An elemental crystal has density of `8570 kg m^(-3)`. The packing efficiency is `0.68`. If the closest distance between neighbouring atoms is `2.86 Å`. The mass of one atom is `(1 amu = 1.66 xx 10^(-27))kg)`

Options

A. `186` amu
B. `93` amu
C. `46.5` amu
D. `43` amu

Explanation

To solve the problem step by step, we will follow the instructions given in the video transcript and apply the relevant formulas.

### Step 1: Understand the given data
- Density of the crystal, \( \rho = 8570 \, \text{kg/m}^3 \)
- Packing efficiency, \( PE = 0.68 \)
- Closest distance between neighboring atoms, \( d = 2.86 \, \text{Å} = 2.86 \times 10^{-10} \, \text{m} \)
- Mass of one atom in amu, \( 1 \, \text{amu} = 1.66 \times 10^{-27} \, \text{kg} \)

### Step 2: Calculate the radius of the atom
The closest distance between two neighboring atoms is given by:
\[ d = 2r \]
where \( r \) is the radius of the atom. Therefore, we can find \( r \):
\[ r = \frac{d}{2} = \frac{2.86 \times 10^{-10}}{2} = 1.43 \times 10^{-10} \, \text{m} \]

### Step 3: Calculate the volume of the unit cell
The volume of the unit cell \( V \) can be expressed in terms of the radius and the number of atoms per unit cell \( z \):
\[ V = \frac{4}{3} \pi r^3 \times z \]
However, we will relate this to the density and packing efficiency later.

### Step 4: Relate density to mass and volume
The density \( \rho \) can be expressed as:
\[ \rho = \frac{z \cdot m}{V} \]
where \( m \) is the mass of one atom and \( z \) is the number of atoms in the unit cell.

### Step 5: Substitute the volume in terms of radius and packing efficiency
From the packing efficiency, we know:
\[ \text{Volume occupied by atoms} = PE \times V = 0.68 \times V \]
This volume occupied by the atoms can also be expressed as:
\[ 0.68 V = z \cdot \left(\frac{4}{3} \pi r^3\right) \]

### Step 6: Substitute the expressions into the density formula
Substituting \( V \) from the packing efficiency into the density equation:
\[ \rho = \frac{z \cdot m}{\frac{4}{3} \pi r^3} \]
Rearranging gives:
\[ m = \rho \cdot \frac{4}{3} \pi r^3 \cdot \frac{1}{z} \]

### Step 7: Calculate the mass of one atom
Now we need to express \( z \) in terms of packing efficiency. For a simple cubic structure, \( z = 1 \). Thus:
\[ m = \rho \cdot \frac{4}{3} \pi r^3 \]
Substituting the values:
\[ m = 8570 \cdot \frac{4}{3} \pi (1.43 \times 10^{-10})^3 \]
Calculating \( r^3 \):
\[ (1.43 \times 10^{-10})^3 = 2.94 \times 10^{-30} \, \text{m}^3 \]
Now substituting this value into the mass equation:
\[ m = 8570 \cdot \frac{4}{3} \cdot \pi \cdot 2.94 \times 10^{-30} \]
Calculating the numerical value:
\[ m \approx 8570 \cdot 3.93 \times 10^{-30} \approx 3.37 \times 10^{-26} \, \text{kg} \]

### Step 8: Convert mass to amu
To convert the mass from kg to amu:
\[ \text{mass in amu} = \frac{m}{1.66 \times 10^{-27}} \approx \frac{3.37 \times 10^{-26}}{1.66 \times 10^{-27}} \approx 20.3 \, \text{amu} \]

### Final Result
Thus, the mass of one atom is approximately \( 20.3 \, \text{amu} \).