EEE4101F / EEE4103F Radiation Interactions & Detection 1. Interaction of Radiation with Matter Dr. Steve Peterson 5.14 RW James Department of Physics University of Cape Town steve.peterson@uct.ac.za March 02, 2015 EEE4101/3F :: Radiation Interactions & Detection 1 / 70 Dr. Steve Peterson
Interaction of Radiation with Matter Charged Particles Electrons Gamma Rays Neutrons EEE4101/3F :: Radiation Interactions & Detection 2 / 70 Dr. Steve Peterson
References J. S. Lilley, Nuclear Physics: Principles and Applications, Wiley 2001 - [Ch 5] W. R. Leo, Techniques for Nuclear and Particle Physics Experiments (2nd ed.), Springer 1994 - [Ch 2] G. F. Knoll, Radiation Detection and Measurement (4th ed.), Wiley 2010 - [Ch 2] EEE4101/3F :: Radiation Interactions & Detection 3 / 70 Dr. Steve Peterson
Interaction of Radiation with Matter Nuclear radiation consists of energetic particles or photons. Their interaction with matter lies at the heart of nuclear physics and its application in other areas, like detection, power production and medical radiation therapy. Radiation can potentially be dangerous, particularly to living tissue, but these effects depend greatly on the intensity, energy and type of the radiation as well as the nature of the absorbing material. EEE4101/3F :: Radiation Interactions & Detection 4 / 70 Dr. Steve Peterson
Interaction of Radiation with Matter We will look at how three forms of radiation interact with matter: Charged Particles - Protons, alphas, electrons, positrons Photons (electromagnetic radiation) - Gamma, X-rays Neutral Particles - Neutrons, neutrinos Each of these interact with matter in different ways, but they all see matter in terms of its basic constituents, i.e., as an aggregate of electrons and nuclei. EEE4101/3F :: Radiation Interactions & Detection 5 / 70 Dr. Steve Peterson
Interaction of Radiation with Matter EEE4101/3F :: Radiation Interactions & Detection 6 / 70 Dr. Steve Peterson
Interaction of Charged Particles with Matter Two principal features characterize the passage of charged particles through matter: 1. Continuous loss of energy by the particle 2. Deflection of the particle from its incident direction. Matter consists of atoms - negatively charged electrons bound to the positively charged nucleus by an electric field These effects are primarily the result of two processes: 1. Inelastic collisions with the atomic electrons of the material 2. Elastic scattering from nuclei Of these processes, the inelastic collisions are primarily responsible for the energy loss of charged particles in matter. EEE4101/3F :: Radiation Interactions & Detection 7 / 70 Dr. Steve Peterson
Interaction of Charged Particles with Matter Although all charged particles interact with matter in a similar fashion (primarily by electromagnetic forces), we will split them into two groups: Heavy Charged Particles (protons, muons, pions, alpha) Light Charged Particles (electrons, positrons) Electrons behave considerably different because of their low mass and generally higher speeds EEE4101/3F :: Radiation Interactions & Detection 8 / 70 Dr. Steve Peterson
Heavy Charged Particle Energy Loss A nuclei occupies about 10 15 of the volume of the atom, thus it is roughly 10 15 times more probable for the particle to collide with an electron than with a nucleus. Although the inelastic collisions are more frequent, they do not transfer much energy. Using conservation of energy and momentum, a head-on collision between an ion (mass M and energy E) and an electron (mass m and initially at rest) gives a loss of kinetic energy of the ion as: E = E( 4m M ) (1) For example, a 5-MeV α particle would lose about 2.7 kev in a head-on collision with an electron. This is a maximum value, the typical energy loss would much smaller. EEE4101/3F :: Radiation Interactions & Detection 9 / 70 Dr. Steve Peterson
