Chapter 7. Radiation, Detectors, Accelerators and Reactors

Theory, worked examples and 21 previous-year questions with full solutions — IIT JAM, GATE, JEST, TIFR and CSIR NET-JRF, 2020 onwards.

Theory

7.1 Interaction of charged particles with matter

(a) Energy loss and the Bethe formula

Every detector works by converting the energy that radiation transfers to a material into ionisation, excitation or light that can be turned into an electrical signal. Charged particles lose energy continuously, in very many small Coulomb collisions with atomic electrons; photons and neutrons travel freely until a single interaction hands most of their energy to a charged particle, which is then detected in the same way. A heavy particle of mass \(M\gg m_e\) can give a free electron at most \(T_{\max}\approx2m_ev^2\) in one collision (about 2.7 keV for a 5 MeV \(\alpha\) particle), so it makes tens of thousands of collisions, is hardly deflected and follows a nearly straight track. The mean energy spent per ion pair is almost independent of the radiation, \(W\approx34\) eV in air and 26 eV in argon (more than the 15.8 eV ionisation energy of argon, because part goes into excitation), and about 3 eV per electron–hole pair in a semiconductor. A particle that deposits energy \(E\) therefore produces \[\boxed{\,N = \frac{E}{W}\,}\] ion pairs. Summing the energy \((\Delta p)^2/2m_e\), with impulse \(\Delta p = 2ze^2/(4\pi\varepsilon_0bv)\), over impact parameters gives a loss \(\propto z^2n/v^2\); the full result (Bethe) for the stopping power is \[\boxed{\,-\frac{dE}{dx} = \frac{4\pi n z^2}{m_ec^2\beta^2}\left(\frac{e^2}{4\pi\varepsilon_0}\right)^2\left[\ln\frac{2m_ec^2\beta^2\gamma^2}{I} - \beta^2\right]\,}\] with \(n = N_AZ\rho/A\) the electron density and \(I\approx10Z\) eV the mean excitation energy. The loss depends on the particle only through \(z\) and \(v\) (an \(\alpha\) particle loses four times as fast as a proton of the same speed), varies as \(z^2/v^2\) at low speed, and is nearly the same per g cm\(^{-2}\) in all materials since \(Z/A\approx\tfrac12\). It passes through a broad minimum near \(\beta\gamma\approx3\) (minimum-ionising particles, about 2 MeV per g cm\(^{-2}\)) and then rises slowly. Because the loss increases as the particle slows, a heavy particle deposits most of its energy at the end of its path, in the Bragg peak (Figure 7.1), which is exploited in proton therapy; the statistical spread of the end point is range straggling.

Figure 7.1 Bragg curve: stopping power \(-dE/dx\) (solid) and number \(N(x)\) of particles remaining in a monoenergetic beam (dashed) as functions of depth. \(R\) is the mean range.

(b) Range

The range is \(R = \int_0^{T_0}dT/(-dT/dx)\). Writing \(-dT/dx = z^2f(v)\) and \(dT = Mv\,dv\) gives \(R = (M/z^2)F(v_0)\), so particles with the same initial speed (the same \(T/M\)) have \[\boxed{\,\frac{R_1}{R_2} = \frac{M_1}{M_2}\,\frac{z_2^2}{z_1^2}\,}\] For \(\alpha\) particles of 4–7 MeV in air, Geiger’s rule is \(R_{\rm air}\,(\text{cm}) = 0.318\,E^{3/2}\,(\text{MeV})\): a 5 MeV \(\alpha\) particle travels about 3.5 cm in air and only a few tens of micrometres in a solid, so it is stopped by paper, skin or the glass wall of a counter.

Example 7.1 A proton of 3.0 MeV has range \(R_0\) in aluminium. Find the ranges of a 6.0 MeV deuteron and a 12 MeV \(\alpha\) particle (\(M_d = 2m_p\), \(M_\alpha = 4m_p\)).

Solution. All three have the same \(T/M\), hence the same speed, so \(R\propto M/z^2\). The deuteron has \(M/z^2 = 2m_p\), giving \(R_d = 2R_0\); the \(\alpha\) particle has \(4m_p/4 = m_p\), giving \(R_\alpha\approx R_0\) despite its fourfold energy.

