Question 2 of 7: X-rays, Attenuation and Absorption in Matter, Dosimetry

Question 2: X-rays, Attenuation and Absorption in Matter, Dosimetry (20 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. Tube voltage 60 kV rising to 75 kV; path length 20 mm muscle + 3.5 mm fat + 6.5 mm bone; exposure time fixed at 0.5 s in both cases; mean photon energy assumed to be one-third of the peak (accelerating) energy; the attenuation-coefficient table above, which conveniently tabulates exactly the two required mean energies (60/3 = 20 keV and 75/3 = 25 keV).

Find. (a) the maximum photon energy; (c) the ratio of applied currents that keeps the film exposure (photon count) unchanged; (d) whether the resulting patient dose rises, falls, or stays the same; (e)–(g) short conceptual answers on dosimetry and attenuation.

Approach. Bremsstrahlung's maximum photon energy equals the electron's kinetic energy at the target, $E_{\max}=eV$, giving (a) directly. For (c)–(d), take the standard radiographic simplification that photon quantity (the number of photons leaving the target) is controlled by mAs (current × time) essentially independent of kVp, while photon quality (mean energy and hence penetration) is set by kVp — this is the textbook "mAs controls quantity, kVp controls quality" convention, flagged explicitly since the exam's own instructions invite a stated assumption. The exponential attenuation law and the given coefficient table then convert between photon counts at the film for the two mean energies.

1. Part (a) — maximum photon energy. Bremsstrahlung photons cannot carry more energy than the accelerated electron that produces them, so the spectrum's upper edge is set by the tube voltage: $$\boxed{E_{\max} = eV = 60\text{ keV}}$$
2. Part (b) — spectrum sketch. The bremsstrahlung spectrum, as filtered by the tube window, is continuous, rising from zero intensity at low photon energy (the softest photons are absorbed in the window), peaking at roughly one-third to one-half of $E_{\max}$, and falling to zero intensity at $E_{\max}=60$ keV (no photon can exceed the electron's kinetic energy); with no target material specified, no discrete characteristic lines are superimposed.
   Figure 1 — bremsstrahlung spectrum (as emerging through the tube window): continuous, peaking near 20–25 keV and falling to zero intensity at the 60 keV cut-off.
3. Part (c) — current change to hold film exposure constant. Under the "mAs sets quantity" assumption, the photon count produced in time $t$ is $N_0 \propto i\,t$, independent of $V$; the count reaching the film after traversing the tissue stack is $N_{\text{film}} = N_0\,T(E)$ with transmission $T(E)=e^{-\sum \mu_j(E)x_j}$. Summing the table's coefficients over the given thicknesses ($x_{\text{muscle}}=2.0$ cm, $x_{\text{fat}}=0.35$ cm, $x_{\text{bone}}=0.65$ cm) at the two mean energies: $$\textstyle\sum \mu x\,(20\text{ keV}) = 0.808(2.0)+0.488(0.35)+4.542(0.65) = 4.739$$ $$\textstyle\sum \mu x\,(25\text{ keV}) = 0.594(2.0)+0.379(0.35)+3.058(0.65) = 3.308$$ $$T(20) = e^{-4.739} = 8.75\times10^{-3}, \qquad T(25) = e^{-3.308} = 3.66\times10^{-2}$$ Equating film counts $i_1 t\,T(20) = i_2 t\,T(25)$ (exposure time cancels since it is the same 0.5 s in both cases): $$\boxed{\frac{i_2}{i_1} = \frac{T(20)}{T(25)} = \frac{8.75\times10^{-3}}{3.66\times10^{-2}} \approx 0.239}$$ The applied current must be reduced to about 24% of its original value (roughly a 76% cut) — the 75 kV beam is so much more penetrating through this tissue stack that far fewer photons need to be produced to deliver the same count to the film.
4. Part (d) — effect on patient dose. Dose is proportional to the number of photons absorbed by the patient (not transmitted) times their energy. The photon count entering the patient scales with $i\,t$ (same law as part (c)), so with $i_2/i_1 \approx 0.239$ from above and the absorbed fraction $1-T(E)$: $$\frac{\text{Dose}_2}{\text{Dose}_1} = \frac{(i_2 t)\,[1-T(25)]\,E_2}{(i_1 t)\,[1-T(20)]\,E_1} = \frac{0.239\times(1-0.0366)\times25}{1\times(1-0.00875)\times20}$$ $$\boxed{\frac{\text{Dose}_2}{\text{Dose}_1} \approx 0.29}$$ The dose at 75 kV (with the current reduced to keep the film count fixed) is lower — about 29% of the original 60 kV dose. Even though each photon of the 75 kV beam (25 keV mean) carries 25% more energy than one of the 60 kV beam (20 keV mean), roughly four times fewer photons now enter the patient, and that reduction dominates.
5. Part (e) — measuring absorbed dose in practice. Directly, with a calibrated dosimeter placed at the point of interest: an ionization chamber or a thermoluminescent dosimeter (TLD) reads exposure/air kerma, which is then converted to tissue absorbed dose via known conversion (f-)factors; in diagnostic radiography this is usually reported as entrance skin dose or dose-area product from a calibrated chamber built into the collimator, cross-checked against the known kVp/mAs/distance technique factors.
6. Part (f) — dominant interaction mode. At 20–30 keV, the photoelectric effect dominates. The table itself is the evidence: bone's coefficient (Z of Ca, P far exceeds soft tissue) is about 4–6× higher than muscle's at the same energy, and every coefficient falls steeply with increasing energy — both signatures of a cross-section scaling roughly as $Z^{3\text{-}4}/E^3$, which is the photoelectric effect; Compton scattering's cross-section is nearly energy- and Z-independent per electron and would not produce either trend.
7. Part (g) — why bone attenuates more. Bone's mineral content (Ca, P) gives it a much higher effective atomic number than the light elements (H, C, O, N) dominating muscle and fat, and the photoelectric cross-section — dominant at these energies per part (f) — scales steeply with atomic number ($\propto Z^{3\text{-}4}$). Bone's higher physical density also raises its electron density per unit volume. Both effects compound, which is exactly why bone appears bright on a radiograph.