The angle between the two planes represented by the Miller indices (110) and (111) in a simple cubic lattice is:

Options

A. `30^@`
B. `45^@`
C. `60^@`
D. none

Explanation

To find the angle between the two planes represented by the Miller indices (110) and (111) in a simple cubic lattice, we can follow these steps:

### Step 1: Identify the Normal Vectors
The Miller indices (hkl) represent planes in a crystal lattice, and the normal vector to a plane (hkl) can be represented as \( \vec{n} = (h, k, l) \).
- For the plane (110), the normal vector \( \vec{n_1} = (1, 1, 0) \).
- For the plane (111), the normal vector \( \vec{n_2} = (1, 1, 1) \).

### Step 2: Calculate the Dot Product of the Normal Vectors
The dot product of two vectors \( \vec{a} \) and \( \vec{b} \) is given by:
\[ \vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z \]
Calculating the dot product \( \vec{n_1} \cdot \vec{n_2} \):
\[ \vec{n_1} \cdot \vec{n_2} = (1)(1) + (1)(1) + (0)(1) = 1 + 1 + 0 = 2 \]

### Step 3: Calculate the Magnitudes of the Normal Vectors
The magnitude of a vector \( \vec{a} = (a_x, a_y, a_z) \) is given by:
\[ |\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} \]
Calculating the magnitudes:
- For \( \vec{n_1} \):
\[ |\vec{n_1}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{1 + 1 + 0} = \sqrt{2} \]
- For \( \vec{n_2} \):
\[ |\vec{n_2}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

### Step 4: Use the Dot Product to Find the Cosine of the Angle
The cosine of the angle \( \theta \) between the two vectors is given by:
\[ \cos \theta = \frac{\vec{n_1} \cdot \vec{n_2}}{|\vec{n_1}| |\vec{n_2}|} \]
Substituting the values:
\[ \cos \theta = \frac{2}{\sqrt{2} \cdot \sqrt{3}} = \frac{2}{\sqrt{6}} = \frac{2\sqrt{6}}{6} = \frac{\sqrt{6}}{3} \]

### Step 5: Calculate the Angle \( \theta \)
To find the angle \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{\sqrt{6}}{3}\right) \]
Calculating this gives approximately:
\[ \theta \approx 35.27^\circ \]

### Final Answer
The angle between the two planes represented by the Miller indices (110) and (111) in a simple cubic lattice is approximately \( 35.27^\circ \).

---