A metallic crystal cystallizes into a lattice containing a sequence of layers `ABABAB…`. Any packing of spheres leaves out voids in the lattice. What percentage by volume of this lattice is occupied?

Options

A. `74%`
B. `26%`
C. `50%`
D. none of these

Explanation

To find the percentage by volume of a metallic crystal lattice that crystallizes into a sequence of layers `ABABAB…`, we can follow these steps:

### Step 1: Identify the type of lattice
The given arrangement `ABABAB…` corresponds to a cubic close-packed (CCP) structure, which is also known as face-centered cubic (FCC) lattice.

### Step 2: Understand the arrangement of atoms in FCC
In an FCC lattice:
- Atoms are located at each of the eight corners of the cube and at the centers of each of the six faces.
- Each corner atom is shared by eight adjacent unit cells, and each face-centered atom is shared by two unit cells.

### Step 3: Calculate the number of atoms per unit cell
The total contribution of atoms in one FCC unit cell is:
- Corner atoms: \( \frac{1}{8} \times 8 = 1 \) atom
- Face-centered atoms: \( \frac{1}{2} \times 6 = 3 \) atoms
- Total = \( 1 + 3 = 4 \) atoms per unit cell.

### Step 4: Calculate the volume of the unit cell
Let the edge length of the cube be \( a \). The volume of the unit cell (cube) is given by:
\[ \text{Volume of unit cell} = a^3 \]

### Step 5: Relate the edge length to the radius of the atoms
In an FCC lattice, the relationship between the edge length \( a \) and the radius \( r \) of the atoms can be derived from the face diagonal:
- The face diagonal length can be expressed as \( a\sqrt{2} \).
- The face diagonal consists of 4 radii (2 radii from each of the two atoms touching along the diagonal):
\[ 4r = a\sqrt{2} \implies a = \frac{4r}{\sqrt{2}} = 2\sqrt{2}r \]

### Step 6: Calculate the volume of the spheres
The volume of one atom (considered as a sphere) is given by:
\[ \text{Volume of one sphere} = \frac{4}{3}\pi r^3 \]
Thus, the volume of 4 spheres (since there are 4 atoms in the unit cell) is:
\[ \text{Volume of 4 spheres} = 4 \times \frac{4}{3}\pi r^3 = \frac{16}{3}\pi r^3 \]

### Step 7: Calculate the volume of the unit cell in terms of \( r \)
Substituting \( a = 2\sqrt{2}r \) into the volume of the unit cell:
\[ \text{Volume of unit cell} = (2\sqrt{2}r)^3 = 16\sqrt{2}r^3 \]

### Step 8: Calculate the fraction of the volume occupied
The fraction of the volume occupied by the spheres in the unit cell is:
\[ \text{Fraction occupied} = \frac{\text{Volume of 4 spheres}}{\text{Volume of unit cell}} = \frac{\frac{16}{3}\pi r^3}{16\sqrt{2}r^3} = \frac{\pi}{3\sqrt{2}} \]

### Step 9: Convert fraction to percentage
To find the percentage of the volume occupied:
\[ \text{Percentage occupied} = \left(\frac{\pi}{3\sqrt{2}}\right) \times 100 \]
Calculating this gives approximately:
\[ \text{Percentage occupied} \approx 74\% \]

### Final Answer
Thus, the percentage by volume of this lattice that is occupied is **74%**.

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