LECTURE 6: INTERACTION OF RADIATION WITH MATTER

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1 LCTUR 6: INTRACTION OF RADIATION WITH MATTR All radiation is detected through its interaction with matter! INTRODUCTION: What happens when radiation passes through matter? Interlude The concept of cross-section For a thin target: I N = ( I)( σ )( t) N = # of interactions in the target per unit time I = # of incident particles per unit time σ = cross-section or interaction probability. It is usually expressed in cm 2. t = target thickness expressed in # of target nuclei/cm 2 1 barn = 1 b = 1x10-24 cm 2 xercise A target of 197 Au 52 nm thick is bombarded by a 20 MeV proton beam with an intensity of 1 x 10 8 p/s. If the total reaction cross-section is 1.5 b, what is the number of interactions occurring in the target per second? The density of Au is 19.3 g/cm 3. How does the total cross-section compare with the geometric cross-section of the 197 Au nucleus?

2 III. lectromagnetic Radiation -- Photons A. Sources: lectromagnetic Spectrum 1. Rearrangement of nuclear orbitals: γ-rays 2. Rearrangement of atomic and molecular orbitals: x-rays, uv 3. Annihilation radiation ; e.g., e + e two MeV γ s 4. Bremsstrahlung: electron deceleration 5. Cosmic ray showers B. Interactions 1. Photon: Carriers of lectromagnetic force must interact with electric charge Medium: a. electrons b. protons in nucleus γ - e most probable size argument again 2. Mechanisms a. Photoelectric ffect: γ b. Compton Scattering: γ γ e photon disappears photon scatters c. Pair Production: γ e + e e ± pair produced C. Photoelectric ffect 1. Mechanism: Photon is completely absorbed by a charged particle; all energy γ is transferred to an atomic electron, which is ejected from the atom 2. ON COLLISION STOPS PHOTON γ e (photoelectron); monoenergetic

3 e = γ B ( n ) ; i.e., electron is monoenergetic where B (n ) is electron binding energy for n orbital Photopeak Detector sees electron (instein Nobel Prize) 3. When γ B (n ), λ γ λ e - resonance-like situation; large wave function overlap leads to high absorption probability 4. For γ B P P 5 Z 7 2 γ / Best absorbers: ; heavy elements (Pb) (MeV) D. Compton Scattering: γ > > B (n ) γ e - γ Compton Photon Compton lectron The photon energy is too large to be completely absorbed by the electron

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6 Derivation of Compton scattering: From above we have: m c = p + m0 c m c = p c + m c This is the Relativistic energy momentum- relation: = p c + m c Photons have no rest mass m = 0 0 γ = pγ c = hν = ω Incident photon p γ Scattered photon p γ ' Scattered electron p e From momentum balance: p e 2 p e c 2 = p γ p γ ' = 2 γ + nergy balance: 2 γ ' 2 γ γ ' 2 γ + me c = γ ' + e + Scattered photon: cos ( θ ) ( p c m c ) 1/ 2 e

7 γ ' = γ 1+ 2 mec γ ( 1 cos( θ )) Recall that c = h so that λ hc λ = What is the change in the wavelength of the scattered photon as compared to the incident photon? 1 1 λ' λ h = hc = ( 1 cosθ ) γ ' γ mec Notice that when cos(θ) = 1 that is θ=0, there is no difference between the scattered wavelength and the original wavelength. In this case no momentum is transferred to the electron. The maximum change in wavelength occurs when cos(θ) = -1 that is the gamma ray (photon) is scattered to 180 degrees. The quantity h mec is called the Compton wavelength of the electron. 1. ffects of Scattering Process a. Photon energy is degraded, but photon survives b. lectron is propagated in medium c. Multiple collisions required to stop photon d. Scattering angle can become large 2. nergy spectrum P()

8 3. Probability: P c P c Z/ i.e., P c is a function of the number of electrons (Z) in an element (e sea); Again, high Z is best.. Pair Production: For γ > MeV only e e + Mechanism: Same as in γ-decay, except photon originates outside nucleus 1. Result: e ± is formed ; primary γ disappears 2. e ± is thermalized in medium e e cation -e pairs e + + e annihilation two, MeV γs 3. Net: photon energy is degraded, but two lower-energy photons are produced (0.511 MeV), these then undergo Compton scattering and photoelectric effect. 4. Probability: P pp P pp Z 2 log γ (MeV) ; γ > MeV a. High Z best b. Pair production is only stopping mechanism that becomes more probable as energy increases.

