Transcript Document 7814951

NEEP 541 – Damage and
Displacements
Fall 2003
Jake Blanchard
Outline

Damage and Displacements

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
Definitions
Models for displacements
Damage Efficiency
Definitions
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Displacement=lattice atom knocked from its lattice
site
Displacement per atom (dpa)=average number of
displacements per lattice atom
Primary knock on (pka)=lattice atom displaced by
incident particle
Secondary knock on=lattice atom displaced by pka
Displacement rate (Rd)=displacements per unit
volume per unit time
Displacement energy (Ed)=energy needed to displace
a lattice atom
Formal model
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To first order, an incident particle with
energy E can displace E/Ed lattice atoms
(either itself or through knock-ons)
Details change picture
Let (E)=number of displaced atoms
produced by a pka
Formal Model


Rd   (T ) N    ( E ) ( E , T )dE  dT
Ed
0

Tm
Tm

Rd  N   ( E )   (T ) ( E , T )dT  dE
 Ed

0


Rd  N   ( E ) d ( E )dE
0
Tm

 d    (T ) ( E , T )dT 
 Ed

What is (E)
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For T<Ed there are no displacements
For Ed <T<2Ed there is one
displacement
Beyond that, assume energy is shared
equally in each collision because =1
so average energy transfer is half of the
incident energy
Schematic
tka
ska
pka
Energy per atom
E
E/2
E/4
E/2N
displacements
1
2
4
2N
Displacement model
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Process stops when energy per atom
drops below 2Ed (because no more net
displacements can be produced)
So
T
 2 Ed
N
2
or
T
N
 (E)  2 
2 Ed
Kinchin-Pease model

Ed
2Ed
Ec
T
More Rigorous Approach
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Assume binary collisions
No displacements for T>Ec
No electronic stopping for T<Ec
Hard sphere potentials
Amorphous lattice
Isotropic displacement energy
Neglect Ed in collision dynamics
Kinchin-Pease revisited
 ( E )   ( E  T )   (T )
E
E
0
0
  ( E , T ) ( E )dT    ( E , T ) ( E  T )  (T )dT
 (E, T ) 
E

0
 (E)
E
 (E)
E
;   1; hard sphere
E
 ( E )dT  
0
E
 (E)
E
 ( E  T )   (T )dT
1
 ( E )    ( E  T )   (T )dT
E0
Kinchin-Pease revisited
E
0
 ( E  T )dT   (w)dw;
0
w  E T
E
E
E
0
0
 ( E  T )dT   (T )dT
E
2
 ( E )   (T )dT
E0
Ed
2
2
 ( E )   0dT 
E 0
E
E
2 Ed
E
2
E dT  E 2E (T )dT
d
d
2 Ed 2
 (E) 
  (T )dT
E
E 2 Ed
Kinchin-Pease revisited
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Solution is:
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For power law potential, result is:
E
 (E) 
2 Ed
sE
 (E) 
2 Ed
 11s 
 2  1




E

2  ( E )  0.52 2 E
d
s
E
3  ( E )  0.57

2 Ed
Electronic Stopping
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Repeat with stopping included
Hard sphere potentials
 dE 

 k E
 dx  e
E
 (E) 
2 Ed

4k
1 
 N 2 E
d





Don’t need
cutoff energy
any more
  Hard sphere collision cross section
(independent of E)
Comprehensive Model
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Include all effects (real potential,
electronic stopping)
Define damage efficiency:
 E
 ( E )   ( E )
 2 Ed
 (E) 




1
1  0.13 3.4 1/ 6  0.4 3 / 4  
E
 2 2
2Z e / a
0.88aB
a
Z 1/ 3

Damage Efficiency