上海交通大学：《Structural and Chemical Characterization of Materi》教学资源_homework assignment-tem

1.The resolution of a TEM is limited by diffraction(Rd)and by the spherical aberration which is expressed as:Rsph=Csa3,where Cs is the spherical aberration coefficient and a is the half-angle of the cone of light entering the objective lens. The TEM resolution is the quadratic sum of the diffraction resolution and spherical aberration resolution,given by the expression:(R=Ra2+Rp2).The minimum R is obtained when RRhEstimate the optimum value of a and resolution limit of TEM at 100 and 200 kV respectively.C 1 mm for an electromagnetic objective lens. 0.6110.617 Rd Nsinc h Ri-Rsph 061-C,a= C (+ Acc.Voltage (kv) (nm) Cs (mm) dopt (mrad) Rmin (nm) 2 5.26 0.412 200 0.00251 0.2 9.36 0.232 (relativistic) 0.02 16.65 0.131

1.The resolution of a TEM is limited by diffraction (Rd) and by the spherical aberration which is expressed as: Rsph=Cs3, where Cs is the spherical aberration coefficient and  is the half-angle of the cone of light entering the objective lens. The TEM resolution is the quadratic sum of the diffraction resolution and spherical aberration resolution, given by the expression: ( ). The minimum R is obtained when Rd =Rsph. Estimate the optimum value of  and resolution limit of TEM at 100 and 200 kV respectively. Cs 1 mm for an electromagnetic objective lens. Rd = 0.61𝜆 𝑁𝑠𝑖𝑛α ≈ 0.61𝜆 α Rsph = Cs α 3 Rd =Rsph 0.61𝜆 α = Cs α 3 α = 0.61𝜆 Cs 1 4 Acc. Voltage (kV) 𝝀（nm） 𝐶𝑠（mm） αopt（mrad） Rmin（nm） 200 0.00251 (relativistic) 2 5.26 0.412 0.2 9.36 0.232 0.02 16.65 0.131

1.Below is a selected area electron diffraction pattern of a-Fe (bcc.a=0.2866nm).Find AL for the diffraction pattern,index the pattern and determine the zone axis. R (R/R2 (R:/R)2X2 BCC(hkl) Ri 1.6 1 2 110 R3 2.8 3.06 6.12 211 R 4.2 6.89 13.78 321 R> R2 Ra 1.6 1 2 110 cos中= hh2+k k2 +1l2 Vh+k+Vn好+k好+ Lλ=Rd

1.Below is a selected area electron diffraction pattern of -Fe (bcc. a=0.2866nm). Find L for the diffraction pattern, index the pattern and determine the zone axis. R (Ri /R1 ) 2 (Ri /R1 ) 2X2 BCC(hkl) R1 1.6 1 2 110 R3 2.8 3.06 6.12 211 R2 4.2 6.89 13.78 321 R4 1.6 1 2 110 R1 R2 R R3 4

2. 4.9341/nm

2. 4.934 1/nm

3.Below is a selected area electron diffraction pattern from a polycrystalline region,taken from a cast steel block.There is concern that silicon may have been introduced into the casting as an impurity. Solve the pattern and index the rings.Is there any evidence for the presence of crystalline Si? Fe+Si Si Fe Si da=0.3135 nm a-Fe (bcc a=0.2866 nm)d(o)=0.203 nm

3. Below is a selected area electron diffraction pattern from a polycrystalline region, taken from a cast steel block. There is concern that silicon may have been introduced into the casting as an impurity. Solve the pattern and index the rings. Is there any evidence for the presence of crystalline Si?

1.When dynamic effects are considered,the solution for the intensity of the diffracted beam under two-beam conditions is given by: 1g= ((seei) (πtSeffective)Z Where Seffective=s2+(1/5g2),g is the extinction distance for the selected reflection g. The following picture is a dark-field TEM micrograph of the edge of an aluminium sample recorded under two-beam diffracting conditions,and a selected area diffraction (SAD)pattern of the grain which contains the reflection used to form the two-beam image.Solve the diffraction pattern. Use the thickness fringes in the image and the data in Table to estimate the thickness profile of the sample.How would your results be changed if the material was gold rather than aluminium? ≈185nm (b) (a) 02 200nm B=[011]

