Question

gold has a closed packed structure which can be viewed as spheres occupying 0.74 of the total volume . if the density of the gold is 19.7 g/ml ,the atomic radius of the gold in pm is

given (1.43)³x4pi=37

Answer 1

We know the density , So in order to determine the volume , we find mass of the unit cell . Each unit cell have eight corner and six faces.
The total number of atoms with in such a cell ,
(8×18)+(6×12)(8×18)+(6×12)=4=4
The mass of a unit cell in grams is
m=1.31×1021gunitcellm=1.31×1021gunitcell
From the definition of density (d=m/v)(d=m/v)
We calculated the volume of the unit cell
V=md=1.31×10−21g19.3g/cm3V=md=1.31×10−21g19.3g/cm3
=6.79×10−23cm3=6.79×10−23cm3
the volume is length cubed , we take the cube root of the volume of the unit cell
a=V−−√3a=V3
=6.79×10−23cm3−−−−−−−−−−−−−−√3=6.79×10−23cm33
=4.08×10−8cm=4.08×10−8cm
We see that the radius of an Au sphere (r) is related to the edge length by
a=8–√ra=8r
=1.44×10−8cm=1.44×10−8cm
=1.44×10−8cm×1×10−2m1cm×1pm1×10−12m=1.44×10−8cm×1×10−2m1cm×1pm1×10−12m
=144pm