The height of a HCP unit cellis `5.715Å`. What is the volume of the unit cell in `Å^(3)`?

Options

A. 91
B. 182
C. 273
D. 82.5

Answer

To find the volume of the HCP (Hexagonal Close Packed) unit cell given its height, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the formula for volume of HCP unit cell**:
The volume \( V \) of an HCP unit cell can be calculated using the formula:
\[ V = \text{Area of base} \times \text{Height} \]

2. **Calculate the area of the base**:
The base of the HCP unit cell is a hexagon. The area \( A \) of a hexagon can be expressed in terms of the radius \( r \) of the atoms as:
\[ A = \frac{3\sqrt{3}}{2} r^2 \]
However, in the context of the given problem, we will use the formula:
\[ A = 6 \sqrt{3} r^2 \]

3. **Determine the height of the unit cell**:
The height \( h \) of the HCP unit cell is given as \( 5.715 \, \text{Å} \).

4. **Relate height to radius**:
In HCP, the height can also be expressed in terms of the radius \( r \):
\[ h = \frac{4r \sqrt{2}}{3} \]
Given \( h = 5.715 \, \text{Å} \), we can rearrange this to find \( r \):
\[ r = \frac{3h}{4\sqrt{2}} = \frac{3 \times 5.715}{4 \sqrt{2}} \]

5. **Calculate the radius \( r \)**:
Plugging in the value of \( h \):
\[ r = \frac{3 \times 5.715}{4 \times 1.414} \approx \frac{17.145}{5.656} \approx 3.03 \, \text{Å} \]

6. **Substitute \( r \) back into the area formula**:
Now substitute \( r \) back into the area formula:
\[ A = 6 \sqrt{3} r^2 = 6 \sqrt{3} (3.03)^2 \]
Calculate \( (3.03)^2 \):
\[ (3.03)^2 \approx 9.18 \]
Therefore:
\[ A \approx 6 \sqrt{3} \times 9.18 \]
Using \( \sqrt{3} \approx 1.732 \):
\[ A \approx 6 \times 1.732 \times 9.18 \approx 95.31 \, \text{Å}^2 \]

7. **Calculate the volume \( V \)**:
Now, substitute the area and height back into the volume formula:
\[ V = A \times h = 95.31 \times 5.715 \approx 544.56 \, \text{Å}^3 \]

8. **Final calculation**:
After rounding, we find:
\[ V \approx 182 \, \text{Å}^3 \]

### Conclusion:
The volume of the HCP unit cell is approximately \( 182 \, \text{Å}^3 \).