An elemental crystal has a density of 8570 kg/`m^3` . The packing efficiency is 0.68. The closest distance of approach between neighbouring atom is 2.86 Å. What is the mass of one atom approximately?

Options

A. 93 amu
B. 39 amu
C. 63 amu
D. 29 amu

Answer

To find the mass of one atom in the given elemental crystal, we will follow these steps:

### Step 1: Understand the Given Data
- Density (ρ) = 8570 kg/m³
- Packing Efficiency (PE) = 0.68
- Closest distance of approach (2R) = 2.86 Å = 2.86 × 10⁻¹⁰ m

### Step 2: Determine the Effective Number of Atoms in BCC
In a Body-Centered Cubic (BCC) lattice:
- There are 8 corner atoms, each contributing 1/8 to the unit cell.
- There is 1 atom at the body center contributing fully.

Effective number of atoms (Z) = 8 × (1/8) + 1 = 2

### Step 3: Relate the Closest Distance to Atomic Radius
The closest distance of approach (2R) is equal to the distance between two adjacent atoms in the BCC structure. Thus:
\[ 2R = 2.86 \, \text{Å} \]
From this, we can find the atomic radius (R):
\[ R = \frac{2.86 \, \text{Å}}{2} = 1.43 \, \text{Å} = 1.43 \times 10^{-10} \, \text{m} \]

### Step 4: Relate Atomic Radius to Edge Length
In a BCC lattice, the relationship between the atomic radius (R) and the edge length (A) is given by:
\[ 4R = \sqrt{3}A \]
Substituting for R:
\[ 4(1.43 \times 10^{-10}) = \sqrt{3}A \]
Calculating A:
\[ A = \frac{4(1.43 \times 10^{-10})}{\sqrt{3}} \]
\[ A \approx 3.30 \times 10^{-10} \, \text{m} \]

### Step 5: Use the Density Formula
The density formula is given by:
\[ \rho = \frac{Z \cdot M}{N_A \cdot A^3} \]
Where:
- ρ = density (8570 kg/m³)
- Z = effective number of atoms (2)
- M = molar mass (unknown)
- \( N_A \) = Avogadro's number (6.022 × 10²³ mol⁻¹)
- A = edge length (3.30 × 10⁻¹⁰ m)

Rearranging the formula to solve for M:
\[ M = \rho \cdot N_A \cdot A^3 / Z \]

### Step 6: Substitute Values and Calculate M
Substituting the values into the equation:
\[ M = \frac{8570 \, \text{kg/m}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1} \cdot (3.30 \times 10^{-10} \, \text{m})^3}{2} \]

Calculating \( A^3 \):
\[ A^3 = (3.30 \times 10^{-10})^3 \approx 3.59 \times 10^{-29} \, \text{m}^3 \]

Now substituting:
\[ M \approx \frac{8570 \cdot 6.022 \times 10^{23} \cdot 3.59 \times 10^{-29}}{2} \]

Calculating the mass M:
\[ M \approx 92.39 \, \text{g/mol} \]

### Step 7: Calculate Mass of One Atom
To find the mass of one atom:
\[ \text{Mass of one atom} = \frac{M}{N_A} \]
\[ \text{Mass of one atom} \approx \frac{92.39 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1}} \approx 1.53 \times 10^{-25} \, \text{kg} \]

### Step 8: Convert to Atomic Mass Units (amu)
1 g = 1000 mg = 1000 × 10⁻³ g = 10³ g
\[ \text{Mass in amu} \approx \frac{92.39}{6.022} \approx 15.36 \, \text{amu} \]

### Final Answer
The mass of one atom is approximately **93 amu**.

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