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**Unformatted text preview: **MSE 4020 HW10-SOLUTION 11/11/2011
Problem 1
(a) Burger’s vector a
b = [ 1 1 1], i.e., !D in T hom pson Tetrahedron
3
is a Frank dislocation and perpendicular to plane (111), thus, it is immobile and
does not have a glide plane.
(b)
One possible dissociate is,
D! " D# + #!
i.e.,
2 2 2 (a / 3
(
(
!"# !"6
$a /$ !"9
# $a /$
#
a$ & a$
a$
111 " 211& + %011&
'6
'
3% ' 6%
a2 / 3 > a2 / 6 + a2 / 9 Thus, energy favorable 1 Problem 2 After cutting, two additional surfaces are created (in green color) with width x.
The green area is approximately,
A = 2!r " x The surface energy is provided by the work done by Fc , i.e., A ! "S = Fc ! x
# 2$r ! x ! "S = k ! r ! x
# k = 2$ ! " S
At optimal radius,
Gb2
= Gb2
2
2
= Gb Fc = 2T = 2 !
" k ! ropt 2 " ropt = 2 Gb
Gb
=
=
k
2#$s ( 1011 ! 2.5 ! 10%10
2 ! 3.142 ! 1 ) 2 = 9.946 ! 10%10 m " 4ropt = 39.784 ! 10%10 m 2 From volume ratio,
2 " 2r %
r
f
f !$ ' ( =
$l '
l0
2
#0& In the regime of cutting(r < ropt), !c = (F ) 3/ 2 c 2 b " l0 " G = (k " r )
2 3/ 2 b " l0 " G = k 3/ 2 " r " r1/ 2
2 b " l0 " G = k 3/ 2 " f 1/ 2 " r1/ 2
2 2b " G # r1/ 2 In the regime of Orowan loop (r > ropt)
!c = G " b G " r " b G " f 1/ 2 " b 1
=
=
#
l0
l0 " r
2r
r Using the attached Matlab script, the plot asked is given below: 3 4 Problem 3 Ln = n!G!b
" ! # applied (1) G
(2)
30
(1) $ (2), giving
n!G!b
G
! n ! # applied = Ln !
" ! # applied
30 n ! # applied = n2 ! b Ln
=
"
30
n2 ! b 30 ! 72 ! 3.1! 10&10
% Ln = 30 !
=
= 1.451! 10&7 m
"
3.142
% Problem 4 & 5
Solutions can be found in Prof. Ast’s powerpoint file(also attached here). 5 The puzzler answers Nov 1st, 2011 Stress applied Disloca;on sort themselves out, get stuck in twin boundaries Forming twin boundaries with superimposed small angle grain boundaries TEM measures ;lt angle which gives disloca;on content of original grain When the twin boundary has collected a cri;cal number of disloca;ons, a small angle grain boundary will split of the twin boundary, and the boundary will return to the coherent twin forma;on The next puzzler Spiral Disloca;on The spiral, in extremis, is, hold on, a Frank loop! The Frank Disloca;on loop Which means we can determine the number of point defects (number of missing atoms) by measuring the area and divide it by area of an atom! By balancing line force (increase in disloca;on length due to spiral) versus number of point defects stored, we can measure the super satura;on ra;o of point defects. If we are good at TEM, we can decide if we store vacancies or inters;;als (curvature of laMce plane!) The point defects explains why the diﬀusion proﬁle of P in Silicon is by now means Erfc(x/Sq(2Dt) The diﬀusion in Si is very complex, as e.g. the vacancy appears in four ﬂavors Neutral Single posi;ve Single nega;ve Double nega;ve Each of which has its own diﬀusion parameters and the concentra;on of which (except the neutral) depends on the Fermi level (which in turn changes with the doping proﬁle) ...

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