To calculate the density of the unit cell of `MgTiO3` after removing the atoms from one of the face diagonals in the perovskite structure, we can follow these steps:
### Step 1: Understand the Structure
In the perovskite structure, the unit cell consists of:
- Corner atoms: `Mg^(2+)`
- Face center atoms: `O^(2-)`
- Body center atom: `Ti^(4+)`
### Step 2: Determine the Effect of Removing Atoms
When we remove all the atoms from one of the face diagonals, we remove:
- 2 corner `Mg^(2+)` atoms (since there are 2 corners on the face diagonal)
- 1 face center `O^(2-)` atom
### Step 3: Calculate the Number of Atoms Remaining
Initially, the unit cell has:
- 8 corner `Mg^(2+)` atoms (each contributes 1/8 to the unit cell)
- 1 body center `Ti^(4+)` atom (contributes 1)
- 6 face center `O^(2-)` atoms (each contributes 1/2)
After removing the atoms:
- Remaining `Mg^(2+)` atoms: \(8 - 2 = 6\) corners, contributing \(6 \times \frac{1}{8} = \frac{6}{8} = \frac{3}{4}\)
- Remaining `Ti^(4+)` atoms: \(1\)
- Remaining `O^(2-)` atoms: \(6 - 1 = 5\) face centers, contributing \(5 \times \frac{1}{2} = \frac{5}{2}\)
### Step 4: Write the New Formula
The new formula after removing the atoms is:
\[ \text{Mg}_{\frac{3}{4}} \text{Ti}_{1} \text{O}_{\frac{5}{2}} \]
### Step 5: Calculate the Mass of the Unit Cell
Using the atomic masses:
- Mass of `Mg = 24 g/mol`
- Mass of `Ti = 48 g/mol`
- Mass of `O = 16 g/mol`
The total mass of the unit cell can be calculated as:
\[
\text{Mass} = \left( \frac{3}{4} \times 24 \right) + \left( 1 \times 48 \right) + \left( \frac{5}{2} \times 16 \right)
\]
Calculating each term:
- Mass of `Mg`: \( \frac{3}{4} \times 24 = 18 \, \text{g/mol} \)
- Mass of `Ti`: \( 1 \times 48 = 48 \, \text{g/mol} \)
- Mass of `O`: \( \frac{5}{2} \times 16 = 40 \, \text{g/mol} \)
Total mass:
\[
\text{Total Mass} = 18 + 48 + 40 = 106 \, \text{g/mol}
\]
### Step 6: Calculate the Volume of the Unit Cell
Given that the corner ions are touching each other, we can calculate the side length of the unit cell:
\[
\text{Side length} (A) = 2 \times \text{radius of } Mg^{2+} = 2 \times 0.7 \, \text{Å} = 1.4 \, \text{Å}
\]
Convert to cm:
\[
1.4 \, \text{Å} = 1.4 \times 10^{-8} \, \text{cm}
\]
Volume \( V \) of the unit cell:
\[
V = A^3 = (1.4 \times 10^{-8})^3 \, \text{cm}^3
\]
### Step 7: Calculate the Density
Density \( \rho \) is given by:
\[
\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{106 \, \text{g/mol}}{(1.4 \times 10^{-8})^3}
\]
Calculating the volume:
\[
(1.4 \times 10^{-8})^3 = 2.744 \times 10^{-24} \, \text{cm}^3
\]
Now substituting into the density formula:
\[
\rho = \frac{106 \times 1.66 \times 10^{-24}}{2.744 \times 10^{-24}} \approx 65 \, \text{g/cm}^3
\]
### Final Answer
The density of the unit cell after removing the atoms from one of the face diagonals is approximately **65 g/cm³**.
