Question

Suppose the charge of a proton and an electron differ slightly. One of them is $- e$, the other is ($e + \Delta e $). 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 $\Delta e$ is of the order of [Given mass of hydrogen $m_h = 1.67 \times 10^{-27} \, kg$]

• A. $10^{-23} C $
• B. $10^{-37} C $
• C. $10^{-47} C $
• D. $10^{-20} C $

Answer 1

The Correct Option is B

To solve this problem, we need to consider both electrostatic and gravitational forces acting between two hydrogen atoms placed at a distance \( d \). The condition given is that the net force between the hydrogen atoms is zero. Hence, the electrostatic force must balance out the gravitational force.

1. \text{Electrostatic Force, } F_e = \frac{k (q_1)(q_2)}{d^2}, where \( k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \) is the Coulomb's constant, \( q_1 = -e \), and \( q_2 = e + \Delta e \).
2. Compute \( F_e \): F_e = \frac{k (-e)(e + \Delta e)}{d^2} = \frac{-k e^2 - k e \Delta e}{d^2}.
3. \text{Gravitational Force, } F_g = \frac{G m_h^2}{d^2}, where \( G = 6.674 \times 10^{-11} \, \text{m}^3/\text{kg s}^2 \) is the gravitational constant and \( m_h = 1.67 \times 10^{-27} \, \text{kg} \).
4. For the net force to be zero, we equate the magnitudes of \( F_e \) and \( F_g \): \left| \frac{-k e^2 - k e \Delta e}{d^2} \right| = \frac{G m_h^2}{d^2}.
5. Simplifying, the \( d^2 \) cancels out: k e \Delta e = G m_h^2.
6. Solve for \( \Delta e \): \Delta e = \frac{G m_h^2}{k e}.
7. Substitute the given values: \Delta e = \frac{6.674 \times 10^{-11} \times (1.67 \times 10^{-27})^2}{8.99 \times 10^9 \times 1.6 \times 10^{-19}}.
8. Calculate each term: \Delta e = \frac{6.674 \times 10^{-11} \times 2.7889 \times 10^{-54}}{1.4384 \times 10^{-9}}.
9. Simplify the calculation: \Delta e \approx 1.28 \times 10^{-37} \, \text{C}.

Thus, the order of \(\Delta e\) is \(10^{-37} \, \text{C}\). The correct answer is:

• 10^{-37} \, \text{C}