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Correspondence
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CALCULATING CRYSTAL DIMENSIONS I have read the excellent article, "Some Elementary Principles of XRay Crystal Analysis" in the January issue of the JOURNAL May I make a suggestion that I believe will make one calculation used there more understandable t o the average student? This refers to the calculation of the distance between the atomic planes of the sodium chloride crystal. The method of calculation I suggest is not new, of course, but I offer i t because I believe it will be followed by more students because they are more familiar with density in grams per cubic centimeter than in terms of the volume of unit cubes and the atoms associated with it. Since a mole of NaCl is 58.455 grams this weight represents the weight of Z(6.06 X 10fZS)ions of Na and C1. Since these are arranged symmetrically in a cube we may let y represent the number of ions on an edge of a cube one centimeter on an edge. The number of ions in the cubic centimeter of the crystal of NaCl can be found by proportion as follows: 58.455 : 12.12 X 1OtZ8 = 2.163 : x z = 44.84 X loP1ions of Na and CL in one cc. of NaCl crystal. Then, ys = 44.84 X 1011and y = 3.553 X 107ionson an edge of the cube.
The distance between the atomic planes will be the reciprocal of the number along an edge (in reality the distance i&/y - 1but the value of l/y is accurate enough here). The distance between planes in the salt cm. crystal is 1/3.553 X lo7 = 2.816 X R. S. HOWARD AND JUNIOR COLLEGE, LYONS Tow~snwHIGHSCHOOL LA GRANGE, ILLINOIS
