Suppose the mass of a single Ag atoms is 'm' Ag metal crystallises in fcc lattice with unit cell edge length 'a' The density of Ag metal in terms of 'a' and 'm' is:

Options

A. `(4m)/a^(3)`
B. `(2m)/a^(3)`
C. `m/a^(3)`
D. `m/(2a^(3))`

Explanation

To find the density of silver (Ag) metal crystallizing in a face-centered cubic (FCC) lattice in terms of the mass of a single Ag atom (m) and the unit cell edge length (a), we can follow these steps:

### Step 1: Determine the number of Ag atoms in the unit cell
In an FCC lattice, there are 4 atoms per unit cell. This is because:
- Each corner atom is shared among 8 unit cells (1/8 contribution per unit cell).
- Each face-centered atom is shared between 2 unit cells (1/2 contribution per unit cell).
- Therefore, the total number of atoms in an FCC unit cell is:
\[ \text{Total atoms} = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4 \]

### Step 2: Calculate the mass of the unit cell
The mass of the unit cell can be calculated by multiplying the number of atoms in the unit cell by the mass of a single Ag atom:
\[ \text{Mass of unit cell} = \text{Number of atoms} \times \text{Mass of single atom} = 4 \times m \]

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

### Step 4: Calculate the density
Density (ρ) is defined as mass per unit volume. Therefore, we can express the density of Ag metal as:
\[ \rho = \frac{\text{Mass of unit cell}}{\text{Volume of unit cell}} = \frac{4m}{a^3} \]

### Final Answer
Thus, the density of Ag metal in terms of 'a' and 'm' is:
\[ \rho = \frac{4m}{a^3} \]

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