the unit cell of metallic silver is focused if radius of Ag atoms is 144.4 pm calculate
A edge length of unit cell
B volume of atoms
C that percentage of the volume of a unit cell,that is occupied by Ag atoms
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Given,
Atomic mass of silver
=
108u
Atomic cell length,
a=
4.077×
10 −8
Atomic radius in FCC,
r=
4
a
2
=
4
4.077×
10
−8
2
=
1.44×
10 −8
\
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
and density is
10589.25kgm -3 please mark me brilliant
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