Problem 4 - University Physics with Modern Physics

Question

(a) How much work would it take to push two protons very slowly from a separation of \(2.00 \times 10^{-10}\) m (a typical atomic distance) to \(3.00 \times 10^{-15}\) m (a typical nuclear distance)? (b) If the protons are both released from rest at the closer distance in part (a), how fast are they moving when they reach their original separation?

Step-by-Step Solution

Answer

(a) It takes about 4.8 MeV of work. (b) The speed is approximately 1.4 x 10^7 m/s.

Step 1: Calculate Initial Potential Energy

The electric potential energy between two protons at initial separation can be calculated using the formula for electrostatic potential energy: \[ U_i = \frac{k \cdot e^2}{r_i} \]where:- \( k = 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \) is Coulomb's constant,- \( e = 1.6 \times 10^{-19} \text{ C} \) is the charge of a proton,- \( r_i = 2.00 \times 10^{-10} \text{ m} \) is the initial separation.Substitute the values to find \( U_i \).

Step 2: Calculate Final Potential Energy

Calculate the potential energy at the final separation using the same formula but with the new distance:\[ U_f = \frac{k \cdot e^2}{r_f} \]where \( r_f = 3.00 \times 10^{-15} \text{ m} \).Substitute to find \( U_f \).

Step 3: Determine Work Done

The work done to push the protons is given by the change in potential energy:\[ W = U_f - U_i \]Substitute the values of \( U_i \) and \( U_f \) to calculate the work done.

Step 4: Calculate Potential Energy Difference

Use the previously found values of \( U_i \) and \( U_f \) to confirm the calculation:Ensure that \( \Delta U = U_f - U_i \) matches the work calculated.

Step 5: Find Final Kinetic Energy

If the protons are released, they convert potential energy to kinetic energy.The initial kinetic energy is zero, and so the change in potential energy becomes kinetic energy:\[ \Delta U = K_f = \frac{1}{2} m v^2 \]Where \( m \) is the mass of a proton. Solve for the final speed \( v \).

Step 6: Calculate Proton Speed

Each proton will have half of the total kinetic energy, so divide the energy by 2 and solve for velocity:\[ v = \sqrt{\frac{2 \cdot (\Delta U)}{m}} \]where \( m = 1.67 \times 10^{-27} \text{ kg} \) is the mass of a proton.Substitute values to find \( v \).

Key Concepts

Coulomb's Law

Coulomb's Law is a fundamental principle in physics that helps us understand the electrostatic interaction between two charged particles. It states that the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. This can be represented mathematically as:

• F = \( \frac{k \, |q_1 \, q_2|}{r^2} \)

where:

• \( F \) is the force between the charges,
• \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
• \( r \) is the separation between the charges,
• \( k = 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \) is Coulomb's constant.

Coulomb's Law is pivotal in calculating the electrostatic potential energy between charged particles. In our exercise, we applied Coulomb's Law to determine the potential energy difference when two protons are moved closer.
Understanding this law lays the groundwork for exploring how charged particles interact, as seen with protons that carry equal positive charges.

Proton Interaction

Protons are subatomic particles with a positive charge, commonly found in the nuclei of atoms. When discussing proton interactions, it is crucial to consider their electrostatic repulsion due to like charges.
In the exercise, we analyzed how the potential energy changes when two protons are moved from an atomic distance to a nuclear distance.
Given their identical positive charges, protons repel each other, and overcoming this force requires work input.
As the distance decreases:

• The potential energy increases because the repulsive force grows stronger.
• This increase in potential energy results from the electrical force exerted by each proton on the other.

These interactions are essential for understanding nuclear forces. Despite the repulsion, protons can still coexist in atomic nuclei due to the presence of the nuclear force, which is not covered in this specific problem.

Kinetic Energy Conversion

Kinetic energy conversion involves the transformation of potential energy into kinetic energy and vice versa.
In part (b) of the exercise, we dealt with the conversion of electrostatic potential energy into kinetic energy for two protons.
Initially, the protons are held close together, resulting in a high potential energy. Once they are released:

• The potential energy decreases as they move back to their original separation.
• The lost potential energy is converted into kinetic energy, giving the protons speed.

We computed how fast the protons travel by equating the change in potential energy to kinetic energy using the formula:

• \( \Delta U = K_f = \frac{1}{2} m v^2 \)

where:

• \( \Delta U \) is the change in potential energy,
• \( K_f \) is the final kinetic energy,
• \( m \) is the mass of a proton,
• \( v \) is the velocity of the protons.

By calculating this conversion, we highlight the intricate dance of forces and energy that governs particle motion at microscopic scales.