Last Name:
First Name:
Student #:
MSE 231H1F Characterization of Materials
2004 / 2005
Quiz # 2
Time: 50 minutes
Date: November 29, 2004
____________________________________________________________________________
  2d sin 
p  mv 

h



eV  
 2m0 eV 1 
2 
 2m0 c  

100  m
M

2.76 102 AV 1.67
 Z 0.89
m  m0 
1.226
Rd   L
0.5
1  0.9788 10 V 
6
eV
c2
nm
0.5
[UVW ] (hkl )  hU  kV  lW  0
R    R  0 cos
   Cii
  0.0254  0.016Z  1.86 104 Z 2  8.3 107 Z 3
i
nSE iSE

nB
iB
E  h
h
2meV
V
R
 tan   
WD
nBSE iBSE

nB
iB
h  k2  l2
2

1/ 2
RKO   m  


h
PES 
a
d
T  1   2
E
hc

swl 
hc
V
T   SE   SE
1
 1
1 
E  13.6  Z 2  2  2 
 nf
ni 

2
 in eV 
c = 2.998x108 m/s
e = 1.602x10-19 C
m0 =9.11x10-31 kg
-34
-15
h = 6.626x10 J∙s
h = 4.136x10 eV∙s
1 eV = 1.602x10-19 J
_____________________________________________________________________________
Answer all questions. The marks for each part are shown. There is a total of 30 marks. You
have 50 minutes to complete this quiz.
Page 1 of 4
Last Name:
First Name:
Student #:
Gallium (Z=31) has the oC8 crystal structure at room temperature. Gallium is a unique metal in
that its melting temperature (Tmelt = 30 C) is between room temperature and body temperature –
solid Ga will melt in your hand. Elemental Ga has found use in high temperature thermometers
because it has one of the largest liquid ranges of any metal (from 30 C to 2205 C at 1
atmosphere). Gallium is also responsible for liquid metal embrittlement in aluminum alloys, in
which Ga atoms diffuse along the grain boundaries and rapidly lead to intergranular fracture.
1)
swl
A 10 kV electron beam is incident on a sample of crystalline Ga. Use the 1 electron
approximation to draw a schematic diagram of the expected X-ray spectrum of intensity
vs X-ray wavelength. Indicate clearly both the background and characteristic X-ray
signals. Give numerical values for the short wavelength limit and the following
characteristic X-ray wavelengths: K, K, L, and L.
(10 marks)
hc 4.13 1015 eV s(3 108 )m / s


 0.124nm
V
10000eV
1
1
 2)
2
nf
ni
1 1
 13.6(31) 2 ( 2  2 )  9.8keV  11.6keV  0.127 nm
1 2
 11.6keV  0.107 nm (not present )
E  13.6( Z ) 2 (
Ek
Ek 
Kα
Kβ
Lα
Lβ
nf
1
1
2
2
EL  1.8keV  0.685nm
EL  2.5keV  0.507 nm
Page 2 of 4
ni
2
3
3
4
Last Name:
2)
First Name:
Consider the case where the incident electron beam has ionized (removed) a 1s electron
from a Ga atom in the ground state (i.e. 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4s1). Give the electron
configurations of the permissible transitions, i.e. which electrons could transfer into the
1s state and release an X-ray photon? (4 marks)
n  0
l  1
3)
Student #:
therefore:
4 p  1s
3 p  1s
2 p  1s
Consider the following case: you are investigating the intergranular fracture surface of
aluminum (Z=13) which has suffered liquid metal embrittlement due to exposure to Ga
(Z=31). The gallium is present as discrete islands on an aluminum substrate, i.e. there are
regions of the fracture surface which are 100% Ga and there are regions of the fracture
surface which are 100% Al. What relative atomic number contrast could be expected
using backscattered electrons? (6 marks)
 Al  0.0254  0.016(13)  1.86 104 (13) 2  8.3 107 (13)3  0.153
Ga  0.0254  0.016(31)  1.86 104 (31) 2  8.3 10 7 (31)3  0.317
2.07 Al  Ga
Page 3 of 4
Last Name:
4)
First Name:
Student #:
Assuming that the gallium islands are only 10 nm thick, show by calculation whether
backscattered electrons from a 10 kV electron beam would be a useful signal for imaging
this structure (HINT: estimate the size of the interaction volume). (6 marks)
AAl = 26.98 g/mol
AGa = 69.72 g/mol
Al = 2.7 g/cm3
Ga = 5.91 g/cm3
RKO (um) 
2.76 102 AV 1.67
 Z 0.89
RKO , Al (um) 
2.76 102 (26.98)(10)1.67
 1.316  m
(2.7)(13)0.89
RKO ,Ga (um) 
2.76 102 (69.72)(10)1.67
 0.717  m
(5.91)(31)0.89
Interaction volume is approximately 2 orders of magnitude larger than gallium island
thickness. Therefore backscattered electrons would not be useful because the vast
majority of signal would come from the aluminum substrate. Need a surface sensitive
technique – e.g. imaging XPS, AES, or SIMS.
5)
Briefly explain the mechanisms of formation of secondary electrons if a 10 kV electron
beam is incident on a sample. How can electron detectors in the SEM differentiate
between backscattered and secondary electron signals? (4 marks)
Secondary electrons are generated by the interaction of high energy electrons (i.e.
incident or backscattered) with loosely bound conduction band electrons. Only those
secondary electrons formed near the surface will be able to escape from the sample.
Secondary electrons are therefore emitted from both the entrance point of the incident
beam into the sample and the exit points of the backscattered electrons.
Secondary and backscattered electrons are differentiated/collected on the basis of their
kinetic energies. A small positive bias (e.g. +200V) is applied to collect the low energy
secondary electrons. A small negative bias (e.g. -50V) is used to repel the secondary
electrons so that only the higher energy backscattered electrons will enter the detector.
Page 4 of 4