A Tutorial on Radiation Dose and Dose Rate Kurt Sickafus Dose = - PowerPoint PPT Presentation

A Tutorial on Radiation Dose and Dose Rate Kurt Sickafus

Dose = Absorbed Energy Density Absorbed energy normalized by weight, volume, atoms, etc. 1 Gy = 1 J kg SI units 2

Water: heat to boiling point J H 2 O = 4.1813 c p g ⋅ K (@ 25°C) specific heat of water Δ T = 80 K g × 10 3 g H 2 O Δ T = 334.5 J c p kg = 3.345 ⋅ 10 5 J kg = 0.3345 MGy Absorbed Energy 3

Projectile-Target Interactions # events <volume> or <weight> = ρ σ ϕ t • • •

Projectile-Target Interactions atomic cross- • • flux • time density section ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ # events atoms area projectiles [ ] volume = ρ a ⎦ σ ⎦ ϕ ⎦ t time ⎣ ⎣ ⎣ area i time volume atom ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ # events atoms area projectiles [ ] weight = ρ w ⎦ σ ⎦ ϕ ⎦ t time ⎣ ⎣ ⎣ area i time weight atom

Projectile-Target Interactions fluence = flux • time ⎡ ⎤ ⎡ ⎤ Φ projectiles ⎥ = ϕ projectiles [ ] ⎥ t time ⎢ ⎢ ⎣ ⎦ ⎣ ⎦ area i time area

Projectile-Target Interactions atomic cross- • • fluence density section ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ # events atoms area projectiles volume = ρ a ⎦ σ ⎦ Φ ⎣ ⎣ ⎣ ⎦ volume atom area ⎡ ⎤ ⎡ ⎤ ⎡ ⎤ # events atoms area projectiles weight = ρ w ⎦ σ ⎦ Φ ⎣ ⎣ ⎦ ⎣ weight atom area

Projectile-Target Interactions cross- • fluence section # events ⎡ ⎤ ⎡ ⎤ area projectiles = σ ⎦ Φ volume ⎣ ⎣ ⎦ ⎡ ⎤ atoms atom area ρ a ⎣ ⎦ volume

Projectile-Target Interactions Leading to Atomic Displacements displacement dpa = • fluence cross- section # atomic displacements ⎡ ⎤ ⎡ ⎤ area projectiles = σ ⎦ Φ volume ⎣ ⎣ ⎦ ⎡ ⎤ atoms atom area ρ a ⎣ ⎦ volume Ballistic ⎡ ⎤ ⎡ ⎤ displacements area projectiles = σ ⎦ Φ ⎣ ⎣ ⎦ Dose atom atom area

Electron irradiation-induced amorphization of powellite (CaMoO 4 ) 300 keV electrons room-temperature irradiation conditions

Electron irradiation-induced amorphization of powellite (CaMoO 4 ) Two components of damage: 1. electronic component (electron excitation/ionization; radiolysis) 2. nuclear component (ballistic or displacement damage)

1 . Electronic Stopping

Electron Excitation/Ionization Bethe-Ashkin expression for ionization energy loss per unit length H. A. Bethe, and J. Ashkin, in Experimental Nuclear Physics. Volume I , edited by E. Segrè (John Wiley & Sons, Inc., New York, 1953), pp. 166-357.

Electron Excitation/Ionization Bethe-Ashkin expression for ionization energy loss per unit length relativistic expression ⎧ ⎫ ⎛ ⎞ E 0 β 2 E ⎪ Ln ⎪ ⎜ ⎟ 2 J 2 (1 − β 2 ) ⎝ ⎠ ⎪ ⎪ ( ) Ln2 ⎪ ⎪ − 2 1 − β 2 − 1 − β 2 dx = 2 π e 4 ρ e ⎪ ⎪ − dE ⎨ ⎬ β 2 E 0 ⎪ ⎪ + 1 − β 2 ⎪ ⎪ ( ) ⎪ ⎪ + 1 2 8 1 − 1 − β 2 ⎪ ⎪ ⎩ ⎭

E 0 = m e c 2 = rest energy of the electron m e = rest mass of the electron c = speed of light e 2 = 14.4 eV ⋅ Å

β = v c v = velocity of electron c = speed of light 2 ⎛ ⎞ E 0 β = 1 − ⎜ ⎟ E 0 + E ⎝ ⎠ E 0 = rest energy of the electron E = kinetic energy of the electron

ρ e = Z ⋅ ρ a ρ e = electron density Z = atomic number ρ a = atomic density

J = 9.76 Z + 58.5 Z − 0.19 (eV) = mean electron excitation potential M. J. Berger, and S. M. Seltzer, Nat. Acad. Sci. / Nat. Res. Council Publ. 1133 (Washington, 1964), p. 205.