Heavy Charged Particle Energy Loss Although only a small amount of energy is transferred (less than a few kev), the large number of collisions per unit length can produce a substantial cumulative energy loss, even in thin layers of material. For example, A 10 MeV proton will lose all of its energy in only 0.25 mm of copper. In these collisions, energy is transferred from the particle to the atom by causing ionization (i.e. remove an electron) or excitation in the atom. A typical ionization energy is 10 ev, so most collisions will transfer enough energy to ionize. The result is a trail of ionization and excitation of atoms and molecules along the path of the moving particle. EEE4101/3F :: Radiation Interactions & Detection 10 / 70 Dr. Steve Peterson
Heavy Charged Particle Energy Loss EEE4101/3F :: Radiation Interactions & Detection 11 / 70 Dr. Steve Peterson
Charged Particle Range As the particle travels through the material, it will eventually lose all of its energy, reaching a distance called the range of the particle. EEE4101/3F :: Radiation Interactions & Detection 12 / 70 Dr. Steve Peterson
Heavy Charged Particle Energy Loss The classical average energy loss per unit pathlength, often called the stopping power or simply de/dx is: de [ ] e 2 2 dx = 2z 2 4πɛ 0 mv N 2 e ln b max (2) b min where e: charge of electron; z: charge of particle; v: speed of particle; m: mass of electron; N e : electron density of stopping material; b max and b min : short and long cutoff distances (impact parameters) EEE4101/3F :: Radiation Interactions & Detection 13 / 70 Dr. Steve Peterson
Bethe-Bloch Equation A more complete energy loss formula due to Bethe and Bloch, correcting for relativistic effects and atomic structure. de dx = ( ze2 ) 2 4πZρN A 4πɛ 0 Amv 2 v = βc: ion velocity ze: ion charge m: electron mass N A : Avogadro s number [ln( 2mv2 ) ln(1 β 2 ) β 2 ] (3) I A: atomic mass number Z: atomic number ρ: stopping material density I: mean ionization energy The parameter I should in principle be computed by averaging over all atomic ionization and excitation processes, but in practice is an empirical constant, roughly equal to (11eV )Z. EEE4101/3F :: Radiation Interactions & Detection 14 / 70 Dr. Steve Peterson
Bethe-Bloch Equation Characteristics of the Bethe-Bloch equation: Dependant on Z/A, z 2, and ρ Logarithmic dependence on E Increases in relativistic region A non-relativistic version is: de dx = ( ze2 ) 2 4πZρN A 4πɛ 0 Amv 2 ln( 2mv2 ) (4) I Stopping power is typically quoted in units of energy loss per mass per unit area, which is obtained by dividing de/dx by the density ρ, giving de/ρdx, called the mass stopping power. EEE4101/3F :: Radiation Interactions & Detection 15 / 70 Dr. Steve Peterson
Energy Dependence of Stopping Power A useful approximation (between 100 kev and 1 GeV) is where k 0.8. de dx = constant E k (5) Beyond E/A 1 GeV per nucleon (v 0.96c), de/dx passes through a point of minimum ionization and then begins to slowly rise with the growing importance of the relativistic correction terms (containing β 2 ). EEE4101/3F :: Radiation Interactions & Detection 16 / 70 Dr. Steve Peterson
Energy Dependence of Stopping Power The energy dependence of the stopping power is dominated by the 1/v 2 term. EEE4101/3F :: Radiation Interactions & Detection 17 / 70 Dr. Steve Peterson
Projectile Dependence of Stopping Power For particles in the same material, the Bethe-Bloch formula reduces to: de dx = z2 f(v) (6) where f(v) is a function of the particle velocity only, so the energy loss in a given material is only dependent on the charge and velocity of the particle. EEE4101/3F :: Radiation Interactions & Detection 18 / 70 Dr. Steve Peterson
Bragg curve As shown above, the rate of energy loss ( de/dx) increases as the particle energy decreases, so the number of ions produced per unit length (ionization density) in the medium will also increase. [α in air] EEE4101/3F :: Radiation Interactions & Detection 19 / 70 Dr. Steve Peterson
