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Gold is FCC and has a lattice constant of 0.40788nm Calculate a value for its atomic radius of gold in nanometers?
Wiki User
∙ 14y ago
The unit cell of an fcc lattice has right angles, and each face has gold atoms touching each other along the diagonal (usually the diagonal is depicted as running from the center of one atom, through the center of a second atom, to the center of a third atom). Thus, one can draw a right triangle whose legs both have length of a = 0.40788 nm and whose hypoteneuse is 4r, where r is the radius of a gold atom. By the Pythagorean theorem: a2 + a2 = (4r)2 2a2 = 16r2 r = (21/2/4)a Substituting in a = 0.40788 nm, r = 0.14421 nm, which is the listed covalent atomic radius of gold. Please note that this method only works when considering lattices composed of a single element. When multiple elements are involved, their radii change due to interaction with the other elements.
Wiki User
∙ 14y ago
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