If first ionisation potential of a hypothetical atom is `16 V`, then the first excitation potential will be :

Options

A. `10.2 V`
B. `12 V`
C. `14 V`
D. `16 V`

Explanation

To solve the problem of finding the first excitation potential of a hypothetical atom given its first ionization potential, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Definitions**:
- The first ionization potential (IP) is the energy required to remove the most loosely bound electron from an atom in its gaseous state.
- The first excitation potential (EP) is the energy required to excite an electron from the ground state to the first excited state.

2. **Use the Formula for Ionization Energy**:
- The ionization energy (IE) for a hydrogen-like atom is given by the formula:
\[ IE = -E_1 = \frac{13.6Z^2}{n^2} \]
- For the ground state (n=1), this becomes:
\[ IE = 13.6Z^2 \]

3. **Given Data**:
- The first ionization potential is given as 16 V. Therefore, we can set up the equation:
\[ 13.6Z^2 = 16 \]

4. **Calculate the Atomic Number (Z)**:
- Rearranging the equation to find Z:
\[ Z^2 = \frac{16}{13.6} \]
- Calculating Z:
\[ Z^2 \approx 1.1765 \quad \Rightarrow \quad Z \approx \sqrt{1.1765} \approx 1.084 \]

5. **Calculate the First Excitation Potential**:
- The excitation potential from n=1 to n=2 can be calculated using the formula:
\[ EP = E_2 - E_1 = \frac{13.6Z^2}{1^2} - \frac{13.6Z^2}{2^2} \]
- Substitute Z into the equation:
\[ EP = 13.6Z^2 \left(1 - \frac{1}{4}\right) = 13.6Z^2 \cdot \frac{3}{4} \]
- Substitute the value of \( Z^2 \):
\[ EP = 13.6 \cdot \frac{3}{4} \cdot \frac{16}{13.6} = 16 \cdot \frac{3}{4} = 12 \]

6. **Final Result**:
- The first excitation potential is:
\[ EP = 12 V \]

### Summary:
The first excitation potential of the hypothetical atom is **12 V**.