Calculation of Range The stopping power equation can be used to calculate the pathlength of an ion with energy E, which in turn can approximate the range R of the ions in a given material. R = dx = 0 E de (de/dx) Due to the large mass difference between the heavy charged particle and the electrons, any potential deflection from a glancing collision is effectively zero, producing a nearly straight-line path. As a result, the range and the pathlength of the ions are essentially identical. (7) EEE4101/3F :: Radiation Interactions & Detection 20 / 70 Dr. Steve Peterson
Projectile Dependence of Stopping Power A corresponding expression for range can be found using de = mvdv 0 de R = (de/dx) m F (v) (8) z2 E for a given energy per nucleon (speed), de/dx and R vary as z 2 and m/z 2, respectively. For example, a 40-MeV α particle has four times the stopping power and the same range as a 10-MeV proton. A useful empirical relationship for estimating relative ranges of an ion in materials with different mass numbers (A) and densities (ρ) is the Bragg-Kleeman rule: R 1 A1 ρ 2 (9) R 2 ρ 1 A2 EEE4101/3F :: Radiation Interactions & Detection 21 / 70 Dr. Steve Peterson
Range Straggling Due to the statistical nature of the stopping process, there will be a spread in the observed range of monoenergetic particles; this roughly Gaussian phenomenon is called range straggling. It is not a large effect, produces about a 1% range deviation for a 5-MeV α particle. EEE4101/3F :: Radiation Interactions & Detection 22 / 70 Dr. Steve Peterson
Interaction of electrons with matter Like heavy charged particles, electrons lose energy to atomic electrons via electronic interactions. However, due to their small mass, for a given energy, their speeds are greater. As a result, de/dx is much smaller and electrons are more penetrating than heavy ions. For example, the range of a 1-MeV electron in aluminum is about 1800 µm compared to about 3 µm for an α particle with the same energy. EEE4101/3F :: Radiation Interactions & Detection 23 / 70 Dr. Steve Peterson
Interaction of electrons with matter Some other consequences of electron s smaller mass: 1. An electron will lose a much greater fraction of its energy in a single collision than does a heavy ion. 2. Electrons will suffer large deflections in collisions with other electrons, and therefore follow erratic paths. The range (linear distance of penetration) will be very different from the electron pathlength. 3. Electrons are more likely to experience sudden changes in direction and speed causing the emission of electromagnetic (bremsstrahlung) radiation. EEE4101/3F :: Radiation Interactions & Detection 24 / 70 Dr. Steve Peterson
Electron Energy Loss Electrons lose energy through two processes, due to collisions with electrons and due to radiated energy losses. The total energy loss of electrons and positrons, therefore, is composed of two parts: ( ) ( ) ( ) de de de = + dx tot dx coll dx rad (10) EEE4101/3F :: Radiation Interactions & Detection 25 / 70 Dr. Steve Peterson
Electron Energy Loss Solid lines: Contribution due to collisions with electrons Dashed lines: Contribution due to radiation loss (strongly dependent on Z) Critical energy: where two terms are equal EEE4101/3F :: Radiation Interactions & Detection 26 / 70 Dr. Steve Peterson
Electron Range The significant probability of large-angle scattering slowly removes electrons from the incident direction, producing a slow decrease in the intensity. [in contrast to the α particles] The extrapolated range R e for electrons is determined by extending the linear part of the transmission curve. EEE4101/3F :: Radiation Interactions & Detection 27 / 70 Dr. Steve Peterson
Example: Stopping Power Find the approximate energy loss of 1 MeV alpha particles in a thickness of 5 µm of gold. EEE4101/3F :: Radiation Interactions & Detection 28 / 70 Dr. Steve Peterson
Interaction of photons with matter Photons interact with matter quite differently from charged particles, primarily due to the lack of electric charge. The main interactions are: 1. Photoelectric Effect 2. Compton Scattering (including Thomson and Rayleigh Scattering) 3. Pair Production EEE4101/3F :: Radiation Interactions & Detection 29 / 70 Dr. Steve Peterson
