Respuesta :
Answer:
a)CN=12
b)APF=74 %
c)a=0.35 nm
d)ρ=9090.9 [tex]Kg/m^3[/tex]
Explanation:
Given that
Nickel have FCC structure
We know that in FCC structure ,in FCC 8 atoms at corner with 1/8 th part in one unit cell and 6 atoms at faces with 1/2 part in one unit cell .
Z=8 x 1/8 + 1/2 x 6 =4
Z=4
Coordination number (CN)
The number of atoms which touch the second atoms is known as coordination number.In other word the number of nearest atoms.
CN=12
Coordination number of FCC structure is 12.These 12 atoms are 4 atoms at the at corner ,4 atoms at 4 faces and 4 atoms of next unit cell.
APF
[tex]APF=\dfrac{Z\times \dfrac{4}{3}\pi R^3}{a^3}[/tex]
We know that for FCC
[tex]4R=\sqrt{2}\ a[/tex]
Now by putting the values
[tex]APF=\dfrac{4\times \dfrac{4}{3}\pi R^3}{\left(\dfrac{4R}{\sqrt2}\right)^3}[/tex]
APF=0.74
APF=74 %
[tex]4R=\sqrt{2}\ a[/tex]
[tex]4\times 0.124=\sqrt{2}\ a[/tex]
a=0.35 nm
Density
[tex]\rho=\dfrac{Z\times M}{N_A\times a^3}[/tex]
We know that M for Ni
M=58.69 g/mol
a=0.35 nm
[tex]N_A=6.023\times 10^{23}\ atom/mol[/tex]
[tex]\rho=\dfrac{Z\times M}{N_A\times a^3}[/tex]
[tex]\rho=\dfrac{4\times 58.69}{6.023\times 10^{23}\times( 0.35\times 10^{-9})^3}\ g/m^3[/tex]
ρ=9090.9 [tex]Kg/m^3[/tex]
