Question

Two protons are kept at a separation of 10 nm. Let F_n and F_e the nuclear force and the electromagnetic force between them

• A. F_e = F_n
• B. F_e >> F_n
• C. F_e << F_n
• D. F_e and F_n differ only slightly

Answer 1

The Correct Option is B

At a separation of 10 nm, the electromagnetic force \( F_e \) between two protons is much larger than the nuclear force \( F_n \). The nuclear force is short-range and is effective only at distances on the order of femtometers (fm), typically around \( 10^{-15} \) meters, while the electromagnetic force acts over much larger distances. At 10 nm (\( 10^{-8} \) meters), the electromagnetic force between the protons is much stronger than the nuclear force, so \( F_e \gg F_n \).

The correct answer is (B) : F_e >> F_n.

Answer 2

The nuclear force \( F_n \) and the electromagnetic force \( F_e \) between two protons at a separation of 10 nm can be calculated using the following formulas:
1. Electromagnetic force (Coulomb force): \[ F_e = \frac{1}{4 \pi \epsilon_0} \frac{e^2}{r^2} \] where: - \( e \) is the charge of a proton (\( e = 1.6 \times 10^{-19} \, \text{C} \)), - \( r = 10 \, \text{nm} = 10 \times 10^{-9} \, \text{m} \), - \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N} \, \text{m}^2 \) (permittivity of free space). Substituting the values: \[ F_e = \frac{(9 \times 10^9) \times (1.6 \times 10^{-19})^2}{(10 \times 10^{-9})^2} \approx 2.3 \times 10^{-10} \, \text{N} \]
2. Nuclear force: The nuclear force is a short-range force, typically acting at distances on the order of \( 10^{-15} \, \text{m} \). At a separation of \( 10 \, \text{nm} \), which is much larger than the typical range of the strong force, the nuclear force between the protons is negligible compared to the electromagnetic force. Thus, the nuclear force \( F_n \) is much smaller than the electromagnetic force \( F_e \).

Therefore, the correct answer is \({B} \) — \( F_e \gg F_n \).