1. Electron irradiation of sapphire (Al2O3) results in both electronic and nuclear damage. 2. Electronic damage includes electron excitation and ionization via interactions between the incident 1 MeV electrons and the aluminum and oxygen atoms in Al2O3. 3. Nuclear damage, or displacement damage, occurs when electrons transfer sufficient energy to atomic nuclei to displace them from their lattice sites, calculated using models of electron-atom cross sections and displacement thresholds.

1. Electron irradiation effect on Al2O3
Kurt Sickafus
Younes Sina

2. Ionization vs. Excitation
 Excitation transfers enough energy to an orbital electron to displace it
further away from the nucleus.
IONISATION
EXCITATION
Incident electron with a specific energy
Atomic electron absorbs energy and moves into a higher orbit
High energy incident electron
Ejected electron
In ionization the electron is removed, resulting in an ion pair.

3. Bremsstralung (or Braking) Radiation
•High speed electrons may lose energy in the form of X-
rays when they quickly decelerate upon striking a heavy
material.

4. Bremsstrahlung
 Probability of bremsstrahlung production per atom is
proportional to the square of Z of the absorber
 Energy emission via bremsstrahlung varies inversely with the
square of the mass of the incident particle
Protons and alpha particles produce less than one-millionth
the amount of bremsstrahlung radiation as electrons of the
same energy

5. Bremsstrahlung
Ratio of electron energy loss by bremsstrahlung production to
that lost by excitation and ionization = EZ/820
E = kinetic energy of incident electron in MeV
Z = atomic number of the absorber
Energy loss for Al: Brem./ (Exc. & Ion.) = 1×13/820 = 1.58%

6. Charged Particle Tracks
 Electrons follow tortuous paths in matter as the result of multiple
scattering events
• Ionization track is sparse and nonuniform
 Larger mass of heavy charged particle results in dense and usually linear
ionization track
 Path length is actual distance particle travels; range is actual depth of
penetration in matter

7. Particle interactions
Energetic charged particles interact with matter by
electrical forces and lose kinetic energy via:
Excitation
Ionization
Radiative losses
~ 70% of charged particle energy deposition leads
to nonionizing excitation

8. Dose = Absorbed Energy Density
Absorbed energy normalized by weight, volume, atoms, etc.
J
1 Gy = 1
kg
SI units
8

9. Water: heat to boiling point
H2O J
cp = 4.1813 (@ 25°C)
gK
specific heat of water
T  80 K
3
J 10 g
c H2O
p T = 334.5 
g kg
5 J
 3.345 10
kg
 0.3345 MGy Energy
Absorbed
9

11. Projectile-Target Interactions
# events
• • • t
<volume> or <weight>

12. Projectile-Target Interactions
atomic cross-
• • flux • time
density section
# events
 a  atoms    area    projectiles  t  time 
volume  volume   atom   areagtime 
# events  atoms    area    projectiles  t  time 
 w 
weight  weight   atom   areagtime 


13. Projectile-Target Interactions
fluence = flux • time
 projectiles   projectiles 

 area      areag
 time   t time 

14. Projectile-Target Interactions
atomic cross-
• • fluence
density section
# events
 a  atoms    area    projectiles 
volume  volume   atom   area 
# events  atoms    area    projectiles 
 w 
weight  weight   atom   area 


15. Projectile-Target Interactions
cross-
• fluence
section
# events
volume
  area    projectiles 
 atoms 
a volume  atom   area 
 

16. Projectile-Target Interactions
Leading to Atomic Displacements
displacement
dpa = cross- • fluence
section
# atomic displacements
volume
  area    projectiles 
 atoms 
a volume  atom   area 
 
Ballistic displacements
  area    projectiles 
Dose
atom  atom   area 

17. Electron irradiation-induced amorphization
of sapphire (Al2O3)
1 MeV electrons
room-temperature irradiation conditions

18. Electron irradiation-induced amorphization
of sapphire (Al2O3)
Two components of damage:
1. electronic component
(electron excitation/ionization; radiolysis)
2. nuclear component
(ballistic or displacement damage)

19. 1. Electronic Stopping

20. 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.

