The unit cell of metallic silver Volume of Ag atom is 144.4pm (1) The percent of the empty space
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Answered by
0
Answer:
Given,
Atomic mass of silver =108u
Atomic cell length, a=4.077×10
−8
cm
Atomic radius in FCC, r=
4
a
2
=
4
4.077×10
−8
2
=1.44×10
−8
cm\
Each silver unit cell contains 4 atom.
Density =
N
A
×a
3
4×atomicmassofsilver
=
6.02×10
23
×(4.077×10
−10
)
3
4×108
=10589.25kgm
−3
Hence, atomic radius 1.44×10
−8
cm and density is 10589.25kgm
−3
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