The distance between `H^(+)` and `CI^(-)` ions in `HCI` molecule is `1.28 Å`. What will be the potential due to this dipole at a distance of `12 Å` on the axis of dipole ?

Options

A. `0.13 V`
B. `1.3 V`
C. `13 V`
D. `130 V`

Explanation

To solve the problem of finding the electric potential due to a dipole at a certain distance, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Distance between the charges (dipole length), \( d = 1.28 \, \text{Å} = 1.28 \times 10^{-10} \, \text{m} \)
- Distance from the dipole where the potential is to be calculated, \( R = 12 \, \text{Å} = 12 \times 10^{-10} \, \text{m} \)
- Charge of the ions, \( q = 1.6 \times 10^{-19} \, \text{C} \)

2. **Calculate the Dipole Moment (p):**
The dipole moment \( p \) is given by the formula:
\[ p = q \cdot d \]
Substituting the values:
\[ p = (1.6 \times 10^{-19} \, \text{C}) \cdot (1.28 \times 10^{-10} \, \text{m}) = 2.048 \times 10^{-29} \, \text{C m} \]

3. **Use the Formula for Electric Potential (V) due to a Dipole:**
The potential \( V \) at a distance \( R \) from a dipole is given by:
\[ V = \frac{k \cdot p}{R^2} \]
where \( k \) is Coulomb's constant, \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

4. **Substituting the Values into the Formula:**
\[ V = \frac{(9 \times 10^9) \cdot (2.048 \times 10^{-29})}{(12 \times 10^{-10})^2} \]

5. **Calculate \( R^2 \):**
\[ R^2 = (12 \times 10^{-10})^2 = 144 \times 10^{-20} = 1.44 \times 10^{-18} \]

6. **Substituting \( R^2 \) into the Potential Formula:**
\[ V = \frac{(9 \times 10^9) \cdot (2.048 \times 10^{-29})}{1.44 \times 10^{-18}} \]

7. **Calculating the Potential:**
\[ V = \frac{(9 \cdot 2.048)}{1.44} \times 10^{9 - 29 + 18} = \frac{18.432}{1.44} \times 10^{-2} \]
\[ V \approx 12.8 \times 10^{-2} = 0.128 \, \text{V} \]

### Final Answer:
The potential due to the dipole at a distance of 12 Å on the axis of the dipole is approximately \( 0.128 \, \text{V} \).