Calculate Bravais Lattice Primitive Cell Volume

The volume $ V $ of a primitive cell defined by primitive vectors $ \vec{a_1}, \vec{a_2}, \vec{a_3} $ is given by the scalar triple product:

$ V = |(\vec{a_1} \times \vec{a_2}) \cdot \vec{a_3}| $

Given the primitive vectors:

• $ \vec{a_1} = a\hat{i} $
• $ \vec{a_2} = a\hat{j} $
• $ \vec{a_3} = \frac{a}{2}(\hat{i} + \hat{j} + \hat{k}) $

First, calculate the cross product $ \vec{a_1} \times \vec{a_2} $:

$ \vec{a_1} \times \vec{a_2} = (a\hat{i}) \times (a\hat{j}) = a^2 (\hat{i} \times \hat{j}) = a^2 \hat{k} $

Next, calculate the dot product with $ \vec{a_3} $:

$ (\vec{a_1} \times \vec{a_2}) \cdot \vec{a_3} = (a^2 \hat{k}) \cdot \left(\frac{a}{2}(\hat{i} + \hat{j} + \hat{k})\right) $

$ = \frac{a^3}{2} (\hat{k} \cdot \hat{i} + \hat{k} \cdot \hat{j} + \hat{k} \cdot \hat{k}) $

Since $ \hat{i}, \hat{j}, \hat{k} $ are orthogonal unit vectors ($ \hat{k} \cdot \hat{i} = 0 $, $ \hat{k} \cdot \hat{j} = 0 $, $ \hat{k} \cdot \hat{k} = 1 $):

$ = \frac{a^3}{2} (0 + 0 + 1) = \frac{a^3}{2} $

The volume of the primitive cell is $ V = \left|\frac{a^3}{2}\right| = \frac{a^3}{2} $.

Identify Bravais Lattice Type

The given primitive vectors $ \vec{a_1} = a\hat{i} $, $ \vec{a_2} = a\hat{j} $, and $ \vec{a_3} = \frac{a}{2}(\hat{i} + \hat{j} + \hat{k}) $ are characteristic of a Body-Centered Cubic (BCC) lattice.

Specifically, for a BCC lattice with a conventional cubic lattice parameter $ a_{conv} $, the primitive vectors can be expressed, for example, as:

• $ \vec{p_1} = \frac{a_{conv}}{2}(\hat{i} + \hat{j} - \hat{k}) $
• $ \vec{p_2} = \frac{a_{conv}}{2}(\hat{i} - \hat{j} + \hat{k}) $
• $ \vec{p_3} = \frac{a_{conv}}{2}(-\hat{i} + \hat{j} + \hat{k}) $

By taking linear combinations of these vectors, we can show that the given vectors $ \vec{a_1}, \vec{a_2}, \vec{a_3} $ correspond to the BCC lattice if $ a_{conv} = a $:

• $ \vec{a_1} = a\hat{i} = \vec{p_1} + \vec{p_2} $ (using $ a_{conv}=a $)
• $ \vec{a_2} = a\hat{j} = \vec{p_1} + \vec{p_3} $ (using $ a_{conv}=a $)
• $ \vec{a_3} = \frac{a}{2}(\hat{i} + \hat{j} + \hat{k}) = \vec{p_1} + \vec{p_2} + \vec{p_3} $ (using $ a_{conv}=a $)

Furthermore, the volume of the primitive cell for a BCC lattice is $ V_{BCC} = \frac{a_{conv}^3}{2} $. Since our calculated volume is $ \frac{a^3}{2} $, this implies $ a_{conv} = a $, consistent with the BCC identification. For FCC, the primitive cell volume is $ V_{FCC} = \frac{a_{conv}^3}{4} $, which would not match our calculated volume $ \frac{a^3}{2} $ for any reasonable $ a_{conv} $ relative to the given vectors.

Therefore, the crystal lattice is BCC and the volume of its primitive cell is $ \frac{a^3}{2} $.