ENG 2002 Lecture Notes - Diamond Cubic, Monochrome, Diffractometer
2 Apr 2014
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Atomic structure relates to the number of protons and neutrons in the nucleus of an atom, as well as the number and probability distributions of the constituent electrons. On the other hand, crystal structure pertains to the arrangement of atoms in the crystalline solid material. 3. 2 if the atomic radius of aluminum is 0. 143 nm, calculate the volume of its unit cell in cubic meters. For this problem, we are asked to calculate the volume of a unit cell of aluminum. Aluminum has an fcc crystal structure (table 3. 1). The fcc unit cell volume may be computed from equation 3. 4 as. Vc = 16r3 2 = (16)(0. 143 10-9 m)3( 2) = 6. 62 10-29 m3. 3. 3 show for the body-centered cubic crystal structure that the unit cell edge length a and the atomic radius r are related through a =4r/ 3 . And then for triangle npq, (np)2 = a2 + a2 = 2a2 (nq)2 = (qp)2 + (np)2.
