In a cubic lattice each edge of the unit cell is 400 pm. Atomic weigth of the element is 60 and its density is` 625g//c.c.`Avogadro number =`6times10^(23)`.The crystal lattice is

Options

A. Face centred
B. Primitive
C. Body centred
D. End centred

Explanation

To determine the type of crystal lattice from the given parameters, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Edge length of the unit cell (a) = 400 pm = \(400 \times 10^{-10}\) cm
- Atomic weight (m) = 60 g/mol
- Density (d) = 6.25 g/cm³
- Avogadro's number (N_A) = \(6 \times 10^{23}\) molecules/mol

2. **Use the Density Formula:**
The formula for density in terms of the number of atoms per unit cell (z) is given by:
\[ d = \frac{z \cdot m}{N_A \cdot a^3} \]
Rearranging this formula to solve for z gives:
\[ z = \frac{d \cdot N_A \cdot a^3}{m} \]

3. **Convert Edge Length to Centimeters:**
Since the edge length is given in picometers, we convert it to centimeters:
\[ a = 400 \text{ pm} = 400 \times 10^{-10} \text{ cm} = 4 \times 10^{-8} \text{ cm} \]

4. **Calculate \(a^3\):**
Now, calculate \(a^3\):
\[ a^3 = (4 \times 10^{-8} \text{ cm})^3 = 64 \times 10^{-24} \text{ cm}^3 = 6.4 \times 10^{-23} \text{ cm}^3 \]

5. **Substitute Values into the Formula:**
Now substitute the values into the rearranged density formula:
\[ z = \frac{6.25 \text{ g/cm}^3 \cdot (6 \times 10^{23} \text{ mol}^{-1}) \cdot (6.4 \times 10^{-23} \text{ cm}^3)}{60 \text{ g/mol}} \]

6. **Calculate z:**
Perform the calculation:
\[ z = \frac{6.25 \cdot 6 \cdot 6.4}{60} \]
\[ z = \frac{240}{60} = 4 \]

7. **Determine the Type of Crystal Lattice:**
The value of z = 4 indicates that there are 4 atoms per unit cell. This corresponds to a Face-Centered Cubic (FCC) lattice structure.

### Conclusion:
The crystal lattice is **Face-Centered Cubic (FCC)**.