1.When dynamic effects are considered, the solution for the intensity of the diffracted beam under two-beam conditions is given by: 𝐼𝑔 = ( 𝜋𝑡 𝜉𝑔 ) 2 𝑠𝑖𝑛2 (𝜋𝑡𝑠𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒) (𝜋𝑡𝑠𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒) 2 Where 𝑠𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 = 𝑠 2 + (1/𝜉𝑔 2 ), 𝜉𝑔 is the extinction distance for the selected reflection g. The following picture is a dark-field TEM micrograph of the edge of an aluminium sample recorded under two-beam diffracting conditions, and a selected area diffraction (SAD) pattern of the grain which contains the reflection used to form the two-beam image. Solve the diffraction pattern. Use the thickness fringes in the image and the data in Table to estimate the thickness profile of the sample. How would your results be changed if the material was gold rather than aluminium? ≈185nm

Table 4.1Extinction distance for prominent lattice planes in alminium and gold at 100kV. Material Extinction distance for reflection hk/(nm) S=0 110 111 200 220 400 N Au 二 56.3 68.5 473 76.4 period 18.3 20.2 27.8 43.5 =11s2+/5g)-5o Seff t=6*tperiod=6g (a) 200nm

s = 0 t period= 1 𝑠𝑒𝑓𝑓 = 1/ 𝑠 2 + (1/𝜉𝑔 2 ) = 𝜉𝑔 t=6* t period = 6 𝜉𝑔 𝜉𝑔

2.In a TEM study of polycrystalline molybdenum(BCC,a=0.3147 nm),a low- angle grain boundary was found.The boundary geometry is sketched in figure (a).In order to identify the dislocations composing the boundary,two-beam diffraction images were recorded.The first image was obtained after orienting the sample to align a [1 0 1]zone axis parallel to the incident electron beam.The SAD pattern is shown in figure (b).Use of reflection 1 resulted in the sub-boundary dislocations going out of contrast.The sample was then tilted by 18.43 into a new zone axis,from which a two-beam image was recorded using the reflection 2 shown in the second SAD pattern (figure (c)).Again,no diffraction contrast was detected from the sub-boundary dislocations under these diffracting conditions.The sample was then tilted to bring the boundary parallel to the incident electron beam. (b) (c 211 21 g● 101 121 51 ● d=0.2225nm d=0.1573nm ● 000 020 000 020 ● ● B=[101] B=[102] b 2[111

2.In a TEM study of polycrystalline molybdenum (BCC, a=0.3147 nm), a low￾angle grain boundary was found. The boundary geometry is sketched in figure (a). In order to identify the dislocations composing the boundary, two-beam diffraction images were recorded. The first image was obtained after orienting the sample to align a [1 0 1] zone axis parallel to the incident electron beam. The SAD pattern is shown in figure (b). Use of reflection 1 resulted in the sub-boundary dislocations going out of contrast. The sample was then tilted by 18.43o into a new zone axis, from which a two-beam image was recorded using the reflection 2 shown in the second SAD pattern (figure (c)). Again, no diffraction contrast was detected from the sub-boundary dislocations under these diffracting conditions. The sample was then tilted to bring the boundary parallel to the incident electron beam. B=[101] 101 000 020 121 B=[102] 020 211 211 000 b = 1 2 [111]

(b) ● 21-1 2-1-1 121 10i 12i (c) 岛 ● ● P号 ● d=0.2225nm d=0.1573nm ● ● ● 0 020 000 020 020 000 020 ● ● ● ● 211 -2-11 ● [101] b=1/2[111] [102]

c)The distance between the dislocations in the sub-boundary (h) depends on the Burgers vector(b)and the relative misorientation of the two crystals (0): b h= 2sin(0/2) Sketch the SAD pattern that you would expect from the boundary region when aligned with a [00 1]zone axis parallel to the electron beam. 200 110 ● (a) 110 Grain A 020 Grain B h=156am B=[001]

c)The distance between the dislocations in the sub-boundary (h) depends on the Burgers vector (b) and the relative misorientation of the two crystals (): ℎ = 𝒃 2sin(𝜃/2) . Sketch the SAD pattern that you would expect from the boundary region when aligned with a [0 0 1] zone axis parallel to the electron beam. B=[001] 200 110 020 110 𝜃

3.Note that both the mass-density and the diffraction contrast require only a transmitted beam to pass through the objective aperture.How can you know you have the diffraction contrast without checking selected area diffraction (SAD)? Tilt the sample

3.Note that both the mass-density and the diffraction contrast require only a transmitted beam to pass through the objective aperture. How can you know you have the diffraction contrast without checking selected area diffraction (SAD)? Tilt the sample