The number of Schottky defects (n) present in an ionic compound containing `N` ions at temperature `T` is given by `n = Ne^(-E//2KT)`, where `E` is the energy required to create `n` Schottky defects and `K` is the Boltzmann constant, If the mole fraction of Schottky defect in `NaCl` crystal at `2900 K` is `X`, then calculate `-ln(x)`,
Given: `DeltaH` of Schottky defect `= 2 eV` and
`K = 1.38 xx 10^(-23) J K^(-1)`
`1 eV = 1.608 xx 10^(-19) J`

Explanation

To solve the problem, we will follow the steps outlined in the video transcript, but we will break them down into clearer steps for better understanding.

### Step-by-Step Solution:

1. **Understanding the Formula**:
The number of Schottky defects \( n \) is given by the formula:
\[ n = N e^{-\frac{E}{2kT}} \]
where:
- \( N \) = total number of ions
- \( E \) = energy required to create Schottky defects (given as \( \Delta H = 2 \, \text{eV} \))
- \( k \) = Boltzmann constant \( = 1.38 \times 10^{-23} \, \text{J/K} \)
- \( T \) = temperature in Kelvin (given as \( 2900 \, \text{K} \))

2. **Convert Energy from eV to Joules**:
Since \( 1 \, \text{eV} = 1.608 \times 10^{-19} \, \text{J} \), we convert \( E \):
\[ E = 2 \, \text{eV} = 2 \times 1.608 \times 10^{-19} \, \text{J} = 3.216 \times 10^{-19} \, \text{J} \]

3. **Calculate \( \frac{E}{2kT} \)**:
Substitute \( E \), \( k \), and \( T \) into the equation:
\[ \frac{E}{2kT} = \frac{3.216 \times 10^{-19}}{2 \times (1.38 \times 10^{-23}) \times 2900} \]
Calculate \( 2kT \):
\[ 2kT = 2 \times 1.38 \times 10^{-23} \times 2900 = 8.004 \times 10^{-20} \, \text{J} \]
Now calculate \( \frac{E}{2kT} \):
\[ \frac{E}{2kT} = \frac{3.216 \times 10^{-19}}{8.004 \times 10^{-20}} \approx 4.015 \]

4. **Calculate Mole Fraction \( X \)**:
The mole fraction \( X \) can be expressed as:
\[ X = \frac{n}{N + n} = \frac{N e^{-\frac{E}{2kT}}}{N + N e^{-\frac{E}{2kT}}} = \frac{e^{-\frac{E}{2kT}}}{1 + e^{-\frac{E}{2kT}}} \]
Substitute \( e^{-\frac{E}{2kT}} \):
\[ e^{-\frac{E}{2kT}} = e^{-4.015} \approx 0.0183 \]
Now calculate \( X \):
\[ X = \frac{0.0183}{1 + 0.0183} = \frac{0.0183}{1.0183} \approx 0.0180 \]

5. **Calculate \( -\ln(X) \)**:
Now, we need to find \( -\ln(X) \):
\[ -\ln(0.0180) \approx 4.0 \]

### Final Answer:
Thus, the value of \( -\ln(X) \) is approximately \( 4 \).