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Gold has a close packed structure which can be reviewed as spheres occupying 0.74of the total volume. If the density of gold is 19.3 gg/cc, calculate the apparent radius of a gold atom in the solid (Au=197 amu) |
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Answer» Solution :As the packing FRACTION is 0.74 , packing can be hcp or FCC. Knowing that gold has fcc lattice, Z=4 `rho=(ZxxM)/(a^3xxN_0) "" THEREFORE 19.3=(4xx197)/(a^3xx6.023xx10^23) "or" a=4.07xx10^(-8)` CM For fcc, `r=a/(2sqrt2)=(4.07xx10^(-8))/(2xx1.414)=1.439xx10^(-8)` cm |
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