Question

Nickel has an FCC structure and an atomic radius of 0.124 nm. (12 points) a. What is the coordination number? b. Calculate the APE C. Calculate the lattice parameter, "a" d. Calculate the theoretical density

Answer 1

Answer:

a)CN=12

b)APF=74 %

c)a=0.35 nm

d)ρ=9090.9 $Kg/m^3$

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

$APF=\dfrac{Z\times \dfrac{4}{3}\pi R^3}{a^3}$

We know that for FCC

$4R=\sqrt{2}\ a$

Now by putting the values

$APF=\dfrac{4\times \dfrac{4}{3}\pi R^3}{\left(\dfrac{4R}{\sqrt2}\right)^3}$

APF=0.74

APF=74 %

$4R=\sqrt{2}\ a$

$4\times 0.124=\sqrt{2}\ a$

a=0.35 nm

Density

$\rho=\dfrac{Z\times M}{N_A\times a^3}$

We know that M for Ni

M=58.69 g/mol

a=0.35 nm

$N_A=6.023\times 10^{23}\ atom/mol$

$\rho=\dfrac{Z\times M}{N_A\times a^3}$

$\rho=\dfrac{4\times 58.69}{6.023\times 10^{23}\times( 0.35\times 10^{-9})^3}\ g/m^3$

ρ=9090.9 $Kg/m^3$