EN of the element (A) is `E_(1)` and EA is `E_(2)` hence IP will be :

Options

A. `2E_(1) - E_(2)`
B. `E_(1) - E_(2)`
C. `E_(1) - 2E_(2)`
D. `(E_(1) + E_(2)) //2`

Answer

To find the ionization potential (IP) of element A given its electronegativity (EN) as \( E_1 \) and its electron affinity (EA) as \( E_2 \), we can use the relationship derived from the video transcript.

### Step-by-Step Solution:

1. **Understanding the Definitions**:
- Electronegativity (EN) of an element is a measure of its ability to attract electrons in a chemical bond.
- Electron Affinity (EA) is the amount of energy released when an electron is added to a neutral atom to form a negative ion.
- Ionization Potential (IP) or Ionization Energy (IE) is the energy required to remove an electron from a neutral atom.

2. **Using the Relationship**:
- According to the video transcript, the relationship between these quantities is given by:
\[ IP = 2E_1 - E_2 \]
- Here, \( E_1 \) is the electronegativity and \( E_2 \) is the electron affinity.

3. **Substituting the Values**:
- Substitute the values of \( E_1 \) and \( E_2 \) into the equation to find the ionization potential (IP):
\[ IP = 2 \times E_1 - E_2 \]

4. **Final Expression**:
- This gives us the final expression for the ionization potential in terms of electronegativity and electron affinity:
\[ IP = 2E_1 - E_2 \]

### Conclusion:
The ionization potential (IP) of element A can be calculated using the formula:
\[ IP = 2E_1 - E_2 \]