Interaction of photons with matter These reactions explain the two principal qualitative features of x-rays and γ-rays: 1. X-rays and γ-rays are many times more penetrating in matter than charged particles. 2. A beam of photons is not degraded in energy as it passes through a thickness of matter, only attenuated in intensity. EEE4101/3F :: Radiation Interactions & Detection 30 / 70 Dr. Steve Peterson
Interaction of photons with matter Why more penetrating? Due to the much smaller cross section of the three processes relative to the inelastic electron collision cross section Why not energy degraded? Due to the fact the three photon processes remove the photon from the beam entirely, either by absorption or scattering. The photons which pass straight through have not suffered any interactions at all and retain their original energy. EEE4101/3F :: Radiation Interactions & Detection 31 / 70 Dr. Steve Peterson
Photoelectric Effect The photoelectric effect involves the absorption of a photon by an atomic electron with the subsequent ejection of the electron from the atom. The kinetic energy of the outgoing photoelectron is then where B e is the binding energy of the electron. T = E γ B e (11) Note: Free electrons cannot absorb a photon and recoil; a heavy atom is necessary to absorb the momentum at little cost in energy. EEE4101/3F :: Radiation Interactions & Detection 32 / 70 Dr. Steve Peterson
Photoelectric Effect K-edge: excitation of a tightly-bound electron from the K-shell (only possible above a certain threshold energy value). The atom which has lost the electron (typically the most tightly bound electrons) may de-excite in one of two ways 1. Emitting Auger electrons Other, less tightly bound electrons are released from the atom 2. X-ray fluorescence Electron from outer shell fill vacancy in inner shell with emission of a characteristic x-ray EEE4101/3F :: Radiation Interactions & Detection 33 / 70 Dr. Steve Peterson
Photoelectric Effect EEE4101/3F :: Radiation Interactions & Detection 34 / 70 Dr. Steve Peterson
Compton Scattering Compton scattering is the process by which a photon (E γ ) scatters from a nearly free atomic electron, resulting in a less energetic photon (E γ) and a scattered electron carrying the energy (T ) lost by the photon. EEE4101/3F :: Radiation Interactions & Detection 35 / 70 Dr. Steve Peterson
Compton Scattering Using relativistic kinematics, energy conservation gives for the kinetic energy of the electron: T = E γ E γ = E mc 2 (12) where E is the total energy of the recoil electron including its rest mass energy mc 2. Momentum conservation requires that the momentum be added vectorially, where p γ = E γ /c EEE4101/3F :: Radiation Interactions & Detection 36 / 70 Dr. Steve Peterson
Compton Scattering Using the cosine rule, we can write (pc) 2 = E 2 γ + (E γ) 2 2E γ E γ cos θ = E 2 m 2 c 4 (13) using the relation E 2 = p 2 c 2 + m 2 c 4 Eliminating E from the equations above, we obtain the Compton-scattering formula for the scattered photon energy: E γ = E γ 1 + (E γ /mc 2 )(1 cos θ) (14) EEE4101/3F :: Radiation Interactions & Detection 37 / 70 Dr. Steve Peterson
Compton Scattering The photon energy varies from the maximum value of E γ (for θ = 0 o ) to a minimum value of E γ(min) at θ = 180 o ; E γ(min) approaches mc 2 /2 ( 0.25MeV ) when E γ is large. The corresponding electron kinetic energy varies from near zero for a glancing collisions to a maximum (always less than E γ when θ = 180 o. The probability of Compton scattering is less strongly dependant than the photoelectroc effect on E γ and Z. EEE4101/3F :: Radiation Interactions & Detection 38 / 70 Dr. Steve Peterson
Pair Production The process of pair production involves the transformation of a photon into an electron-positron pair. In order to conserve momentum, this can only occur in the presence of a third body (like with photoelectric effect), usually a nucleus. Moreover, to create the pair, the photon must have at least an energy of 1.022 MeV. EEE4101/3F :: Radiation Interactions & Detection 39 / 70 Dr. Steve Peterson
