Suppose the charge of a proton and an electron differ slightely. One of them is `-e`, the other is `(e+Deltae)`. If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance `d` (much greater than atomic size) apart is zero. Then `Deltae` is of the order of [Given mass of hydrogen `m_(h)=1.67xx10^(-27)kg`]

Options

A. `10^(-20)C`
B. `10^(-23)C`
C. `10^(-37)C`
D. `10^(-47)C`

Answer

To solve the problem, we need to find the value of \(\Delta e\) given that the net electrostatic force and gravitational force between two hydrogen atoms placed at a distance \(d\) apart is zero.

### Step-by-Step Solution:

1. **Understanding the Charges**:
- The charge of the electron is \(-e\).
- The charge of the proton is given as \(e + \Delta e\).
- Therefore, the total charge of a hydrogen atom (which consists of one proton and one electron) can be expressed as:
\[ Q = (e + \Delta e) + (-e) = \Delta e \]

2. **Forces Acting Between Two Hydrogen Atoms**:
- The gravitational force (\(F_G\)) between two hydrogen atoms can be expressed using Newton's law of gravitation:
\[ F_G = \frac{G m_h^2}{d^2} \]
- The electrostatic force (\(F_E\)) between two hydrogen atoms can be expressed using Coulomb's law:
\[ F_E = \frac{k Q^2}{d^2} \]
- Here, \(Q = \Delta e\), so:
\[ F_E = \frac{k (\Delta e)^2}{d^2} \]

3. **Setting the Forces Equal**:
- According to the problem, the net force is zero, which means:
\[ F_G = F_E \]
- Substituting the expressions for \(F_G\) and \(F_E\):
\[ \frac{G m_h^2}{d^2} = \frac{k (\Delta e)^2}{d^2} \]
- Since \(d^2\) is common on both sides, we can cancel it out:
\[ G m_h^2 = k (\Delta e)^2 \]

4. **Solving for \(\Delta e\)**:
- Rearranging the equation gives:
\[ (\Delta e)^2 = \frac{G m_h^2}{k} \]
- Taking the square root:
\[ \Delta e = \sqrt{\frac{G m_h^2}{k}} \]

5. **Substituting Known Values**:
- Given:
- \(G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2\)
- \(m_h = 1.67 \times 10^{-27} \, \text{kg}\)
- \(k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2\)
- Substituting these values into the equation:
\[ \Delta e = \sqrt{\frac{6.67 \times 10^{-11} \times (1.67 \times 10^{-27})^2}{9 \times 10^9}} \]

6. **Calculating \(\Delta e\)**:
- First, calculate \(m_h^2\):
\[ m_h^2 = (1.67 \times 10^{-27})^2 = 2.7889 \times 10^{-54} \, \text{kg}^2 \]
- Now substitute this back into the equation:
\[ \Delta e = \sqrt{\frac{6.67 \times 10^{-11} \times 2.7889 \times 10^{-54}}{9 \times 10^9}} \]
- Calculate the numerator:
\[ 6.67 \times 10^{-11} \times 2.7889 \times 10^{-54} = 1.858 \times 10^{-64} \]
- Now divide by \(9 \times 10^9\):
\[ \frac{1.858 \times 10^{-64}}{9 \times 10^9} = 2.064 \times 10^{-74} \]
- Finally, take the square root:
\[ \Delta e = \sqrt{2.064 \times 10^{-74}} \approx 2.06 \times 10^{-37} \, \text{C} \]

### Conclusion:
Thus, \(\Delta e\) is of the order of \(2.06 \times 10^{-37} \, \text{C}\).