Interaction of Radiation with Matter

(a) $$x_{1/2} = \frac{\ln 2}{\mu} = \frac{0.693}{0.68\ \mathrm{cm^{-1}}} = 1.02\ \mathrm{cm}$$

(b) $I/I_{0} = 0.01 = e^{-\mu x}$ gives $$x = \frac{\ln 100}{\mu} = \frac{4.605}{0.68\ \mathrm{cm^{-1}}} = 6.77\ \mathrm{cm}$$

So ~1 cm of lead halves Co-60 γ intensity; ~6.8 cm cuts it to a hundredth. This is why γ shielding is quoted in half-value layers, not ranges.

(a) Threshold: $\beta_{th} = 1/n = 0.7519$, $\gamma_{th} = 1/\sqrt{1-\beta_{th}^{2}} = 1.5168$.

$$T_{th} = m_{e}c^{2}(\gamma_{th}-1) = 0.511\ \mathrm{MeV}\times 0.5168 = 0.264\ \mathrm{MeV}\approx 0.26\ \mathrm{MeV}$$

(b) $$\cos\theta = \frac{1}{n\beta} = \frac{1}{1.33\times 0.90} = 0.8354\;\;\Longrightarrow\;\;\theta = 33.4^{\circ}$$

Note the angle depends only on $\beta$ and $n$ — measuring $\theta$ measures the particle's velocity, the basis of Cerenkov detectors.

$$R_{\mathrm{Al}} = R_{\mathrm{air}}\,\frac{\rho_{\mathrm{air}}\sqrt{A_{\mathrm{Al}}}}{\rho_{\mathrm{Al}}\sqrt{A_{\mathrm{air}}}}$$

$$R_{\mathrm{Al}} = 3.63\ \mathrm{cm}\times\frac{1.226\times 10^{-3}}{2.70}\times\sqrt{\frac{27}{14.5}} = 3.63\times 4.541\times 10^{-4}\times 1.3646 = 2.25\times 10^{-3}\ \mathrm{cm}$$

≈ 0.022 mm — a few hundredths of a millimetre. α-particles are stopped by a sheet of paper: this is the quantitative reason.

From the Bethe formula, $-dE/dx \propto z^{2}M/E$ (since $1/\beta^{2} \propto M/E$). Hence $R = \int dE/(-dE/dx) \propto E^{2}/(z^{2}M)$.

$$\frac{R_{p}}{R_{\alpha}} = \frac{z_{\alpha}^{2}\,M_{\alpha}}{z_{p}^{2}\,M_{p}} = \frac{4\times 4}{1\times 1} = 16$$

At the same velocity, by contrast, the ratio is 1 (same $z^{2}/M$ per unit mass) — a standard exam trap.

$v_{0} = c\sqrt{2E/Mc^{2}} = (3.0\times 10^{8})\sqrt{10/3727} = 1.55\times 10^{7}\ \mathrm{m\,s^{-1}}$. Taking an average speed of ~$v_{0}/2$,

$$t \approx \frac{R}{\bar{v}} = \frac{3.6\times 10^{-2}\ \mathrm{m}}{7.8\times 10^{6}\ \mathrm{m\,s^{-1}}}\approx 5\times 10^{-9}\ \mathrm{s}$$

Slowing down is complete in a few nanoseconds in gases (picoseconds in solids) — effectively instantaneous on any laboratory timescale.

Current mode: the charge liberated by a large number of individual ionizing events is collected as a steady (time-averaged) current. The current is proportional to the average rate of energy deposition — i.e. to the dose rate or exposure. This is the mode of the ionization chamber used as a dosimeter/beam monitor: it works at high intensities where individual pulses would hopelessly pile up, but it gives no information about individual particles.

Pulse mode: the charge from each single ionizing event is collected as a separate electrical pulse, whose height is proportional to the energy deposited in that event. Counting the pulses gives the event rate; analysing pulse heights gives the energy spectrum; pulse timing gives coincidence information. This is the mode of counters and spectrometers (proportional counters, GM counters, scintillation and semiconductor detectors). Its weakness is dead time — at high rates pulses overlap and the detector saturates.

One line: current mode → dose rate (average); pulse mode → counts, energy and timing (individual events).

Stopping power $-dE/dx$ is the mean energy lost by a charged particle per unit path length — it describes the particle. Unit: $\mathrm{MeV\,cm^{-1}}$ (linear) or $\mathrm{MeV\,cm^{2}\,g^{-1}}$ (mass stopping power, $-dE/\rho\,dx$).

LET (linear energy transfer) is the energy locally absorbed in the medium per unit path length — the restricted stopping power, excluding energy carried away by long-range δ-rays. It describes the medium's local dose. Unit: $\mathrm{keV\,\mu m^{-1}}$.

I of water by the Bragg additivity rule, $\ln I = \sum n_{i}Z_{i}\ln I_{i}/\sum n_{i}Z_{i}$: for $\mathrm{H_{2}O}$, H: $n=2, Z=1, I=19.0\ \mathrm{eV}$; O: $n=1, Z=8, I=11.2+11.7(8)=104.8\ \mathrm{eV}$:

$$\ln I = \frac{2(1)\ln 19.0 + 8\ln 104.8}{10} = \frac{2(2.9444)+8(4.6522)}{10} = 4.3107$$ $$I = e^{4.3107} = 74.5\ \mathrm{eV}$$

Definitions and units as above: stopping power $-dE/dx$ in $\mathrm{MeV\,cm^{-1}}$ (mass form $\mathrm{MeV\,cm^{2}\,g^{-1}}$); LET in $\mathrm{keV\,\mu m^{-1}}$.

