The shortest distance between `I^(st) and V^(th)` layer of HCP arrangement is:

Options

A. `8 sqrt((2)/(3)) r`
B. `4sqrt((3)/(2))r`
C. `16(sqrt12)/(3)r`
D. `8sqrt((3)/(2))r`

Answer

To find the shortest distance between the first and fifth layer of a Hexagonal Close-Packed (HCP) arrangement, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding HCP Arrangement**:
- In HCP, the layers are arranged in a specific manner where the first layer is above the second layer, and the third layer is directly above the first layer again, followed by the fourth and fifth layers.

2. **Identifying the Layers**:
- The first layer (I) and the fifth layer (V) are separated by several layers in between. To find the distance between these layers, we need to calculate the height of the layers.

3. **Height of the Tetrahedron**:
- In HCP, the distance between adjacent layers can be represented using the tetrahedral arrangement. The height (h) of a tetrahedron formed by the atoms can be calculated.
- The height of the tetrahedron is given by the formula:
\[ h = \frac{2 \sqrt{2} r^2}{\sqrt{3}} \]

4. **Calculating the Distance Between Layers**:
- The distance between the first and fifth layers involves four heights (since there are four layers between them).
- Therefore, the total height (H) between the first and fifth layers can be calculated as:
\[ H = 4h = 4 \left(\frac{2 \sqrt{2} r^2}{\sqrt{3}}\right) \]
- Simplifying this gives:
\[ H = \frac{8 \sqrt{2} r^2}{\sqrt{3}} \]

5. **Final Expression**:
- Thus, the shortest distance between the first and fifth layers of the HCP arrangement is:
\[ H = \frac{8 \sqrt{2} r}{\sqrt{3}} \]