ELECTRON TRANSMISSION THROUGH THIN METAL SECTIONS 873 



radius of curvature R by 



Aa = -^ (4) 



If the situation is such that R can be determined from equation (3), then 

 the misorientation Aa can be calculated from (4). This combination of cir- 

 cumstances is at present one of chance and does not occur frequently in 

 images of thin sections. An example will be shown. 



The image contrast between adjacent regions of a thin metal section com- 

 posed of small misoriented domains as depicted in Fig. 1 is determined by 

 the rocking curve of equation (1). This is simply a plot of the intensity 

 given by (1) against Ag or A^. As an example, a rocking curve for the (200) 



GRAIN BOUNDARY 



Fig, 4 — Displacement (/) at a grain boundary of extinction contours due to bending 

 as predicted by equations (3) and (4). 



reflection of aluminum is shown in Fig. 5. The Fourier coefficient Vg is 

 computed from the relation.* 



Vm = 300 ^' X) (^y - 0)^'"' (h^i + kvj + Iwj) volts (5) 



TTU j 



with e = charge on the electron = 4.80 X 10~^° esu 



Q = volume of the unit cell = 70.4 A^ for aluminum 

 df.ki = interplanar spacing 

 Zj = atomic number of atom species j 

 fj = atom form factor 

 (uj, Vj, wj) = atomic coordinates of atom species j 



For aluminum, F(2oo) = 5.13 volts. Using 50KV electrons: £ = 5 X 10'*, 

 X = 0.055A, sin 2^o = 0.0272 radian and a reasonable value oi D = 250A, 

 the intensity can be computed from (1) as a function of Ag as shown in Fig. 5. 



* Reference 1, Appendix I. 