I of methane $\mathrm{CH_{4}}$: C: $n=1, Z=6, I=11.2+11.7(6)=81.4\ \mathrm{eV}$; H: $n=4, Z=1, I=19.0\ \mathrm{eV}$:

$$\ln I = \frac{6\ln 81.4 + 4(1)\ln 19.0}{6+4} = \frac{6(4.3992)+4(2.9444)}{10} = 3.8173$$ $$I = e^{3.8173} = 45.5\ \mathrm{eV}$$

Positron threshold (Ch1, in brief): creating a positron from a photon requires creating its antiparticle partner too (charge and lepton-number conservation): $\gamma \to e^{+}+e^{-}$, so at least the rest energy of two electrons, $2m_{e}c^{2} = 1.022\ \mathrm{MeV}$, must be supplied.

Neutrons with matter: no charge → no Coulomb interaction with electrons; they interact only with nuclei. Fast neutrons lose energy by elastic scattering (best on light nuclei — moderation by hydrogen); neutrons are captured by nuclei — radiative capture $(n,\gamma)$, or reactions ejecting charged particles $(n,p), (n,\alpha)$ — and heavy nuclei may fission. Detection is indirect: the detector responds to the charged secondaries — recoil protons from elastic scattering on hydrogen, or the proton/α from $^{3}\mathrm{He}(n,p)$ / $^{10}\mathrm{B}(n,\alpha)$ conversion reactions.

Charged particles with matter: they interact directly through the Coulomb field with atomic electrons, producing ionization (ion pairs) and excitation (later de-excitation photons) along a dense track. Heavy particles travel nearly straight and stop after a definite range; electrons also lose energy by bremsstrahlung. Detection is direct: the ion pairs are collected (gas-ionization detectors — ionization chamber, proportional counter, GM counter) or the de-excitation light is converted to an electrical pulse (scintillation detectors).

Neutrons: interact only with nuclei — elastic scattering (moderation, most effective on hydrogen), inelastic scattering, radiative capture $(n,\gamma)$, and charged-particle emission $(n,p), (n,\alpha)$ / fission. Because they create no ionization themselves, detection is indirect: fast neutrons are seen via recoil protons in hydrogenous detectors, slow neutrons via conversion reactions such as $^{3}\mathrm{He}(n,p)^{3}\mathrm{H}$ or $^{10}\mathrm{B}(n,\alpha)^{7}\mathrm{Li}$, whose energetic charged products then ionize the detector medium.

Charged particles: lose energy continuously by Coulomb ionization and excitation of atomic electrons (plus bremsstrahlung for electrons), producing a dense, nearly straight track ending in a definite range. Detection is direct: collect the ion pairs (gas-filled detectors, semiconductor detectors) or the scintillation light (scintillators) — each interaction gives a countable pulse (pulse mode) or contributes to a steady current (current mode).

Stopping power $-dE/dx$: mean energy lost by the charged particle per unit path length — a property of the particle's energy loss. Units: linear $\mathrm{MeV\,cm^{-1}}$; mass stopping power $-dE/\rho\,dx$ in $\mathrm{MeV\,cm^{2}\,g^{-1}}$.

LET (linear energy transfer): energy locally absorbed in the medium per unit path length — the restricted stopping power, excluding energy carried off by δ-rays above the cut-off $\Delta$. A measure of local dose density. Unit: $\mathrm{keV\,\mu m^{-1}}$.

Distinguishing line for the examiner: stopping power counts all energy the particle loses; LET counts only what is deposited near the track. α-particles are high-LET ($\sim 10^{2}\ \mathrm{keV\,\mu m^{-1}}$); γ-rays and β-particles are low-LET ($\sim 10^{-1}\ \mathrm{keV\,\mu m^{-1}}$).

For electrons, radiative losses (bremsstrahlung, $\propto Z^{2}E$) grow with energy and overtake collisional losses above the critical energy — the Bethe formula covers only the collisional part. Also, the incident electron is indistinguishable from atomic electrons, changing the $W_{\max}$ kinematics. A modified Berger–Seltzer form is used instead.

The critical energy $E_{c}$ is the electron energy at which radiative and collisional stopping powers are equal. Berger–Seltzer: $E_{c} = 610\,\mathrm{GeV}/(Z+1.24)$ (solids/liquids). For lead ($Z=82$): $E_{c} = 610/83.24\ \mathrm{GeV} \approx 7.3\ \mathrm{MeV}$.

Photons are removed exponentially ($I = I_{0}e^{-\mu x}$) — no photon has a definite penetration depth, so the natural measure is the half-value layer $x_{1/2}=\ln 2/\mu$. Charged particles slow down continuously and stop after a definite range; beyond it essentially none penetrate.

Condition: the charged particle must outrun light in the medium, $v > c/n$ (i.e. $\beta > 1/n$). Emission is on a cone at $\cos\theta = c/nv = 1/n\beta$.

Photoelectric (low E, high Z) in lead at 60 keV; Compton scattering (mid E, low Z) in water at 1 MeV; pair production (high E, high Z) in lead at 10 MeV.

They are nearly equal: an α-particle's secondary electrons (δ-rays) are short-ranged, so almost all lost energy is deposited locally. For fast electrons they differ markedly, because energetic δ-rays carry energy far from the track.

Neutrons carry no charge and produce no ionization directly. A converter turns them into detectable charged particles: hydrogenous material (recoil protons) for fast neutrons; $^{3}\mathrm{He}$ or $^{10}\mathrm{B}$ (via $(n,p)$ / $(n,\alpha)$ reactions) for slow neutrons.

No — at the same velocity the α-particle's stopping power is 4× the proton's, because $-dE/dx \propto z^{2}$ and $z_{\alpha}^{2}/z_{p}^{2} = 4$. (Their ranges at the same velocity are equal, since $R \propto M/z^{2}$.)