9 Aluminum Copper Lead

10 F. Absorbtion of Photons 1. Net effect of mechanisms: Heavy elements are best absorbers Detectors collect electrons (or light converted to electrons). 2. Probability of Absorbtion: µ absorbtion coefficient. Stopping mechanism is random; photon ranges are indeterminate ; e.g., it may take one, or a series of collisions to make photons disappear due to multiple stopping mechanisms µ = µ P + µ c + µ pp di = µ I dx, where I is the photon intensity b. For MONONRGTIC PHOTONS I = I 0 e µd, where d is the absorber thickness (Beer-Lambert Law) c. Half-thickness: d 1/2 d 1/2 = 0.693/µ, where the half-thickness is the amount of material required to reduce the intensity to one-half of its original value. 3. Critical absorbtion of x-rays Bohr atom Quantum #s a. Definitions: K, L, M. Transitions to n = 1, 2, 3 levels α, β, γ. n = 1, 2, 3 1, 2, 3. = 0, 1, 2 b. xample: 2s 1s = K α1 4p 2s = L β2 c. x-ray mission nergetics

11 xray = B (n ) B (n ), where n < n and B is electron binding energy d. Critical absorbtion When γ > B, µ is very high Critical absorber an atom cannot absorb its own x-rays efficiently e. Problem: What is the best absorber for Zn K 1 x-rays? Z Fe Ni Cu Zn B (1s) B (2s) Kα critical absorber kev Z n K α1 x-ray IV. Neutrons A. Sources: Fission, reactors, (< n > ~ 0.5 MeV) Accelerators Thermonuclear explosions (neutron bombs) B. Interactions 1. Properties: M n M H 1u NO LCTROMAGNTIC Z n = 0 INTRACTION & neglibible ionization 2. Collisions: only nuclear collisions are important ; billiard balls σ nucleus 10 8 σ atom

12 C. Stopping Mechanisms Reactor Design 1. STP I: Thermalization lastic and inelastic scattering from nuclei a. nergy Loss: Light nuclei are best materials for reducing energy with fewest collisions b. Average number of collisions n Thermal energy: th < th > = 3/2 kt, where k = ev/k at T = 300 K < th > = 0.04 ev, or ~ 3 km/s THRMAL NUTRONS For H, <n> ~20 U, <n> ~ STP II: Neutron Capture a. σ = f (1/v) i.e., the reaction probability increases as neutron slows down

13 n target low energy (velocity) neutrons captured most effectively Question: Why are low velocity neutrons captured more effectively? b. limination of Neutrons Choose materials with large neutron-capture cross sections: e.g., 10 B(n,α) : σ = 3838 b ; 113 Cd(n,γ) ; σ = b; 157 Gd(n,γ): σ = b c. Preservation of Neutrons Want small σ (n,x) cross sections (reactor moderators). Parafin (~ 100 collisions for thermalization)

14 1 n + 1 H 2 H ; σ = b 1 n + 2 H 3 H ; σ = b (synthesis for nuclear weapons) 1 n + 12 C 13 C; σ = b Water: n + 16 O ; σ = b 4 He: σ = 0 (a bit expensive) 3. Neutron Detection a. Scatter from hydrogenous materials n + 1 H n + H Detect 1 H b. Nuclear reactions (B or Gd loaded plastics) 1 n + 10 B 7 Li + 4 He Detect charged particles