To find the lattice constant of a simple cubic lattice given the interplanar spacing of a (002) plane, we can use the formula for interplanar spacing in a cubic lattice:

d_{hkl} = \frac{a}{\sqrt{h^2 + k^2 + l^2}}

Here, d_{hkl} is the interplanar spacing, a is the lattice constant, and h, k, l are the Miller indices of the plane.

Given:

• Interplanar spacing d_{002} = 3 \, \text{Å}
• Miller indices for the (002) plane: h = 0, k = 0, l = 2

Substitute these values into the formula:

3 = \frac{a}{\sqrt{0^2 + 0^2 + 2^2}}

Simplify the square root term:

3 = \frac{a}{\sqrt{4}} → 3 = \frac{a}{2}

Solving for a, we get:

a = 3 \times 2 = 6 \, \text{Å}

Therefore, the lattice constant for the simple cubic lattice is 6.0 Å.

Hence, the correct answer is:

6.0 Å