Copper crystal has a face-centred cubic lattice structure. Atomic radius of copper atom is 128 pm. Calculate the density of copper. Atomic mass of copper=63.5

Explanation

To calculate the density of copper with a face-centered cubic (FCC) lattice structure, we can follow these steps:

### Step 1: Understand the formula for density
The density (\(d\)) of a crystal can be calculated using the formula:
\[ d = \frac{Z \times M}{N \times A^3} \]
where:
- \(Z\) = number of atoms per unit cell
- \(M\) = molar mass of the substance (in grams per mole)
- \(N\) = Avogadro's number (\(6.022 \times 10^{23}\) mol\(^{-1}\))
- \(A\) = edge length of the unit cell (in cm)

### Step 2: Determine the number of atoms per unit cell for FCC
For a face-centered cubic (FCC) structure, the number of atoms per unit cell (\(Z\)) is:
\[ Z = 4 \]

### Step 3: Find the edge length (\(A\)) using the atomic radius
The relationship between the atomic radius (\(R\)) and the edge length (\(A\)) for an FCC lattice is given by:
\[ A = 2\sqrt{2} \times R \]
Given that the atomic radius of copper is \(R = 128 \, \text{pm} = 128 \times 10^{-12} \, \text{m}\).

Now, substituting the value of \(R\):
\[ A = 2\sqrt{2} \times 128 \, \text{pm} = 2 \times 1.414 \times 128 \, \text{pm} \approx 362 \, \text{pm} \]
Convert \(A\) to centimeters:
\[ A = 362 \, \text{pm} = 362 \times 10^{-12} \, \text{m} = 362 \times 10^{-10} \, \text{cm} \]

### Step 4: Substitute values into the density formula
Now we can substitute the known values into the density formula:
- \(M = 63.5 \, \text{g/mol}\)
- \(N = 6.022 \times 10^{23} \, \text{mol}^{-1}\)
- \(A = 362 \times 10^{-10} \, \text{cm}\)

Substituting these values:
\[ d = \frac{4 \times 63.5}{6.022 \times 10^{23} \times (362 \times 10^{-10})^3} \]

### Step 5: Calculate the density
First, calculate \(A^3\):
\[ A^3 = (362 \times 10^{-10})^3 = 4.74 \times 10^{-29} \, \text{cm}^3 \]

Now substituting back into the density equation:
\[ d = \frac{254}{6.022 \times 10^{23} \times 4.74 \times 10^{-29}} \]
Calculating the denominator:
\[ 6.022 \times 10^{23} \times 4.74 \times 10^{-29} \approx 2.85 \times 10^{-5} \]

Now, calculating the density:
\[ d \approx \frac{254}{2.85 \times 10^{-5}} \approx 8.9 \, \text{g/cm}^3 \]

### Final Answer
The density of copper is approximately \(8.9 \, \text{g/cm}^3\).

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