(c) Electrons and positrons

An electron can lose up to half its energy in one collision and is scattered through large angles, so its path is tortuous and its range ill-defined (about 0.5 g cm\(^{-2}\) per MeV, a hundred times the range of an \(\alpha\) particle of the same energy). Being light, it is also strongly accelerated by nuclei and radiates bremsstrahlung, with a loss \(\propto Z^2E/m^2\); the ratio of radiative to collision loss is about \(ZE/800\) (\(E\) in MeV), so radiation dominates above the critical energy \(E_c\approx800/Z\) MeV (about 10 MeV in lead). Since an electron here gives up energy to create a photon, bremsstrahlung is the inverse photoelectric effect; it gives the continuous X-ray spectrum, with \(\lambda_{\min} = hc/eV\). A positron slows down like an electron and then annihilates, usually into two back-to-back 511 keV photons.

7.2 Interaction of photons with matter

(a) Origin of photons and the attenuation law

Gamma rays (keV to MeV) are emitted in transitions between nuclear levels; X-rays (1–100 keV) in transitions between the inner (core) levels of atoms and in bremsstrahlung, and with wavelengths of order 1 Å they are diffracted by crystals; visible light (2–3 eV, several hundred nm) can eject photoelectrons from a metal of low work function but cannot resolve crystal planes or reach core levels. A photon is removed from a collimated beam in a single event, by the photoelectric effect, Compton scattering or pair production. In a layer \(dx\) containing \(n\) atoms per unit volume, each of total cross-section \(\sigma\), a fraction \(n\sigma\,dx\) interact, so \[\boxed{\,I(x) = I_0\,e^{-\mu x},\qquad \mu = n\sigma = \frac{\rho N_A}{A}\,\sigma\,}\] where \(I(x)\) counts photons that have crossed without any interaction, keeping their energy and direction. The mean free path is \(1/\mu\) (intensity reduced to \(1/e\)), the half-value thickness is \(x_{1/2} = \ln2/\mu\), and \(\mu/\rho\) (cm\(^2\) g\(^{-1}\)) is the mass attenuation coefficient. The same law, with \(\mu\to\Sigma = n\sigma\), describes the attenuation of a neutron beam.

Example 7.2 Photons of 1.25 MeV fall on an iron plate 2.0 cm thick (\(\rho = 7.87\) g cm\(^{-3}\), \(A = 55.85\)), for which the total cross-section is 5.0 b per atom. Find the half-value thickness and the fraction crossing without interaction.

Solution. \(n = 7.87\times6.022\times10^{23}/55.85 = 8.49\times10^{22}\) cm\(^{-3}\), so \(\mu = n\sigma = 8.49\times10^{22}\times5.0\times10^{-24} = 0.424\) cm\(^{-1}\) and \(x_{1/2} = 0.693/0.424 = 1.6\) cm. The unscattered fraction is \(e^{-0.849} = 0.43\).

(b) Photoelectric effect

The photon is absorbed by a bound electron, ejected with \(T_e = h\nu - B\); the vacancy is filled with emission of an X-ray or Auger electron. A free electron cannot absorb a photon (energy and momentum cannot both be conserved), so the atom must take the recoil and the tightly bound K electrons dominate. The cross-section, \(\sigma_{\rm pe}\propto Z^{4\text{–}5}/E_\gamma^{3.5}\), depends strongly on the material, shows absorption edges at the shell binding energies, and dominates at low energy and high \(Z\); it gives the full-energy peak of a detector.

(c) Compton scattering

For photon energies well above the binding energy the electron is effectively free, and conservation of energy and momentum gives \[\boxed{\,E' = \frac{E}{1 + \dfrac{E}{m_ec^2}(1-\cos\theta)},\qquad T_{\max} = \frac{2E^2}{m_ec^2 + 2E}\,}\] equivalently \(\lambda'-\lambda = (h/m_ec)(1-\cos\theta)\). The recoil electrons form a continuum ending at the Compton edge \(T_{\max}\) (at \(\theta = 180^\circ\)); photons backscattered from nearby material give a peak at \(E'_{\min}\to m_ec^2/2\) for \(E\gg m_ec^2\). The cross-section per atom is \(Z\) times that per electron, so Compton attenuation per gram is nearly material-independent; it dominates around 1 MeV.

