The distance between `H^(+) and Cl^(-)` ions in HCl molecules is `1.38Å.` The potential due to this dipole at a sistance of `10Å` on the axis of dipole is

Options

A. `2.1` V
B. `1.8` V
C. `0.2` V
D. `1.2` V

Explanation

To solve the problem of finding the potential due to the dipole formed by the H⁺ and Cl⁻ ions in an HCl molecule at a distance of 10 Å on the axis of the dipole, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Distance between H⁺ and Cl⁻ ions (2a) = 1.38 Å = 1.38 x 10⁻¹⁰ m
- Distance from the dipole at which potential is to be calculated (r) = 10 Å = 10 x 10⁻¹⁰ m = 1 x 10⁻⁹ m
- Charge of an electron (Q) = 1.6 x 10⁻¹⁹ C

2. **Calculate the Dipole Moment (p):**
The dipole moment \( p \) is given by the formula:
\[ p = Q \cdot d \]
where \( d \) is the distance between the charges (which is \( 2a \)).
\[ p = (1.6 \times 10^{-19} \, \text{C}) \cdot (1.38 \times 10^{-10} \, \text{m}) = 2.208 \times 10^{-29} \, \text{C m} \]

3. **Use the Formula for Potential (V) at an Axial Point:**
The potential \( V \) at a distance \( r \) from a dipole on its axial line 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. **Substitute the Values into the Formula:**
\[ V = \frac{(9 \times 10^9) \cdot (2.208 \times 10^{-29})}{(1 \times 10^{-9})^2} \]

5. **Calculate the Denominator:**
\[ (1 \times 10^{-9})^2 = 1 \times 10^{-18} \]

6. **Calculate the Potential:**
\[ V = \frac{(9 \times 10^9) \cdot (2.208 \times 10^{-29})}{1 \times 10^{-18}} = 9 \times 2.208 \times 10^{-1} = 19.872 \times 10^{-1} \, \text{V} = 0.19872 \, \text{V} \]

7. **Round the Result:**
The potential can be approximated as:
\[ V \approx 0.2 \, \text{V} \]

### Final Answer:
The potential due to the dipole at a distance of 10 Å on the axis of the dipole is approximately **0.2 V**.