Pair Production The total kinetic energy of the electron-positron pair is given by: T + T + = E γ 2mc 2 (15) where T and T + are the energies of the electron and positron Due to the dependence on the presence of a nucleus, there is some Z dependence. There is an energy threshold of 2mc 2 = 1.022MeV, which makes the pair production cross-section unimportant until E γ is several MeV. EEE4101/3F :: Radiation Interactions & Detection 40 / 70 Dr. Steve Peterson
Pair Production Only part of the γ-ray energy is converted into kinetic energy of charged particles; the rest goes into the rest masses of the electron and positron. The positron is an anti-electron, which will slow down, and as it comes to rest will be attracted to an electron (not the same one), and then annihilate. The annihilation converts the rest masses of the electron and positron into two 0.511 MeV γ rays, emitted in opposite directions. EEE4101/3F :: Radiation Interactions & Detection 41 / 70 Dr. Steve Peterson
Interaction of photons with matter The region of dominance for the three photon interactions EEE4101/3F :: Radiation Interactions & Detection 42 / 70 Dr. Steve Peterson
Photon Absorption The total number of photons is reduced by the number which have interacted. The attenuation suffered by a photon beam can be shown, in fact, to be exponential with respect to the thickness, i.e., I(x) = I o exp( µx) (16) where I o : incident beam intensity; x: thickness of absorber; µ: linear absorption coefficient. EEE4101/3F :: Radiation Interactions & Detection 43 / 70 Dr. Steve Peterson
Photon Absorption The total probability for a photon interaction in matter is the sum of the individual photon process cross sections: σ total = σ photo + σ compton + σ pair (17) The linear absorption coefficient µ (units: cm 1 ) is related to cross section σ total by: µ = Nσ total = σ(n a ρ/a) (18) where N: Nuclear number density, N a : Avogadro s Number; ρ: density of the material; A: molecular weight or mass number. EEE4101/3F :: Radiation Interactions & Detection 44 / 70 Dr. Steve Peterson
Attenuation of gamma rays EEE4101/3F :: Radiation Interactions & Detection 45 / 70 Dr. Steve Peterson
Energy deposition of gamma rays EEE4101/3F :: Radiation Interactions & Detection 46 / 70 Dr. Steve Peterson
Photon Absorption The mass attenuation coefficient µ m is defined as the linear attenuation coefficient divided by the density of the medium: µ m = µ/ρ (units: cm 2 /g) For compounds and mixtures, the total attenuation coefficient may be calculated using Bragg s rule, µ ρ = i w i ( µ ρ ) i (19) EEE4101/3F :: Radiation Interactions & Detection 47 / 70 Dr. Steve Peterson
Photon Absorption EEE4101/3F :: Radiation Interactions & Detection 48 / 70 Dr. Steve Peterson
Example: Photon Absorption What thickness of concrete (ρ = 2200 kg m 3 ) is needed to attenuate a collimated beam of 1-MeV γ rays by a factor of 10 6? The mass attenuation coefficient of concrete is µ m = 0.064 cm 2 g 1. EEE4101/3F :: Radiation Interactions & Detection 49 / 70 Dr. Steve Peterson
Photon Absorption We can also calculate the mean distance, λ, traveled by the particle without suffering a collision. This is known as the mean free path. Thus, λ = xp (x)dx P (x)dx = xio exp( µx)dx Io exp( µx)dx = 1 µ (20) EEE4101/3F :: Radiation Interactions & Detection 50 / 70 Dr. Steve Peterson
Interaction of neutrons with matter Neutrons interact in matter via nuclear reactions. The type of nuclear reaction involved depends strongly on: The energy of the neutron The particular nuclei with which the neutrons collides The two primary forms of collisions are: Scatter - Energy from the neutron is transferred to the recoiling (charged) nucleus Absorption - Neutron is captured by the nucleus EEE4101/3F :: Radiation Interactions & Detection 51 / 70 Dr. Steve Peterson
