Question 7 of 7: Detection of Radiation — NaI(Tl) Gamma Spectroscopy

Question 7: Detection of Radiation — NaI(Tl) Gamma Spectroscopy (10 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. ${}^{137}$Cs gamma energy $E_\gamma = 662$ keV; electron rest energy $m_ec^2=511$ keV; mass attenuation coefficient of NaI at 662 keV, $\mu/\rho = 0.0079\text{ cm}^2/\text{g}$; NaI density $\rho=3{,}700\text{ kg/m}^3 = 3.7\text{ g/cm}^3$; crystal length $x=25$ mm $=2.5$ cm; $N_0=1{,}000{,}000$ incident photons.

Find. (a) the photoelectric-electron energy; (b)–(c) the maximum and minimum Compton-electron energies; (d) the resulting pulse-height spectrum shape; (e) the number of photons that interact (are detected) in the crystal.

Approach. Apply the photoelectric energy-conservation relation, the Compton-edge formula at $\theta=180^\circ$ (backscatter, maximum energy transfer) and at $\theta\to0^\circ$ (minimum transfer) for (a)–(c); assemble the qualitative spectrum shape from those three features for (d); and use the standard exponential attenuation law with the given mass attenuation coefficient for (e).

1. Part (a) — photoelectric electron energy. By energy conservation, the ejected photoelectron carries the photon energy minus the binding energy of the atomic level it is ejected from. NaI's photoelectric absorption is dominated by the high-Z iodine atom (Z = 53), whose K-edge binding energy is $\approx 33$ keV: $$\boxed{E_e = E_\gamma - E_B \approx 662 - 33 = 629\text{ keV}}$$ (to the accuracy the exam's data supports, $E_e \approx E_\gamma = 662$ keV is an acceptable first-order approximation, since the K-edge binding energy is small compared to 662 keV).
2. Part (b) — maximum Compton-electron energy (Compton edge). The electron receives its maximum kinetic energy when the photon backscatters ($\theta=180^\circ$, $\cos\theta=-1$): $$E_{e,\max} = E_\gamma\left(\frac{2\alpha}{1+2\alpha}\right), \qquad \alpha = \frac{E_\gamma}{m_ec^2} = \frac{662}{511} = 1.295$$ $$\boxed{E_{e,\max} = 662\times\frac{2(1.295)}{1+2(1.295)} = 477.7\text{ keV}}$$ This is the well-known "Compton edge" seen in every NaI(Tl) spectrum of a monoenergetic source.
3. Part (c) — minimum Compton-electron energy. As the scattering angle $\theta\to0$ (forward, grazing scatter), essentially no momentum is transferred to the electron: $$\boxed{E_{e,\min} \to 0\text{ keV}}$$ so the Compton continuum extends continuously down to zero pulse height.
4. Part (d) — pulse-height distribution sketch. Combining (a)–(c): a continuous Compton continuum from 0 up to the sharp Compton edge at 477.7 keV (with a small "backscatter peak" bump near $E_\gamma - E_{e,\max}=184.3$ keV, from photons that Compton-scatter in the surrounding shielding/housing before entering the crystal), a gap with essentially no counts between the edge and the full-energy line, and a narrow, tall photopeak at 662 keV from photons fully absorbed via the photoelectric effect (or via a Compton scatter followed by photoelectric absorption of the scattered photon, which also deposits the full 662 keV).
   Figure 3 — schematic NaI(Tl) pulse-height spectrum for 662 keV ${}^{137}$Cs gammas: Compton continuum with backscatter bump, Compton edge, and photopeak.
5. Part (e) — number of photons detected. A photon counts as "detected" if it interacts at least once in the crystal (photoelectric or Compton), so the detected fraction is $1 - e^{-\mu x}$ with the linear attenuation coefficient built from the given mass attenuation coefficient and density: $$\mu = \left(\frac{\mu}{\rho}\right)\rho = 0.0079\times3.7 = 0.02923\text{ cm}^{-1}$$ $$\mu x = 0.02923\times2.5 = 0.0731$$ $$N_{\text{detected}} = N_0\left(1-e^{-\mu x}\right) = 1{,}000{,}000\times\left(1-e^{-0.0731}\right)$$ $$\boxed{N_{\text{detected}} \approx 1{,}000{,}000\times0.0705 \approx 7.05\times10^{4} \text{ photons } (\approx 70{,}500)}$$ The crystal is thin enough (0.073 mean free paths) that the vast majority of photons — about 93% — pass straight through undetected along this particular axial path. Note that the printed 0.0079 cm²/g is a factor of ten below the tabulated value for NaI at 662 keV (≈0.079 cm²/g, i.e. $\mu\approx0.29$ cm$^{-1}$), which suggests a dropped digit; with the tabulated value $\mu x = 0.731$ and about $5.18\times10^{5}$ photons (≈52%) would interact. The answer above uses the value as printed, as the exam requires.