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 find the first excitation potential of a hypothetical atom given that its first ionization potential is 16 V, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Ionization Potential**: The first ionization potential (IP) is the energy required to remove the most loosely bound electron from an atom in its ground state. In this case, it is given as 16 V.

2. **Relate Ionization Energy to Total Energy**: The total energy of an electron in the ground state (n=1) of a hydrogen-like atom can be expressed as:
\[ E_1 = -\frac{13.6 Z^2}{n^2} \]
For the ground state (n=1), this simplifies to:
\[ E_1 = -13.6 Z^2 \]
Since the ionization potential is the energy required to remove the electron from this state, we have:
\[ -E_1 = 16 \text{ V} \]
Thus, we can write:
\[ 13.6 Z^2 = 16 \]

3. **Calculate Z**: From the equation \(13.6 Z^2 = 16\), we can solve for \(Z^2\):
\[ Z^2 = \frac{16}{13.6} \approx 1.1765 \]
Taking the square root gives:
\[ Z \approx \sqrt{1.1765} \approx 1.084 \]

4. **Find the First Excitation Potential**: The first excitation potential refers to the energy required to excite the electron from n=1 to n=2. The energy difference (ΔE) between these two states can be calculated using the formula:
\[ \Delta E = E_2 - E_1 \]
Where:
\[ E_2 = -\frac{13.6 Z^2}{2^2} = -\frac{13.6 Z^2}{4} \]
Therefore:
\[ \Delta E = -\frac{13.6 Z^2}{4} - (-13.6 Z^2) = 13.6 Z^2 \left(1 - \frac{1}{4}\right) = 13.6 Z^2 \cdot \frac{3}{4} \]

5. **Substitute Z**: Now substitute \(Z^2\) into the equation:
\[ \Delta E = 13.6 \cdot \frac{3}{4} \cdot \frac{16}{13.6} = 16 \cdot \frac{3}{4} \]

6. **Calculate the Final Value**:
\[ \Delta E = 12 \text{ V} \]

### Conclusion:
Thus, the first excitation potential of the hypothetical atom is **12 V**.