Calculate the density of unit cell of gold if crystallizes in fcc structure and its edge length is 407 pm given atomic mass of gold is 197 u
Answers
Answer:
For an FCC crystal, atoms are in contact along the diagonal of the unit cell.
4r=
2
a, where a is the edge length of unit cell and r is the radius of atom.
Substituting values in the above expression, we get
r=
4
2
×407
=143.9 pm
Number of atoms per unit cell (FCC) =4
Mass of 1 unit cell =197×4=788amu=1.66×10
−27
×788kg=1.30808×10
−24
kg
Volume of 1 unit cell =(407×10
−12
)
3
m
3
=6.741×10
−29
m
3
=2.824×10
−28
m
3
Density=
6.741×10
−29
1.30808×10
−24
=19404.83kg/m
3
=19.4g/cm
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Explanation:
CHEMISTRY
Gold crystallizes in a face centered cubic lattice. If the length of the edge of the unit cell is 407 pm, calculate the density of gold as well as its atomic radius assuming it to be spherical. Atomic mass of gold =197 amu.
December 20, 2019avatar
Disha Kashyap
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VIDEO EXPLANATION
ANSWER
For an FCC crystal, atoms are in contact along the diagonal of the unit cell.
4r=
2
a, where a is the edge length of unit cell and r is the radius of atom.
Substituting values in the above expression, we get
r=
4
2
×407
=143.9 pm
Number of atoms per unit cell (FCC) =4
Mass of 1 unit cell =197×4=788amu=1.66×10
−27
×788kg=1.30808×10
−24
kg
Volume of 1 unit cell =(407×10
−12
)
3
m
3
=6.741×10
−29
m
3
=2.824×10
−28
m
3
Density=
6.741×10
−29
1.30808×10
−24
=19404.83kg/m
3
=19.4g/cm
3