Iron (`alpha`-from) crystallises in a body - centred cubic system with edge length 2.86 A. The density of iron is : (At.wt of Fe=56) 

Options

A. `"7.93 gm cm"^(-3)`
B. `"9.15 gm cm"^(-3)`
C. `"0.72 gm cm"^(-3)`
D. `"4.22 gm cm"^(-3)`

Answer

To find the density of iron (α-form) which crystallizes in a body-centered cubic (BCC) structure, we can follow these steps:

### Step 1: Determine the number of atoms per unit cell in BCC
In a body-centered cubic unit cell:
- There are 8 corner atoms, each contributing \( \frac{1}{8} \) of an atom to the unit cell.
- There is 1 atom at the body center contributing 1 atom.

So, the total number of atoms \( z \) in a BCC unit cell is:
\[ z = 8 \times \frac{1}{8} + 1 = 1 + 1 = 2 \]

### Step 2: Calculate the mass of the unit cell
The atomic weight of iron (Fe) is given as 56 g/mol. To find the mass of the iron atoms in the unit cell, we first calculate the mass of one atom of iron:
\[ \text{Mass of one atom} = \frac{\text{Atomic weight}}{\text{Avogadro's number}} = \frac{56 \text{ g/mol}}{6.022 \times 10^{23} \text{ atoms/mol}} \approx 9.3 \times 10^{-23} \text{ g} \]

Now, the mass of the 2 atoms in the unit cell is:
\[ \text{Mass of unit cell} = 2 \times \text{Mass of one atom} = 2 \times 9.3 \times 10^{-23} \text{ g} \approx 1.86 \times 10^{-22} \text{ g} \]

### Step 3: Calculate the volume of the unit cell
The volume \( V \) of the unit cell can be calculated using the edge length \( a \):
\[ V = a^3 \]
Given that the edge length \( a = 2.86 \, \text{Å} = 2.86 \times 10^{-8} \, \text{cm} \):
\[ V = (2.86 \times 10^{-8} \, \text{cm})^3 \approx 2.34 \times 10^{-23} \, \text{cm}^3 \]

### Step 4: Calculate the density
Density \( \rho \) is given by the formula:
\[ \rho = \frac{\text{Mass}}{\text{Volume}} \]
Substituting the values we have:
\[ \rho = \frac{1.86 \times 10^{-22} \text{ g}}{2.34 \times 10^{-23} \text{ cm}^3} \approx 7.93 \, \text{g/cm}^3 \]

### Conclusion
The density of iron (α-form) is approximately \( 7.93 \, \text{g/cm}^3 \).

---