X-RAY DIFFRACTION 95 



To detector 



Figure 47 



Consider the three measurable parameters indicated in the sketch: 

 L, the distance from the crystal to the film ; 

 s h the distance from the undeflected beam to the individual lines ; 

 X, the wavelength of the x-rays. 

 Since in practice the angles are always small, 



sin 20 « 2 sin 

 and 



Si 



L 



X = 2d sin « 2d - S f « ^ • 



Thus, as indicated earlier in this section, the spacing d can readily be 

 computed from the measured quantities. For the simple monatomic 

 crystal we are discussing, it is only moderately difficult to relate the 

 calculated spacings to the distances between atoms, but it is beyond the 

 scope of this presentation. 



It remains to indicate how the intensities can yield the structures of 

 molecules placed at the crystal positions. To this end we return to Fig. 

 44 and, as in Fig. 47, insert atoms midway between the planes. It is true 

 that there will now be some new powder lines, but we shall neglect them 

 for the present and consider only what happens to the already existing 

 powder lines. In the case presented in the sketch, the extra distance 

 traveled by ray III is just one-half that for ray II. Since the latter dis- 

 tance is exactly one whole wave more than for ray I in the case of ray 

 reinforcement, ray III travels exactly one half wave more than ray I. 

 Thus it cancels ray I because its troughs are superposed on the crests of 

 ray I, and vice versa. In this instance, then, the line will be missing. 

 Now, if the extra atoms are not placed so that their contribution is pre- 

 cisely one half wave different, then there will not be a total cancellation 

 of ray I. In general, then, the insertion of the extra atoms will diminish 