Bragg’s Rule for Additivity of Stopping Powers W. H. Bragg, and M. A. Elder, Phil. Mag. 10 , 318 (1905)

Stopping Power ⎛ ⎞ eV ⋅ Å 2 ( ) = 1 dE ε e = S e E ⎜ ⎟ atom ⋅ e − ρ a ⎝ ⎠ dx e

Bragg’s Rule for Additivity of Stopping Powers For binary compound with molecular unit, AmBn : A + n ε A m B n = m ε ε B e e e where m is the number of A atoms in molecule AmBn and n is the number of B atoms in molecule AmBn One can show that: A m B n A B dE A m B n = dE + dE A m B n ε = ρ m dx e dx e dx e e where ρ m A m B n is the molecular density of AmBn molecules in the compound.

Ionization stopping in powellite dE / dx ( E = 300 keV) = -0.0729 dE / dx (eV/Å• e - ) E (eV) electron energy

E = 300 keV β = 0.776526 dE / dx ( E = 300 keV) = -0.0729 eV/Å• e - thickness = 1000 Å TEM sample thickness Total ionization energy loss over sample thickness = 72.9 eV/ e - = 1.17x10 -17 J/ e -

e − e − e − ϕ = 10 7 nm 2 ⋅ s = 10 5 Å 2 ⋅ s = 10 21 cm 2 ⋅ s electron flux t = 5 min. = 300 s irradiation time e − Φ = 3 ⋅ 10 7 Å 2 electron fluence

Areal Energy Density = dE ⋅ Φ dx electronic = 3.504 ⋅ 10 − 10 J Å 2 Total Energy Density = Areal Energy Density thickness = 3.504 ⋅ 10 − 13 J Å 3

ρ w = 4.259 ⋅ 10 − 27 kg Å 3 Dose = 8.2 ⋅ 10 13 J kg = 82 TGy Magnitude of dose: Tens of TeraGray !!

2 . Nuclear Stopping

Electron displacement damage calculation Primary damage cross-section after Seitz & Koehler (1956): F. Seitz, and J. S. Koehler, in Solid State Physics: Advances in Research & Applications , edited by F. Seitz, and D. Turnbull (Academic Press, 1956), pp. 305-448. Based on the relativistic electron cross-section expression derived by McKinley & Feshbach (1948): W. A. McKinley, Jr., and H. Feshbach, Physical Review 74 , 1759 (1948). Total cross-section (primary plus secondaries) after Oen (1973): O. S. Oen, (Oak Ridge National Laboratory, Oak Ridge, TN, 1973), pp. 204.

Differential displacement cross-section, d σ ⎡ ⎤ ⎧ ⎫ d σ ( T ) = π ′ b 2 dT T m 1 − β 2 T T − T ⎢ ⎥ + π ′ α β ⎨ ⎬ 4 T 2 ⎢ T m T m T m ⎥ ⎩ ⎭ ⎣ ⎦ where T is the kinetic energy of the electron 2 ⎛ ⎞ E 0 β = v / c = 1 − ⎜ ⎟ E 0 + E ⎝ ⎠ α = α Z ′ where α is the fine structure constant (~1/137)

2 ⎛ ⎞ e 2 b 2 = 4 Z 2 1 ′ β 4 γ 2 ⎜ ⎟ ⎝ ⎠ E 0 where 1 γ = 1 − β 2

T m = maximum energy transfer from e − to target atom ⎛ ⎞ 4 m e M E T m = 2 E 1 + ⎜ ⎟ ( ) ⎝ ⎠ m e + M 2 E 0 where E is the incident electron energy T m (eV) (maximum energy transfer to target atom) O Ca Mo E (keV) (incident electron energy)