21. Electron Excitation/Ionization
Bethe-Ashkin expression for ionization energy loss per unit length
relativistic expression
  E0  2 E  
Ln  2 2  
  2J (1   )  

dE 2 e e 
dx

4
2 
E0  

 2 1   2  1   2 Ln2 

 
1   2 
 
1
 
2
 1  1   2 
 8
 


22. E0  me c  rest energy of the electron
2
me  rest mass of the electron
c  speed of light
e  14.4 eV  Å
2

23. v

c
v  velocity of electron
c  speed of light
2
 E0 
  1  
 E  E
0
E0  rest energy of the electron
E  kinetic energy of the electron

24. e  Z  a
e  electron density
Z  atomic number
a  atomic density

25. 0.19
J  9.76 Z  58.5 Z (eV)
 mean electron excitation potential
M. J. Berger, and S. M. Seltzer, Nat. Acad. Sci. / Nat. Res. Council Publ. 1133 (Washington,
1964), p. 205.

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

27. Stopping Power
1 dE  eV  Å2 
 e  Se E    atom  e 
a dx e  

28. Bragg’s Rule for Additivity of Stopping Powers
For binary compound with molecular unit, A B :
m n
 Am Bn
e
 m e  n e
A B
where m is the number of A atoms in molecule A B
m n
and n is the number of B atoms in molecule A B
m n
One can show that:
Am Bn A B
dE dE dE
 Am Bn
m  Am Bn
 
dx e
e
dx e dx e
where 
Am Bn
is the molecular density of A B
m m n
molecules in the compound.

29. Ionization stopping in Al2O3

30. E = 1000 keV= 1 MeV
dE/dx (E = 1 MeV) = -0.0377 eV/Å . e-
thickness = 1000 Å
TEM sample thickness
Total ionization energy
= 37.7 eV/e- = 6.032x10-18 J/e-
loss over sample thickness

31. Electron fluence:
Φ=1×1028 e/m2=1×108 e/Ȧ2
Irradiation time= t= 2 hr = 7200 s
φ= 1.38×104 e-/Ȧ2s

32. dE
Areal Energy Density = 
dx electronic
J 11
 3.504 10
=37.7×108 eV/Ȧ2= 3.77×10-10 J/Ȧ2Å 2
Areal Energy Density
Total Energy Density =
thickness
14 J
 3.504 10 3
=3.77×10-13 J/Ȧ3 Å

33. ρAl2O3= 3980 Kg/m3
Dose= 94.72×1012 J/Kg= 94.7 TGy
Magnitude of dose: TeraGray !!

34. 2. Nuclear Stopping

35. 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.

36. Differential displacement cross-section, dσ
 b 2 T  T T   dT
d (T )  T 1  2      2
4 m Tm  Tm Tm   T

where T is the kinetic energy of the electron
2
 E0 
  v / c  1 
 E0 E 

   Z
where  is the fine structure constant (~1/137)

37. 
Tm  maximum energy transfer from e to target atom
4 me M  E 
Tm  E  1
me  M   2 E0 
2

where E is the incident electron energy
O
Ca

41. 2
e  2
1
b  4 Z  
2 2
 E0   4  2
where
1
=
1 2

42. Primary displacement cross-section:
Tm   area  
 p (E)   d  (T ) 
Ed
 atom  
where E d is the displacement threshold energy
Cascade cross-section:
Tm   area  
 tot (E)    (T ) d  (T ) 
Ed
 atom  
where  (T ) is the number of secondary displacements,
given most simply by the Kinchin-Pease expression:
 (T )  0; T < Ed
 (T )  1; Ed  T < 2Ed
T
 (T )  ; T  2Ed
2Ed

43. E = 1000 keV
ZO = 8 TmO =271
ZAl = 13 TmAl =161
ZAve =10 TmAve =227

44. Ed = 20 eV
ZO = 8 EtO = 129,000
ZAl = 13 EtAl = 205,000
Zave =10 EtAve = 159,400

45. Ed = 40 eV
ZO= 8 EO= 238,000
ZAl= 13 EAl= 365,000
ZAve=10

46. Ed = 50 eV
ZO= 8 EO = 290,000
ZAl= 13 EAl = 430,000
ZAve=10

47. E=1 MeV
Ed=40 eV
ZO= 8 EtO= 290,000 eV
ZAl= 13 EtAl= 430,000 eV
ZAve=10
TmAve=227 eV
2Ed=80 eV

49. α-Al2O3
E=1 MeV
Ed=40 eV
σp @ 1 MeV =2.18 barns

50. E  300 keV
powellite (CaMoO4) Ed  25 eV
Z ave
 15.67 Ethreshold  295 keV
ave
Tm  25.54 eV
ave
2Ed  50 eV
2
Å
 tot (E)   p (E)  0.588 barns = 5.88 10 9
atom

