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

Options

A. `F_e = F_n`
B. `F_e gtgt F_n`
C. `F_e ltlt F_n`
D. `F_e and F_n` differ only slightly

Answer

To solve the problem of comparing the nuclear force (F_n) and the electromagnetic force (F_e) between two protons separated by a distance of 10 nm, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Forces Involved**:
- The **nuclear force** (F_n) is the force that acts between protons and neutrons in an atomic nucleus. It is a short-range force, typically effective over distances on the order of 1 femtometer (1 fm = 10^-15 m).
- The **electromagnetic force** (F_e) is the force between charged particles. Since protons are positively charged, they repel each other due to the electromagnetic force.

2. **Convert the Distance**:
- The separation between the two protons is given as 10 nanometers (nm).
- Convert this to meters:
\[ 10 \text{ nm} = 10 \times 10^{-9} \text{ m} = 10^{-8} \text{ m} \]

3. **Compare the Ranges of the Forces**:
- The nuclear force is significant only at distances on the order of 1 fm (10^-15 m).
- The distance of 10 nm (10^-8 m) is much larger than 1 fm (10^-15 m).

4. **Conclusion about the Nuclear Force**:
- Since 10 nm is much greater than the range of the nuclear force, we can conclude that the nuclear force (F_n) between the two protons at this distance is negligible.

5. **Calculate the Electromagnetic Force**:
- The electromagnetic force can be calculated using Coulomb's law:
\[ F_e = k \frac{q_1 q_2}{r^2} \]
where:
- \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),
- \( q_1 \) and \( q_2 \) are the charges of the protons (\( 1.6 \times 10^{-19} \, \text{C} \)),
- \( r \) is the separation between the charges (10 nm = \( 10^{-8} \, \text{m} \)).

6. **Substituting Values**:
- Substitute the values into Coulomb's law:
\[ F_e = 8.99 \times 10^9 \frac{(1.6 \times 10^{-19})^2}{(10^{-8})^2} \]

7. **Calculate the Result**:
- Calculate \( F_e \):
\[ F_e = 8.99 \times 10^9 \frac{(2.56 \times 10^{-38})}{(10^{-16})} = 8.99 \times 10^9 \times 2.56 \times 10^{-22} \approx 2.30 \times 10^{-12} \text{ N} \]

8. **Final Comparison**:
- Since \( F_n \) is negligible at this distance, we conclude that \( F_e \) is much larger than \( F_n \).

### Conclusion:
- The electromagnetic force (F_e) between the two protons at a separation of 10 nm is significantly larger than the nuclear force (F_n), which is negligible at this distance.