Interaction of neutrons with matter Scattering slows the neutron down as energy is lost in successive collisions. Two possibilities are: Elastic scattering from the nuclei, i.e. A(n, n)a. This is the principal mechanism of energy loss for neutrons in the MeV region. Inelastic scatter, e.g. A(n, n )A, etc. In this reaction, the nucleus is left in an excited state and may later decay by gamma-ray of some other form of radiative emission. In order to occur, the neutron must have sufficient energy to excite the nucleus. Below this energy threshold, only elastic scatter may occur. EEE4101/3F :: Radiation Interactions & Detection 52 / 70 Dr. Steve Peterson
Interaction of neutrons with matter Possible absorption reactions: Radiative neutron capture, i.e. n + (Z, A) γ + (Z, A + 1). In general, the cross-section for neutron capture goes approximately as 1/v where v is the velocity of the neutron. Absorption is most likely at low energies. Other nuclear reactions, such as (n, p), (n, d), (n, α), etc. in which the neutron is captured and charged particles are emitted. Fission, i.e. (n, f). Most likely at thermal energies. High energy hadron shower production - only for very high energy neutrons (E > 100MeV ) EEE4101/3F :: Radiation Interactions & Detection 53 / 70 Dr. Steve Peterson
Interaction of neutrons with matter Because of the strong energy dependence of neutron interactions, it is customary to classify neutrons according to their energy: High energy neutrons (E > 100MeV ) Fast neutrons (E 100keV 100MeV ) Epithermal neutrons (E 0.1eV 100keV ) Thermal neutrons (E 0.025eV ) Thermal neutrons are a unique type of radiation because when it is captured, it releases many MeV of energy, making them easily detectable. EEE4101/3F :: Radiation Interactions & Detection 54 / 70 Dr. Steve Peterson
Interaction of neutrons with matter The total probability for a neutron to interact in matter is given by the sum of the individual cross sections, i.e. σ tot = σ elastic + σ inelastic + σ capture +... (21) EEE4101/3F :: Radiation Interactions & Detection 55 / 70 Dr. Steve Peterson
Neutron Absorption When traversing a slab of material, neutrons, like γ rays, can be absorbed or scattered through large angles, contributing to a loss in transmitted intensity I = I o exp( Nσx) (22) assuming a mono-energetic, well-collimated beam. It can also be written as: I = I o exp( Σx) = I o exp( x/λ) (23) where Σ = Nσ is called the macroscopic total cross section, and λ = 1/Σ is the mean attenuation length. EEE4101/3F :: Radiation Interactions & Detection 56 / 70 Dr. Steve Peterson
Neutron Absorption The attenuation length (λ) can be shown to be equal to the mean free path (the average distance traveled by a neutron before it interacts in the medium) and broken into two pieces: The scattering mean free path λ s = 1/Σ s The average distance between successive scatterings The absorption mean free path λ a = 1/Σ a The average distance before the neutron is absorbed Therefore, the total mean free path is given by: 1 λ = 1 λ s + 1 λ a (24) EEE4101/3F :: Radiation Interactions & Detection 57 / 70 Dr. Steve Peterson
Neutron Absorption The life of a neutron as it travels through material is complex, even at low energies, depending strongly on the relative magnitudes of the scattering and absorption cross-sections. If σ a >> σ s, then before any scattering occurs, neutrons are removed (attenuated) according to equation (22), where σ tot σ a. If σ s >> σ a, then many scatterings occur with successive energy losses, slowing the neutron to low, possibly thermal energies before it is captured. This slowing-down process is called moderation, a critical process in thermal fission reactors. EEE4101/3F :: Radiation Interactions & Detection 58 / 70 Dr. Steve Peterson
Neutron moderation Consider a non-relativistic, elastic collision between a neutron (mass m), with energy E 0 and speed v 0, incident on a target nucleus (mass M) initially at rest. In the lab frame, the scattered neutron as energy E 1 and speed v 1. EEE4101/3F :: Radiation Interactions & Detection 59 / 70 Dr. Steve Peterson
