If a crystal a contains a total of `N` atoms and `n` Schottky defects are produced by removing `n` cations and `r` anions from the interior of the crystal, then `n =` ……. .

Answer

To solve the problem regarding the number of Schottky defects in a crystal, we need to derive the expression for `n`, which represents the number of Schottky defects produced by removing `n` cations and `r` anions from the crystal.

### Step-by-Step Solution:

1. **Understanding Schottky Defects**:
- Schottky defects occur when an equal number of cations and anions are removed from the crystal lattice. This maintains charge neutrality in the crystal.

2. **Defining Variables**:
- Let `N` be the total number of atoms in the crystal.
- Let `n` be the number of cations removed.
- Let `r` be the number of anions removed.

3. **Charge Neutrality Condition**:
- For a Schottky defect, the number of cations removed (`n`) must equal the number of anions removed (`r`).
- Therefore, we can write:
\[ n = r \]

4. **Total Number of Defects**:
- The total number of Schottky defects (`n`) produced in the crystal can be expressed in terms of the total number of atoms and the energy required to create these defects.
- The relationship can be given by the equation:
\[ n = N \cdot e^{-\frac{E}{2kT}} \]
- Here, `E` is the energy required to create a Schottky defect, `k` is Boltzmann's constant, and `T` is the temperature in Kelvin.

5. **Final Expression**:
- From the above relationship, we can conclude that:
\[ n = N \cdot e^{-\frac{E}{2kT}} \]

### Final Answer:
Thus, the value of `n` (the number of Schottky defects) is given by:
\[ n = N \cdot e^{-\frac{E}{2kT}} \]