Question

A cubic lattice has A atoms at the body center, B atoms at the corners, C atoms at half of the face centers. The formula of the lattice is

• A. ABC$_2$
• B. AB$_2$C$_4$
• C. A$_2$B$_2$C$_3$
• D. ABC$_3$

Hint:

Always account for atom contribution based on position
Multiply by LCM if fractional atoms are found

Answer 1

The Correct Option is C

- A at body center = 1 atom per unit cell (fully inside) - B at 8 corners = \( \frac{1}{8} \times 8 = 1 \) atom - C at 3 face centers (half of 6) = \( \frac{1}{2} \times 3 = 1.5 \) atoms To remove decimals, multiply all by 2: \[ \text{A} : 1 \Rightarrow 2,\quad \text{B} : 1 \Rightarrow 2,\quad \text{C} : 1.5 \Rightarrow 3 \] Empirical formula = A$_2$B$_2$C$_3$