Volume of HCP unit cell is:

Options

A. `24 sqrt2 r^(3)`
B. `8sqrt2 r^(3)`
C. `16 sqrt2 r^(3)`
D. `24 sqrt3 r^(3)`

Answer

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

### Step 1: Understand the Structure of HCP
The HCP unit cell has a hexagonal base and a certain height. The volume of the unit cell can be calculated using the formula:
\[ \text{Volume} = \text{Area of base} \times \text{Height} \]

### Step 2: Calculate the Area of the Hexagonal Base
The base of the HCP unit cell is a hexagon that can be divided into six equilateral triangles. The area \( A \) of one equilateral triangle with side length \( a \) is given by:
\[ A = \frac{\sqrt{3}}{4} a^2 \]
Thus, the area of the hexagonal base, which consists of six such triangles, is:
\[ \text{Area of base} = 6 \times \frac{\sqrt{3}}{4} a^2 = \frac{3\sqrt{3}}{2} a^2 \]

### Step 3: Determine the Height of the Unit Cell
For the HCP structure, the relationship between the height \( h \) of the unit cell and the side length \( a \) is given by:
\[ \frac{h}{a} = \frac{\sqrt{8}}{3} \]
From this, we can express the height \( h \) as:
\[ h = \frac{\sqrt{8}}{3} a \]

### Step 4: Substitute Area and Height into the Volume Formula
Now, substituting the area of the base and the height into the volume formula:
\[ \text{Volume} = \text{Area of base} \times \text{Height} = \left(\frac{3\sqrt{3}}{2} a^2\right) \times \left(\frac{\sqrt{8}}{3} a\right) \]

### Step 5: Simplify the Expression
Now we simplify the expression:
\[ \text{Volume} = \frac{3\sqrt{3}}{2} a^2 \times \frac{\sqrt{8}}{3} a = \frac{\sqrt{8}}{2} a^3 \sqrt{3} \]
This can be further simplified as:
\[ \text{Volume} = \frac{3 \cdot 2\sqrt{2}}{2} a^3 = 3\sqrt{2} a^3 \]

### Step 6: Relate Side Length to Atomic Radius
In HCP, the side length \( a \) is related to the atomic radius \( r \) by the formula:
\[ a = 2r \]
Substituting this into the volume expression gives:
\[ \text{Volume} = 3\sqrt{2} (2r)^3 = 3\sqrt{2} \cdot 8r^3 = 24\sqrt{2} r^3 \]

### Final Answer
Thus, the volume of the HCP unit cell is:
\[ \text{Volume} = 24\sqrt{2} r^3 \]