Solutions of Chapter 3
3.1.
Generally, the wavelength of electron and the acceleration voltage has a relation
as λ ∝ 1/ Vo . Do the values in Table 3.2 hold the relation exactly? Show the
relations between λ and Vo in Table 3.2 and comparing them with λ ∝ 1/ Vo .
The wavelength of electron λ is determined by its velocity v and mass m，that is
h
(1.1)
λ＝
mv
Where h is Planck’s constant.
While the velocity v and the acceleration voltage has the following relation,
1 2
mv = eV
2
We obtain
v=
2eV
m
(1.2)
From equation (1.1) and (1.2), we can get λ =
h
h
=
/ V0
2emV
2em
If the velocity of electron is relatively low, its mass is almost equal to the static
mass. That is m= m0 . If the velocity of electron is very high, it should be rectified
m0
by theory of relativity. m =
2
⎛v⎞
1− ⎜ ⎟
⎝c⎠
We know Planck’s constant h = 6.63 × 10－34 J − s , e = 1.602 ×10－19 C . Supposing
m = m0 = 9.11× 10－31 kg ; thus
h
6.63 ×10－34
=
= 1.22718 × 10－9 .
－19
－31
2em
2 ×1.602 × 10 × 9.11×10
We can calculate the correlations and compare them with Table 3.2, as shown in
the following table.
Acceleration voltage
(KV)
40
60
80
100
200
500
Electron wavelength
(nm)
0.00614
0.00501
0.00434
0.00388
0.00274
0.00174
Electron wavelength(nm)
in Table3.2
0.00601
0.00487
0.00418
0.00370
0.00251
0.00142
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From the comparison, we can find that the wavelength values are slightly shorter
due to the consideration of the relativity theory.
3.2. The resolution of TEM is limited by the diffraction (Rd) as discussed in Chapter I
(Eq. 1.3) and by the spherical aberration which is expressed as Rsph = Csα 3 . The
TEM resolutions are the quadratic sum of diffraction and spherical aberration
one ( R = Rd 2 + Rsph 2 ).
The minimum R is obtained when Rd ≅ Rsph.
Estimate the optimum value of α and resolution limit of TEM at 100 and 200 kV.
Cs ≅ 1 mm for an electromagnetic objective lens.
With α decreasing, the spherical aberration will decrease but the resolution will
be worse. Considering the diffraction limited resolution and spherical aberration,
the optimum value of α can be obtained as follows.
0.61λ
,
Since Csα 3 =
μ sin α
1
⎛ 0.61λ ⎞ 4
we get α = ⎜
⎟ .
⎝ μ Cs ⎠
At 100KV, as is shown in table 3.2, the corresponding electron wavelength is
0.00370nm. For Cs ≅ 1mm and supposing μ ≈ 1 ,
we can estimate α = 6.228 ×10−3 , and the resolution
R1 =
0.61λ 0.61× 0.0037 nm
=
= 0.362nm
μ sin α 1× 6.228 ×10−3
Similarly, for 200KV; we can get
α = 5.652 × 10−3
and R1 ==
0.61× 0.00251nm
= 0.275nm
1× 5.565 ×10−3
3.3. How to increase contrast of mass-density image? Explain why. Can we obtain a
mass-density image of metal specimen? Can we obtain a diffraction contrast of
a polymer specimen?
Staining with solution with heavy metal elements is common method used to
increase contrast of polymeric and biological specimens. Heavy metal ions may
selectively stain certain constituents in the specimens, and thus increase the
mass-density among different constituents. The staining method does not work
well with metal. However, the mass-contrast exhibits in all types of materials,
including metals. Generally, the diffraction contrast does not work well in a
polymer specimen because only crystalline polymer may generate diffraction
contrast.
3.4. Note that both the mass-density and the diffraction contrast requires only
transmitted beam pass the objective aperture. How can you possibly know you
have a diffraction contrast without checking SAD?
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The diffraction contrast is very sensitive to specimen tilting, but mass-density is
not. We may tilt the specimen and check whether the contrast of the image
changes during tilting specimen.
3.5. Sketch diffraction patterns of single crystals in TEM: a) FCC crystal with
transmitted beam direction (B) parallel to [001]; b) BCC crystal with B = [001].
This question should have same answers as questions 2.5 and 2.6.
(BCC)
(FCC)
3.6. Does a zone axis of single crystal diffraction is exactly parallel to transmitted
beam? Explain your answer graphically.
To obtain a single crystal diffraction pattern, it is not necessary to have a zone
axis exactly parallel to transmitted beam.
As shown in the above figure, the zone axis is not exactly parallel to the
transmitted beam. Due to thin foil effect, the reciprocal point will extend to rod, if
the Edwald sphere cut the rod, we can also get diffraction pattern of single crystal.
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3.7. What kind of diffraction patterns you should expect, if a SAD includes a large
number of grains in a polycrystalline specimen? Why?
A ring pattern will be as shown in Fig. 3.34a. The pattern including a large
number of grains, which randomly orientated in the specimen, is equivalent to
rotate a single crystal pattern along axis perpendicular to the image plane.
3.8. You are asked to obtain maximum contrast of one grain from its neighbors.
Which orientation of crystalline specimen do you wish to obtain? Illustrate it
with a pattern of single crystal diffraction.
The orientation of the specimen should let one set of crystallographic planes in
the grain satisfies the exact Bragg’s condition. It means that the intensity of such
diffraction is maximized. In such case, the dark image of the grain distinguish
itself with surround ones.
3.9. You have been told that thickness fringe and bend contours can be distinguished
How?
Thickness fringes often show as a pattern with rather regularly spacing between
bright and dark fringes. Bend contours do not show much regularity in the
fringe spacing. The bend contours are sensitive to specimen titling, but not the
thickness fringes.
3.10. The reason we see a dislocation is that it bents crystal plane near its core region.
Can you indicate a case we may not be able to see the dislocation in diffraction
contrast in a two-beam condition (the two-beam condition is referred to a crystal
orientation in which intensity of one diffract spot is much higher than those of rest
spots)? Please indicate the either graphically or using the relation of vectors g
(normal of planes which generate the diffraction beam) and b (Burgers vector of
dislocation).
The dislocation contrast can disappear if g hkl ⋅ b = 0 , the dislocations will be
invisible. In other word, when g ⊥ b, we can not see the dislocation.
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