The Solid State

The correct option is (C) : 2

In BCC, the cube's body centre and corners are occupied by lattice atoms. A cube contains 8 corners, each of which contributes of the total. The effective number of atoms per cube = ×8 = 1. Additionally, each cube will contain one body centre, and a body centre will have a contribution of 1. Therefore, the actual number of atoms in a cube is 1. So, the BCC has a total of 1+1=2 atoms.

The vacant space in bcc lattice unit cell is 0.32.

BCC stands for body-centered cubic lattice, which is one of the specific arrangements of atoms in the crystalline state. Eight atoms occupy the corners of the cube in the BCC unit cell. There is one atom at the center of the cube as well. 

Packing efficiency in bcc lattice =
vacant space in bcc lattice =

Therefore, the correct answer is option B.

Body-Centered Cubic (BCC) Calculation

For BCC, the formula to calculate the edge length is:

Where:

• Z = Number of atoms per unit cell (for BCC, Z = 2)
• M = Molar mass (for lithium, M = 6.94 g/mol)
• N_A = Avogadro's number (6.02 × 10²³ atoms/mol)
• a = Edge length of the unit cell

Now, substituting the values into the formula:

Rearranging the equation to solve for :

After solving:

Conclusion:

The edge length of the BCC unit cell is 352 pm.

Calcium Fluoride Crystal Structure

Calcium fluoride crystallizes in a Face-Centered Cubic (FCC) unit cell with an edge length of Angstroms. The unit cell structure consists of:

• Calcium cations ( ) in blue, which occupy the FCC lattice sites.
• Fluoride anions ( ) in green, occupying all tetrahedral sites.

The fluoride anions are surrounded by 4 calcium cations, which are arranged at the vertices of a tetrahedron, creating a tetrahedral coordination geometry. The calcium cations have 8 fluoride anions as neighbors, arranged at the corners of a smaller cube.

The coordination ratio is or , meaning for every two calcium cations, there is one fluoride anion.

In , the coordination number of is 8, as each calcium ion is surrounded by 8 fluoride ions in the structure.

To find the atomic mass of copper, we use the formula for density:

Density (d) = Mass of atoms in the unit cell / Volume of the unit cell

We are given:

• Density, d = 8.92 g cm^-3
• Edge length of unit cell, a = 3.608 x 10^-8 cm

For a face-centered cubic (fcc) unit cell, there are 4 atoms per unit cell. The volume of the unit cell, V, is:

V = a³ = (3.608 x 10^-8)³ cm³

Then, calculate:

V = 3.608³ x 10^-24 = 4.695 x 10^-23 cm³

The mass of the unit cell is the product of the number of atoms per unit cell, the atomic mass (M), and the conversion factor to grams (divided by Avogadro's number, N_A):

Mass = (4 x M) / N_A

Avogadro's number, N_A = 6.022 x 10²³ mol^-1. Substitute the known values in the density equation:

8.92 = [(4 x M) / (6.022 x 10²³)] / (4.695 x 10^-23)

Solving for M:

8.92 = (4M / 6.022) / 4.695

8.92 x 4.695 = 4M / 6.022

41.86644 = 0.664M

M = 41.86644 / 0.664

M = 63.059 g/mol

Thus, the atomic mass of copper is approximately 63.1 u.

In a face-centered cubic (fcc) unit cell, there are several types of voids or interstitial sites where smaller atoms or ions can fit. One of these voids is the octahedral void, which in the context of an fcc lattice, is found at specific locations including the edge centers and the center of the cubic unit cell.

For an octahedral void located at the edge center of a cube, its geometry involves sharing the void among multiple neighboring unit cells. Specifically, an edge center octahedral void is shared by four unit cells.

Given this sharing distribution, the fraction of the void associated with each individual unit cell can be calculated. Since each edge is shared by four cells, the fraction of an edge-centered octahedral void that belongs to one unit cell is .

Thus, the correct answer to the question is the fraction of the octahedral void at the edge center that lies within one unit cell of the fcc lattice.