The distance between a proton and electron both having a charge `1.6xx10^(-19)` coulomb, of a hydrogen atom is `10^(-10)` metre . The value of intensity of electric field produced on electron due to proton will be

Options

A. `2.304xx10^(-10)N//C`
B. `14.4 V//m`
C. `16V//m`
D. `1.44 xx10^(11)N//C`

Answer

To find the intensity of the electric field produced on an electron due to a proton in a hydrogen atom, we can use the formula for the electric field (E) created by a point charge (Q) at a distance (r):

\[ E = \frac{k \cdot |Q|}{r^2} \]

Where:
- \( E \) is the electric field intensity,
- \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \),
- \( Q \) is the charge of the proton,
- \( r \) is the distance between the proton and the electron.

### Step-by-Step Solution:

1. **Identify the values**:
- Charge of the proton, \( Q = 1.6 \times 10^{-19} \, \text{C} \)
- Distance between the proton and electron, \( r = 10^{-10} \, \text{m} \)
- Coulomb's constant, \( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)

2. **Substitute the values into the formula**:
\[ E = \frac{(8.99 \times 10^9) \cdot (1.6 \times 10^{-19})}{(10^{-10})^2} \]

3. **Calculate \( (10^{-10})^2 \)**:
\[ (10^{-10})^2 = 10^{-20} \]

4. **Substitute this back into the equation**:
\[ E = \frac{(8.99 \times 10^9) \cdot (1.6 \times 10^{-19})}{10^{-20}} \]

5. **Simplify the expression**:
\[ E = (8.99 \times 10^9) \cdot (1.6 \times 10^{1}) \quad \text{(since } \frac{1}{10^{-20}} = 10^{20}\text{)} \]

6. **Calculate \( 8.99 \times 1.6 \)**:
\[ 8.99 \times 1.6 = 14.384 \]

7. **Combine the powers of ten**:
\[ E = 14.384 \times 10^{9 + 1} = 14.384 \times 10^{10} \]

8. **Express in scientific notation**:
\[ E = 1.4384 \times 10^{11} \, \text{N/C} \]

9. **Round to two decimal places**:
\[ E \approx 1.44 \times 10^{11} \, \text{N/C} \]

### Final Answer:
The intensity of the electric field produced on the electron due to the proton is approximately:
\[ E \approx 1.44 \times 10^{11} \, \text{N/C} \]