at the greatest thickness (100 mm), are the combination most likely to show significant dose buildup — the largest number of mean free paths combined with Compton-dominated interactions.
6. Part (f) — adequacy of ${}^{241}$Am as a shielding-adequacy check. At 50 mm, the ${}^{241}$Am transmission is 14.2% — not "practically zero" by itself. However, ${}^{241}$Am is deliberately the more penetrating of the two sources (part c): the real X-ray beam's transmission at the same 50 mm is only 0.47%, over an order of magnitude smaller. Using ${}^{241}$Am as the test source therefore gives a conservative (worst-case) check: if the isotopic source's transmitted intensity outside the shield is acceptably low, the true X-ray leakage — which is intrinsically softer and more strongly attenuated — will be lower still. The source is adequate for a pass/fail adequacy check of the shield provided the 14.2% gamma-transmission figure is compared against an appropriately conservative action level (rather than literally requiring "zero"), and provided the surveyor recognizes the two transmitted fractions are not numerically interchangeable. Taken at face value ("practically zero"), 50 mm is not adequate against the ${}^{241}$Am proxy; taken as a conservative bound on the actual X-ray leakage, 50 mm is comfortably adequate for the real beam.