The atomic fraction `(d)` of tin in bronze (fcc) with a density of `7717 kg m^(-3)` and a lattice parameter of `3.903 Å` is `(Aw Cu = 63.54, Sn = 118.7, 1 amu = 1.66 xx 10^(-27 kg))`

Options

A. `0.01`
B. `0.05`
C. `0.10`
D. `3.8`

Explanation

To find the atomic fraction of tin (Sn) in bronze (an alloy of copper and tin) with the given parameters, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**:
We need to find the atomic fraction of tin in bronze, which has a face-centered cubic (FCC) structure, a density of 7717 kg/m³, and a lattice parameter of 3.903 Å.

2. **Convert Lattice Parameter to Meters**:
Convert the lattice parameter from angstroms to meters:
\[ a = 3.903 \, \text{Å} = 3.903 \times 10^{-10} \, \text{m} \]

3. **Calculate the Volume of the Unit Cell**:
The volume \( V \) of the unit cell for an FCC structure is given by:
\[ V = a^3 = (3.903 \times 10^{-10})^3 \, \text{m}^3 \]

4. **Calculate the Volume**:
\[ V = 3.903^3 \times 10^{-30} \approx 5.96 \times 10^{-29} \, \text{m}^3 \]

5. **Use the Density Formula**:
The density \( \rho \) is given by:
\[ \rho = \frac{\text{mass of atoms in unit cell}}{\text{volume of unit cell}} \]
Rearranging gives us:
\[ \text{mass of atoms in unit cell} = \rho \times V \]

6. **Calculate Mass of Atoms in Unit Cell**:
\[ \text{mass} = 7717 \, \text{kg/m}^3 \times 5.96 \times 10^{-29} \, \text{m}^3 \approx 4.59 \times 10^{-25} \, \text{kg} \]

7. **Convert Mass to Atomic Mass Units (amu)**:
Since \( 1 \, \text{amu} = 1.66 \times 10^{-27} \, \text{kg} \), we convert the mass:
\[ \text{mass in amu} = \frac{4.59 \times 10^{-25}}{1.66 \times 10^{-27}} \approx 276.5 \, \text{amu} \]

8. **Set Up the Equation for Atomic Mass**:
Let \( N_{Cu} \) be the number of copper atoms and \( N_{Sn} \) be the number of tin atoms in the unit cell. The total mass can be expressed as:
\[ N_{Cu} \times 63.54 + N_{Sn} \times 118.7 = 276.5 \]
For FCC, \( N_{Cu} + N_{Sn} = 4 \) (since there are 4 atoms per unit cell in FCC).

9. **Solve the System of Equations**:
From \( N_{Cu} + N_{Sn} = 4 \), we can express \( N_{Cu} = 4 - N_{Sn} \). Substitute this into the mass equation:
\[ (4 - N_{Sn}) \times 63.54 + N_{Sn} \times 118.7 = 276.5 \]
Simplifying gives:
\[ 254.16 - 63.54N_{Sn} + 118.7N_{Sn} = 276.5 \]
\[ 55.16N_{Sn} = 22.34 \implies N_{Sn} \approx 0.405 \]

10. **Calculate Atomic Fraction**:
The atomic fraction of tin \( d \) is given by:
\[ d = \frac{N_{Sn}}{N_{Cu} + N_{Sn}} = \frac{0.405}{4} \approx 0.10125 \]

### Final Result:
The atomic fraction of tin in bronze is approximately \( 0.10125 \).