Density of a unit cell is respresented as
`rho = (" Effective no. of atoms (s)" xx "Mass of a unit cell ")/("Volume of a unit cell ")=(Z.M)/(N_(A).a^(3))`
where , Z = effective no . of atoms(s) or ion (s) per unit cell.
M= At . mass// formula
`N_(A) `= Avogadro' s no . `rArr 6.0323 xx 10^(23)`
a= edge length of unit cell
Silver crystallizes in a fcc lattice and has a density of `10.6 g//cm^(3)`. What is the length of a edge of the unit cell ?

Options

A. `40.07 nm`
B. `0.2035 nm`
C. `0.101 nm`
D. `4.07 nm`

Answer

To find the length of the edge of the unit cell for silver, which crystallizes in a face-centered cubic (FCC) lattice, we can follow these steps:

### Step 1: Identify the effective number of atoms (Z) in the FCC unit cell
In a face-centered cubic (FCC) lattice:
- There are 8 corner atoms, each contributing \( \frac{1}{8} \) of an atom to the unit cell.
- There are 6 face-centered atoms, each contributing \( \frac{1}{2} \) of an atom to the unit cell.

Calculating the total contribution:
\[ Z = \left(8 \times \frac{1}{8}\right) + \left(6 \times \frac{1}{2}\right) = 1 + 3 = 4 \]

### Step 2: Use the density formula
The formula for density (\( \rho \)) of a unit cell is given by:
\[ \rho = \frac{Z \cdot M}{N_A \cdot a^3} \]
Where:
- \( \rho \) = density of the unit cell (given as \( 10.6 \, \text{g/cm}^3 \))
- \( Z \) = effective number of atoms per unit cell (calculated as 4)
- \( M \) = atomic mass of silver (given as \( 107 \, \text{g/mol} \))
- \( N_A \) = Avogadro's number (\( 6.022 \times 10^{23} \, \text{atoms/mol} \))
- \( a \) = edge length of the unit cell (unknown)

### Step 3: Rearrange the formula to solve for \( a^3 \)
Rearranging the density formula to find \( a^3 \):
\[ a^3 = \frac{Z \cdot M}{N_A \cdot \rho} \]

### Step 4: Substitute the values into the equation
Substituting the known values:
\[ a^3 = \frac{4 \cdot 107 \, \text{g/mol}}{(6.022 \times 10^{23} \, \text{atoms/mol}) \cdot (10.6 \, \text{g/cm}^3)} \]

### Step 5: Calculate \( a^3 \)
Calculating the numerator:
\[ 4 \cdot 107 = 428 \, \text{g} \]
Calculating the denominator:
\[ (6.022 \times 10^{23}) \cdot (10.6) = 6.39232 \times 10^{24} \, \text{g/cm}^3 \]
Now substituting these into the equation:
\[ a^3 = \frac{428}{6.39232 \times 10^{24}} \approx 6.693 \times 10^{-23} \, \text{cm}^3 \]

### Step 6: Calculate the edge length \( a \)
Taking the cube root to find \( a \):
\[ a = (6.693 \times 10^{-23})^{1/3} \approx 4.0036 \times 10^{-8} \, \text{cm} \]

### Step 7: Convert \( a \) to angstroms
To convert from cm to angstroms (1 cm = \( 10^{10} \) angstroms):
\[ a \approx 4.0036 \times 10^{-8} \, \text{cm} \times 10^{10} \, \text{Å/cm} \approx 4.0036 \, \text{Å} \]

### Final Answer:
The length of the edge of the unit cell for silver is approximately **4.0036 Å**.

---