E = 300 keV Z Ca = 20 Ca = 21.245 eV T m Z Mo = 42 Mo = 8.8756 eV T m Z O = 8 O = 53.219 eV T m Z ave = 15.67 ave = 25.54 eV T m

E d = 40 eV Z Ca = 20 = 493 keV Ca E threshold Z Mo = 42 = 920 keV Mo E threshold Z O = 8 = 237 keV O E threshold Z ave = 15.67

E d = 8 eV Z Ca = 20 = 130 keV Ca E threshold Z Mo = 42 = 275 keV Mo E threshold Z O = 8 = 55.3 keV O E threshold Z ave = 15.67

Primary displacement cross-section: ( T ) < area > ⎡ ⎤ T m ∫ σ p ( E ) = d σ ⎢ ⎥ ⎣ ⎦ atom E d where E d is the displacement threshold energy Cascade cross-section: ( T ) < area > ⎡ ⎤ T m ∫ σ tot ( E ) = ν ( T ) d σ ⎢ ⎥ ⎣ ⎦ atom E d where ν ( T ) is the number of secondary displacements, given most simply by the Kinchin-Pease expression: ν ( T ) = 0; T < E d ν ( T ) = 1; E d ≤ T < 2 E d ν ( T ) = T ; T ≥ 2 E d 2 E d

E = 300 keV E d = 25 eV Z ave = 15.67 = 295 keV ave E threshold ave = 25.54 eV T m 2 E d = 50 eV σ tot ( E ) = σ p ( E ) = 0.588 barn = 5.88 ⋅ 10 − 9 Å 2 atom 1 barn = 10 -24 cm 2 = 10 − 8 Å 2

e − e − e − ϕ = 10 7 nm 2 ⋅ s = 10 5 Å 2 ⋅ s = 10 21 cm 2 ⋅ s electron flux t = 5 min. = 300 s irradiation time e − Φ = 3 ⋅ 10 7 Å 2 electron fluence

Total displacement damage dose displacements per atom = σ tot Φ atom × 3 ⋅ 10 7 e − = 5.88 ⋅ 10 − 9 Å 2 Å 2 = 0.18 dpa Presumably, this magnitude of displacement damage is not sufficient to induce a crystal- to-amorphous phase transformation

Displacement cross-section, σ p ( E , E d ) versus e - beam energy, E , and displacement threshold energy, E d , for oxygen (O) target atoms ( Z = 8) σ p (barns/atom) E d (eV) 4 5 6 7 8 910 11 E (keV) Compare this plot to Fig. 6 in P. S. Bell, and M. H. Lewis, Phil. Mag. 29, 1175 (1974).

Displacement cross-section, σ p ( E , E d ) versus e - beam energy, E , and displacement threshold energy, E d , for vanadium (V) target atoms ( Z = 23) E d (eV) 4 5 σ p (barns/atom) 6 7 8 9 10 11 E (keV) Compare this plot to Fig. 6 in P. S. Bell, and M. H. Lewis, Phil. Mag. 29, 1175 (1974).

Dose Rate Ming’s experiment on CaMoO 4 : Dose Rate = 82 TGy = 0.273 T Gy = 273 GGy 300 s s s e − e − e − ϕ = 10 7 nm 2 ⋅ s = 10 5 Å 2 ⋅ s = 10 21 cm 2 ⋅ s electron flux

Dose Rate Field Emission Gun (FEG) Scanning Transmission Electron Microscope (STEM) probe: 1 nA in 1 nm diam. probe e − ϕ = 6.24 ⋅ 10 9 nm 2 ⋅ s e − 6.24 times = 6.24 ⋅ 10 7 Å 2 ⋅ s greater than Ming’s e − experiment = 6.24 ⋅ 10 23 cm 2 ⋅ s electron flux

Dose Rate Field Emission Gun (FEG) Scanning Transmission Electron Microscope (STEM) probe: 1 nA in 1 nm diam. probe Dose Rate = 6.24 × 273 GGy s = 1.7 ⋅ 10 3 GGy s = 1.70 TGy s