Question

Suppose the charge of a proton and an electron differ slightly. One of them is –e, the other is (e +△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 △e is of the order of [Given mass of hydrogen m_h = 1.67 × 10^–27 kg]

• A. 10^–20 C
• B. 10^–23 C
• C. 10^–37 C
• D. 10^–47 C

Answer 1

The Correct Option is C

 To solve this problem, we need to determine the order of the difference in charges between a proton and an electron when the net electrostatic and gravitational force between two hydrogen atoms placed at a significant distance \( d \) apart is zero.

We understand that each hydrogen atom consists of one proton and one electron. The charge of an electron is \( -e \), and we assume the charge of a proton is \( e + \Delta e \).

Step-by-step Solution:

1. Electrostatic Force Between Two Hydrogen Atoms:
   • The electrostatic force between two protons is given by \(\frac{k(e+\Delta e)^2}{d^2}\).
   • The electrostatic force between two electrons is given by \(\frac{k(e)^2}{d^2}\).
   • The electrostatic force between a proton and an electron is attractive and is given by \(2\frac{k(e)(e+\Delta e)}{d^2}\).
2. Net Electrostatic Force:
3. Gravitational Force Between Two Hydrogen Atoms:
4. Condition for Forces to Cancel Each Other:
5. Substitute Known Values:
   • Gravitational constant \( G = 6.674 \times 10^{-11} \, \text{Nm}^2\text{kg}^{-2} \)
   • Coulomb's constant \( k = 8.9875 \times 10^9 \, \text{Nm}^2\text{C}^{-2} \)
   • Mass of hydrogen atom \( m_h = 1.67 \times 10^{-27} \, \text{kg} \)

Therefore, the correct answer is that \( \Delta e \) is of the order of

10^–37 C

.