52. 52

53. 53

54. 22
28
41

55. where  (T ) is the number of secondary displaceme
given most simply by the Kinchin-Pease expression
 (T )  0; TmT < Ed   area  
 tot (E)    (T ) d  (T ) 
 (T )  1; EdEd  T < 2Ed atom  
where  (TT is the number of secondary displacemen
)
 (T )  ; T  2Ed
given most simply by the Kinchin-Pease expression:
2Ed
 (T )  0; T < Ed
 section Ed  T < for
Cross(T )  1; calculation 2EdAl (Ed=20 eV):
T
 (T )  ; T  2Ed
2Ed
σ =42 barns/atom= 4.2×10-7 Å2/atom
tot
1 barn = 10-24 cm 2  10 8 Å2

56. Electron fluence:
Φ=1×1028 e/m2=1×108 e/Å2
Irradiation time, t = 2 hr = 7200 s
φ= 1.38×104 e-/Å2s
displacements per atom =  tot 
Å2 e
σtot=42 barns/atom= 4.2×10-7 Å2/atom310 6 2
 5.88 10 6 
atom Å
= 0.018 dpa
dpa=(4.2×10-7 Å2/e).(1×108 e/Å2) = 42

57. RADIATION DAMAGE OF α-Al2O3 IN THE HVEM
II. Radiation damage at high temperature and high dose
G.P. PELLS and D.C. PHILLIPS

58. C. L. Chen, H. Furusho and H. Mori
• The decomposition of α- Al2O3 under 200 keV
(Ultra High Vacuum) electron irradiation
• Aluminum precipitated from α- Al2O3 under 200
keV electron irradiation for less than 1 min over
the temperature range 700 to 1273 K.
• φ (electron dose rate)= 1023 e m-2s-1
• Vacuum level < 3×10-8 Pa
Model:
Thermally activated atom movement
 Forced atom displacement ( knock-on collision)

60. RADIATION DAMAGE OF α-Al2O3 IN THE HVEM
II. Radiation damage at high temperature and high dose
G.P. PELLS and D.C. PHILLIPS
 Single-crystal α-Al2O3 irradiated with 1 MeV electrons in a high-voltage
electron microscope at several fixed temperatures in the range 320-
1070 K.
• At 770 K and below the nature of the observed damage could not be
resolved.
• At 870 K and above island-like surface features rapidly formed followed
by dislocations which grew to form a dense network.
• After high doses (>l0 dpa) precipitates were observed.
• The associated diffraction patterns and their temperature dependence
suggested that the precipitates were of aluminum metal.

61. Cryogenic radiation response of sapphire
R. Devanathan, W.J. Weber, K.E. Sickafus, M. Nastasi, L.M. Wang, S.X. Wang
Sapphire (a-Al2O3) irradiated by heavy-ion and electron at cryogenic
temperatures using a high-voltage electron microscope.
1.5 MeV Xe
1 MeV Kr
Dual beam of 1 MeV Kr and 900 keV electrons
T=20 to 100 K
At 20 K, α-alumina is amorphized by 1.5 MeV Xe about 3.8 (dpa)
Critical temperature for amorphization is about 170 K
The material remains crystalline when irradiated at 26 K with a dual beam
of heavy ions and electrons.
Electron irradiation can promote damage annealing, even at cryogenic
temperatures, by causing the migration of point-defects produced in
ceramics by ion irradiation.

62. Effects of ionizing radiation in ceramics
R. Devanathan ,K.E. Sickafus, W.J. Weber, M. Nastasi
α-Al2O3 was irradiated with 1 MeV Kr+ or 1.5 MeV Xe+ and 1
MeV electrons in a high-voltage electron microscope interfaced
to an ion accelerator that enabled the in situ observation of the
structural changes.
The results indicate that simultaneous electron irradiation can
retard or prevent amorphization by heavy ions.
Comparison with similar experiments in metals suggests that
highly ionizing radiation can anneal damage to the crystal lattice
in ceramics by enhancing the mobility of point defects.

63. High flux e-
O2
~1000 Å heat
Al ppt.
Vacuum
>40 dpa
Long time
Surface at high stress