Reciprocal Lattice Vectors Calculation

This solution computes the reciprocal lattice vectors ($b_1, b_2, b_3$) from given direct lattice vectors ($a_1, a_2, a_3$).

Direct Lattice Vectors

The direct lattice vectors are provided as:

• $a_1 = \hat{j} + \hat{k} = (0, 1, 1)$
• $a_2 = 3\hat{i} - 2\hat{k} = (3, 0, -2)$
• $a_3 = 4\hat{i} + 3\hat{j} = (4, 3, 0)$

Reciprocal Lattice Vector Formulas

The reciprocal lattice vectors ($b_1, b_2, b_3$) are defined by:

• $b_1 = \frac{2\pi (a_2 \times a_3)}{V}$
• $b_2 = \frac{2\pi (a_3 \times a_1)}{V}$
• $b_3 = \frac{2\pi (a_1 \times a_2)}{V}$

where $V = a_1 \cdot (a_2 \times a_3)$ is the unit cell volume.

Calculate Unit Cell Volume (V)

First, compute the cross product $a_2 \times a_3$:

$a_2 \times a_3 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 0 & -2 \\ 4 & 3 & 0 \end{vmatrix} = 6\hat{i} - 8\hat{j} + 9\hat{k}$

Next, compute the volume $V$ using the scalar triple product $V = a_1 \cdot (a_2 \times a_3)$:

$V = (0, 1, 1) \cdot (6, -8, 9) = 0(6) + 1(-8) + 1(9) = -8 + 9 = 1$

Calculate Reciprocal Lattice Vectors

Using the volume $V=1$:

Calculate $b_1$

$b_1 = \frac{2\pi (a_2 \times a_3)}{V} = \frac{2\pi (6\hat{i} - 8\hat{j} + 9\hat{k})}{1}$

$b_1 = 2\pi(6\hat{i} - 8\hat{j} + 9\hat{k})$

Calculate $b_2$

Compute the cross product $a_3 \times a_1$:

$a_3 \times a_1 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & 0 \\ 0 & 1 & 1 \end{vmatrix} = 3\hat{i} - 4\hat{j} + 4\hat{k}$

Now, calculate $b_2$:

$b_2 = \frac{2\pi (a_3 \times a_1)}{V} = \frac{2\pi (3\hat{i} - 4\hat{j} + 4\hat{k})}{1}$

$b_2 = 2\pi(3\hat{i} - 4\hat{j} + 4\hat{k})$

Calculate $b_3$

Compute the cross product $a_1 \times a_2$:

$a_1 \times a_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 1 \\ 3 & 0 & -2 \end{vmatrix} = -2\hat{i} + 3\hat{j} - 3\hat{k}$

Now, calculate $b_3$:

$b_3 = \frac{2\pi (a_1 \times a_2)}{V} = \frac{2\pi (-2\hat{i} + 3\hat{j} - 3\hat{k})}{1}$

$b_3 = 2\pi(-2\hat{i} + 3\hat{j} - 3\hat{k})$

Final Reciprocal Lattice Vectors

The calculated reciprocal lattice vectors are:

• $b_1 = 2\pi(6\hat{i} - 8\hat{j} + 9\hat{k})$
• $b_2 = 2\pi(3\hat{i} - 4\hat{j} + 4\hat{k})$
• $b_3 = 2\pi(-2\hat{i} + 3\hat{j} - 3\hat{k})$

This matches option B.