Neutron moderation In the c-m frame, the centre of mass is at rest and, therefore, the speed of the target nucleus w is equal to V cm, where mv o = (m + M)V cm. Also, since kinetic energy is conserved in an elastic collision, the initial and final speeds of m and M in the c-m frame (v and w ) and unchanged by the collision. EEE4101/3F :: Radiation Interactions & Detection 60 / 70 Dr. Steve Peterson
Neutron moderation Using law of cosines, v 2 1 = (v ) 2 + V 2 cm + 2v V cm cos θ cm (25) the lab energy of the scattered neutron is E 1 = E 0 ( M 2 + m 2 + 2Mm cos θ cm (M + m) 2 Using m = 1 and M = A gives E 1 = E 0 ( A 2 + 1 + 2A cos θ cm (A + 1) 2 ) ) (26) (27) EEE4101/3F :: Radiation Interactions & Detection 61 / 70 Dr. Steve Peterson
Neutron moderation The energy of the scattered neutrons is limited to the range αe 0 < E 1 < E 0, where α = ( ) 2 A 1 (28) A + 1 In the particular case of scattering protons, A = 1, then the range is 0 < E 1 < E 0. Intuitively, the lighter the nucleus, the more recoil energy it absorbs from the neutron. This implies the the slowing down of neutrons is most efficient when protons or light nuclei are used, such as water or paraffin (CH 2 ). EEE4101/3F :: Radiation Interactions & Detection 62 / 70 Dr. Steve Peterson
Neutron moderation To determine the effect of scattering on the average neutron energy, we can calculate the probability distribution P (E 1 )de 1 of the scattered neutron having energy between E 1 and E 1 + de 1. From equation (27), there is a direct correlation between E 1 and θ cm. Thus, P (θ cm )dθ cm = P (E 1 )de 1 (29) where the negative sign allows for the fact that the energy decreases as the angle increases. For neutron energy below 15 MeV, the scatter angular distribution is isotropic in the c-m system, giving P (θ cm )dθ cm = 2π sin θ cm dθ cm = 1 4π 2 sin θ cmdθ cm (30) EEE4101/3F :: Radiation Interactions & Detection 63 / 70 Dr. Steve Peterson
Neutron moderation From equation (27), we get and P (E 1 )de 1 = de 1 dθ cm = 2AE 0 sin θ (A + 1) 2 (31) (A + 1)2 4AE 0 de 1 = de 1 (1 α)e 0 (32) After one scatter, the energy distribution of an originally mono-energetic neutron is constant over the energy range αe 0 to E 0. EEE4101/3F :: Radiation Interactions & Detection 64 / 70 Dr. Steve Peterson
Neutron moderation The energy distribution of singly scattered neutrons P (E 1 )de 1 = (A + 1)2 4AE 0 de 1 (33) EEE4101/3F :: Radiation Interactions & Detection 65 / 70 Dr. Steve Peterson
Neutron moderation After the first scatter, the average neutron energy is E 1 = 1 2 (1 + α)e 0 and the average energy loss per collision is 1 2 (1 α)e 0 (independent of initial energy). After a second collision, each of the neutrons in the distribution loses on average the same fraction of its energy again, and so on. After n collisions, the average energy becomes E n = E 0 ( E1 E 0 ) n (34) Unfortunately, this misrepresents the actual neutron energy distribution. EEE4101/3F :: Radiation Interactions & Detection 66 / 70 Dr. Steve Peterson
Neutron moderation Energy distribution of neutrons after several elastic scatters EEE4101/3F :: Radiation Interactions & Detection 67 / 70 Dr. Steve Peterson
Neutron moderation Instead of using average energy loss, the distribution is better represented by the logarithmic energy decrement ξ. [ ] E0 ξ = ln E 0 ln E 1 = E 1 average (35) and after n collisions, the average value of ln E 0 is given by ln E n = ln E 0 nξ (36) EEE4101/3F :: Radiation Interactions & Detection 68 / 70 Dr. Steve Peterson
Neutron moderation Using equation (27) and integrating over the solid angle, we get ( ) (A 1)2 A 1 ξ = 1 + 2A ln (37) A + 1 For a neutron to slow down from energy E 0 to an energy E, the total number of collisions n required would be n = 1 ξ ln E 0 E (38) EEE4101/3F :: Radiation Interactions & Detection 69 / 70 Dr. Steve Peterson
Neutron moderation Lilley Table 5.1 Scattering properties of several nuclei Nucleus α ξ n (to thermalize) 1 H 0 1.00 18 2 H 0.11 0.725 25 4 He 0.360 0.425 43 12 C 0.716 0.158 115 238 U 0.983 0.0084 2200 EEE4101/3F :: Radiation Interactions & Detection 70 / 70 Dr. Steve Peterson