Example 7.3 For the 662 keV \(\gamma\) rays of \({}^{137}\mathrm{Cs}\), find the Compton edge and the backscatter energy.

Solution. \(E/m_ec^2 = 1.296\), so \(E'_{\min} = 662/(1 + 2.591) = 184\) keV and \(T_{\max} = 662 - 184 = 478\) keV.

(d) Pair production and the relative importance of the processes

A photon above \(2m_ec^2 = 1.022\) MeV can create an \(e^+e^-\) pair in the field of a nucleus, which takes up the recoil; a photon alone cannot (its invariant mass is zero). The threshold in the field of a particle of mass \(M\) is \(2m_ec^2(1 + m_e/M)\), that is 1.022 MeV for a nucleus and \(4m_ec^2\) near an electron. The pair shares \(h\nu - 2m_ec^2\), the cross-section grows as \(Z^2\) and dominates above about 5 MeV in lead, and annihilation of the positron gives escape peaks at \(h\nu - 511\) and \(h\nu - 1022\) keV. The total coefficient \(\mu = \mu_{\rm pe} + \mu_{\rm C} + \mu_{\rm pair}\) therefore falls steeply, passes through a minimum (about 3–4 MeV in lead) and rises slowly. In short: the photoelectric effect involves bound electrons and depends on the specific material, Compton scattering involves loosely bound or free electrons, pair production needs at least 1.02 MeV, and bremsstrahlung and annihilation are the inverses of the photoelectric effect and pair production.

7.3 Interaction and detection of neutrons

Neutrons interact only with nuclei, so they are detected through the charged products of nuclear reactions or recoils. Thermal neutrons have \(E\approx k_BT = 0.025\) eV, \(v\approx2200\) m s\(^{-1}\) and de Broglie wavelength \(\lambda = h/m_nv\approx1.8\) Å (\(\lambda\,[\text{Å}] = 0.286/\sqrt{E\,[\text{eV}]}\)), comparable with interatomic spacings, so they are used for crystal and magnetic structure determination; fast (fission) neutrons have about 2 MeV.

A neutron is slowed most effectively by elastic collisions with light nuclei. With a nucleus of mass number \(A\) the energy after a head-on collision is least, \[\boxed{\,E'_{\min} = \alpha E,\qquad \alpha = \left(\frac{A-1}{A+1}\right)^2,\qquad \xi = 1 + \frac{\alpha\ln\alpha}{1-\alpha}\,}\] where \(\xi\) is the mean loss of \(\ln E\) per collision (\(\xi = 1\) for H, 0.725 for D, 0.158 for C), so \(n = \ln(E_0/E)/\xi\) collisions are needed. A good moderator (water, heavy water, graphite, beryllium, paraffin) has a large \(\xi\Sigma_s\) and a small absorption; heavy nuclei such as cadmium cannot moderate.

At low energies capture cross-sections follow the \(1/v\) law, and some nuclei have enormous thermal values: \({}^{113}\)Cd (natural Cd about 2500 b, so a thin sheet absorbs all neutrons below 0.5 eV), \({}^{10}\)B (3840 b), \({}^{6}\)Li (940 b) and \({}^{3}\)He (5330 b). Slow neutrons are detected through \({}^{10}\mathrm{B}(n,\alpha){}^{7}\mathrm{Li}\) (\(Q = 2.31\) or 2.79 MeV), \({}^{6}\mathrm{Li}(n,\alpha){}^{3}\mathrm{H}\) (\(Q = 4.78\) MeV) and \({}^{3}\mathrm{He}(n,p){}^{3}\mathrm{H}\) (\(Q = 0.764\) MeV) in the BF\(_3\) and \({}^3\)He proportional counters and in \({}^6\)Li-loaded scintillators (LiI(Eu), Li glass, Li-loaded plastic), whose large \(Q\)-value makes them the most efficient thermal-neutron detectors; ordinary GM counters, silicon detectors and scintillators without a converter are essentially blind to slow neutrons. Fast neutrons are detected by recoil protons in hydrogenous scintillators or after moderation, and their energies are measured by time of flight.

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Chapter 7 Radiation, Detectors, Accelerators and Reactors