`N_(0)//2` atoms of X(g) are converted into `X^(+)` (g) by energy `E_(1) . N_(0)//2` atoms of X(g) are converted into `X^(-)` (g) by the energy `E_(2)` . Hence ionisation potential and electron affinity of X(g) are :

Options

A. `I.E. = (2E_(1))/(N_(0)), Delta_(eq)H =- (E_(2))/(2N_(0))`
B. `I.E. = (2E_(2))/(N_(0)). Delta_(eq)H = (2E_(1))/(N_(0))`
C. `I.E. = (E_(1))/(2N_(0)), Delta_(eq)H =- (E_(2))/(2N_(0))`
D. `I.E. = (N_(0))/(2E_(0)). Delta_(eq)H =- (2N_(0))/(E_(2))`

Explanation

To solve the problem, we need to determine the ionization potential and electron affinity of the gaseous atom X based on the given energies for the conversion of X atoms into ions.

### Step-by-Step Solution:

1. **Understanding Ionization Energy**:
- Ionization energy (IE) is defined as the energy required to remove an electron from a gaseous atom.
- In the problem, we are given that `N₀/2` atoms of X are converted into `X⁺` by the energy `E₁`.

2. **Calculating Ionization Energy**:
- Since `N₀/2` atoms require energy `E₁`, the energy required for one atom of X to become `X⁺` can be calculated as follows:
\[ \text{Ionization Energy (IE)} = \frac{E₁}{N₀/2} = \frac{2E₁}{N₀} \]

3. **Understanding Electron Affinity**:
- Electron affinity (EA) is defined as the energy released when an electron is added to a gaseous atom.
- In the problem, `N₀/2` atoms of X are converted into `X⁻` by the energy `E₂`.

4. **Calculating Electron Affinity**:
- The energy released when `N₀/2` atoms become `X⁻` is `E₂`. Therefore, the energy for one atom of X to become `X⁻` can be calculated as:
\[ \text{Electron Affinity (EA)} = -\frac{E₂}{N₀/2} = -\frac{2E₂}{N₀} \]

5. **Final Results**:
- The ionization potential of X(g) is:
\[ \text{Ionization Potential} = \frac{2E₁}{N₀} \]
- The electron affinity of X(g) is:
\[ \text{Electron Affinity} = -\frac{2E₂}{N₀} \]

### Summary:
- Ionization potential of X(g): \(\frac{2E₁}{N₀}\)
- Electron affinity of X(g): \(-\frac{2E₂}{N₀}\)