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Sagot :
The question is incomplete. The complete question is :
Iron β is a solid phase of iron still unknown to science. The only difference between it and ordinary iron is that Iron β forms a crystal with an fcc unit cell and a lattice constant, a = 0.352 nm. Calculate the density of Iron β.
Solution :
The density is given by :
[tex]$\rho = \frac{ZM}{a^3N_0} \ \ g/cm^3$[/tex] ..................(i)
Here, Z = number of atoms in a unit cell
M = atomic mass
[tex]$N_0$[/tex] = Avogadro's number = [tex]$6.022 \times 10^{23}$[/tex]
a = edge length or the lattice constant
Now for FCC lattice, the number of atoms in a unit cell is 4.
So, Z = 4
Atomic mass of iron, M = 55.84 g/ mole
Given a = 0.352 nm = [tex]$3.52 \times 10^{-8}$[/tex] cm
From (i),
[tex]$\rho = \frac{ZM}{a^3N_0} $[/tex]
[tex]$\rho = \frac{4 \times 55.84}{(3.52 \times 10^{-8})^3 \times 6.022 \times 10^{23}} $[/tex]
[tex]$= 8.51 \ \ g \ cm ^{-3}$[/tex]
Therefore, the density of Iron β is [tex]$ 8.51 \ \ g \ cm ^{-3}$[/tex].
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