Solid State

The FCC (face-centered cubic) unit cell is a type of crystal structure where each corner and face center of the cube is occupied by atoms or molecules. In an FCC lattice, the simplest way to determine the distance between the centers of molecules, such as fullerenes (C₆₀), adhering to the crystal lattice rules is to use the face diagonal of the cube. Each face diagonal passes through the centers of two opposite face-centered spheres and contains two radii of the spheres.

Given edge length a is 14.14 Å.

In FCC, the face diagonal relates to the edge length as follows:

Face diagonal = √2 * a = √2 * 14.14 Å.

The smallest distance between the centers of two C₆₀ molecules is along the face diagonal and equals half of it because it stretches from the center of one face-centered atom to the other:

Distance = (√2 * 14.14) / 2 = (1.414 * 14.14) / 2 = 9.99358 Å.

Rounding off, we get the smallest distance as 9.99 Å.

The smallest distance between the centers of two C₆₀ molecules in an FCC unit cell is therefore 9.99 Å.

The question pertains to the solid-state structure of hydrogen fluoride (HF). In the solid state, HF forms a zigzag structure due to hydrogen bonding.

Explanation:

• Hydrogen fluoride molecules are highly polar, and they form strong hydrogen bonds.
• In the solid state, these hydrogen bonds lead to the formation of a chain-like zigzag structure. This is because each hydrogen atom forms a hydrogen bond with a fluorine atom of an adjacent HF molecule.
• As a result, HF arranges itself in a non-linear pattern to maximize hydrogen bonding, which stabilizes the structure.

The correct representation of this zigzag structure corresponds to option (B).

To determine the d spacing for the first-order X-ray diffraction event, we use Bragg's Law, which is given by:

nλ=2d sinθ

Where:

• n is the order of diffraction, which is 1 for the first order.
• λ is the wavelength of the X-ray, λ=1.54 Å.
• d is the interplanar spacing, which we need to find.
• θ is the angle of diffraction, 20.2° in this case.

First, we convert the angle 2θ to θ:

θ=20.2°/2=10.1°

Now, plug the values into Bragg’s Law and solve for d:

1×1.54 Å=2d sin(10.1°)

d=1.54 Å/(2sin(10.1°))

Calculate sin(10.1°):

sin(10.1°)≈0.175

Substitute back to find d:

d=(1.54 Å)/(2×0.175)

d≈4.400 Å

Therefore, the d spacing is approximately 4.400 Å.

The face-centered cubic (fcc) lattice structure is one of the most common packing arrangements in crystalline materials. To calculate the volume fraction, we must determine the fraction of space filled by atoms in the lattice. In an fcc lattice, atoms are arranged in layers with each unit cell containing 4 atoms. When analyzing the packing, it's useful to consider the geometric relationship between the lattice parameter (a) and the atomic radius (r).

In an fcc unit cell, the diagonal of the face is equal to 4r, where r is the atomic radius. The face diagonal can also be expressed in terms of the lattice parameter a using the Pythagorean theorem:

a√2=4r

From this equation, we solve for a:

a = 2√2r

The volume of the unit cell (V_cell) is:

V_cell = a³ = (2√2r)³ = 16√2r³

Next, calculate the total volume occupied by the atoms (V_atoms) within the fcc unit cell. Since there are 4 atoms each with volume (4/3)πr³, we have:

V_atoms = 4×(4/3)πr³ = (16/3)πr³

The volume fraction (VF) is the ratio of V_atoms to V_cell:

VF = V_atoms/V_cell = ((16/3)πr³) / (16√2r³)

Simplifying the fraction:

VF = π/(3√2)

Calculating this expression:

VF ≈ 0.7405

Thus, the volume fraction for an element in an fcc lattice, rounded to two decimal places, is:

0.74

The task is to determine the separation, also known as the interplanar spacing, of the {134} planes in an orthorhombic unit cell with given cell parameters. To solve this, we will use the formula for interplanar spacing for an orthorhombic lattice:

Where:

• are the Miller indices of the plane.
• are the lattice parameters.

Given:

• h = 1, k = 3, l = 4
• a = 0.5 nm, b = 0.6 nm, c = 0.8 nm

Substitute these values into the formula:

Calculate each term in the denominator:

Add these values:

Now, calculate the interplanar spacing:

Compute :

Thus:

Therefore, the separation of the {134} planes is approximately 0.136 nm.

The correct answer is: 0.136

In a two-dimensional square lattice, the unit cell is a square with side length a. The four grey circles inside the unit cell have a radius r. The total area occupied by the grey circles is the combined area of the four circles.

The area of one circle is given by the formula:

A_circle =

Since there are four circles in the unit cell, the total area occupied by the circles is:

A_total = 4 ·

The area of the square unit cell is:

A_unitcell =

Since the distance between the centers of adjacent circles is a, the relationship between a and r is:

a = 2r

Therefore, the area occupied by the grey circles as a percentage of the total unit cell area is:

Percentage occupied = × 100 = × 100

Simplifying this:

Percentage occupied = × 100 = × 100 ≈ 77%

Thus, the percentage area occupied by the grey circles is 77%.

• Using Bragg’s law:

• For the first-order diffraction (n = 1):

• Substituting into Bragg’s law:

The angle corresponding to is:

Thus, the angle θ for the first-order Bragg diffraction is 30◦.