Problem 54 - University Physics with Modern Physics

Question

A proton and an alpha particle are released from rest when they are 0.225 nm apart. The alpha particle (a helium nucleus) has essentially four times the mass and two times the charge of a proton. Find the maximum speed and maximum acceleration of each of these particles. When do these maxima occur: just following the release of the particles or after a very long time?

Step-by-Step Solution

Answer

Maximum speeds occur after a long time; maximum accelerations occur immediately upon release.

Step 1: Understanding the System

We have a proton and an alpha particle initially separated by a distance of 0.225 nm. The alpha particle has four times the mass of the proton and twice the charge. The proton has charge +e and the alpha particle has charge +2e. They are released from rest at this distance.

Step 2: Conservation of Energy

The total mechanical energy of this system remains constant. Initially, the kinetic energy is zero, and only electric potential energy exists:\[ U_i = K \frac{q_1 q_2}{r} = K_e \frac{e imes 2e}{0.225} \times 10^{-9} \]where \( K_e \) is Coulomb's constant, \( q_1 = e \), and \( q_2 = 2e \). The initial potential energy equals the final kinetic energy of both particles:\[ U_i = rac{1}{2} m_p v_p^2 + rac{1}{2} m_{\alpha} v_{\alpha}^2 \]Here, \( m_p \) is the proton's mass and \( m_{\alpha} = 4m_p \) is the alpha particle's mass.

Step 3: Using Conservation of Momentum

The initial momentum is zero since both particles start from rest. Thus, the total momentum at any later time must be zero:\[ m_p v_p = -m_{\alpha} v_{\alpha} \]Using \( m_{\alpha} = 4m_p \), we have \( v_p = -4v_{\alpha} \). This relationship will allow us to express one velocity in terms of the other.

Step 4: Solving for Velocities

Substituting \( v_p = -4v_{\alpha} \) into the energy equation:\[ U_i = rac{1}{2} m_p (4v_{\alpha})^2 + rac{1}{2} (4m_p) v_{\alpha}^2 \]Simplify to find:\[ U_i = 8m_p v_{\alpha}^2 + 2m_p v_{\alpha}^2 = 10m_p v_{\alpha}^2 \]Then solve for \( v_{\alpha} \):\[ v_{\alpha} = \sqrt{\frac{U_i}{10m_p}} \]Find \( v_p \):\[ v_p = 4v_{\alpha} = 4 \sqrt{\frac{U_i}{10m_p}} \]

Step 5: Calculating the Potential Energy

Calculate the initial potential energy using:\[ U_i = K_e \frac{2e^2}{0.225 \times 10^{-9}} = (8.99 \times 10^9 \frac{N \cdot m^2}{C^2}) \frac{2(1.6 \times 10^{-19})^2}{0.225 \times 10^{-9}} \]Evaluate this expression to find \( U_i \).

Step 6: Solving Maximum Velocities

Insert value of \( U_i \) back to determine the velocities using the formulas derived. The maximum speed occurs when the particles are infinitely far apart, and kinetic energy is max.

Step 7: Calculating Maximum Accelerations

The maximum acceleration occurs just after release due to maximum force at minimum separation:\[ F = K_e \frac{q_1 q_2}{r^2} \]Calculate the force, then determine acceleration using \( a = F/m \) for each particle.

Key Concepts

Conservation of Energy

In the context of electric forces acting between charged particles, such as a proton and an alpha particle, the principle of conservation of energy plays a crucial role. Initially, when these particles are at rest, the total mechanical energy is all potential, given by the electric potential energy. This energy is calculated using the formula:

• Electric potential energy: \( U_i = \frac{K_e q_1 q_2}{r} \)

Here, \( K_e \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges of the particles, and \( r \) is the initial distance between them.

As the particles move due to the electric forces, their potential energy converts into kinetic energy. According to the conservation of energy, the initial potential energy equals the total kinetic energy when the particles have moved infinitely far apart. The equation for conservation of energy at any given point is:

• \( U_i = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 \)

This way, by knowing the initial potential energy, we can determine the final velocities when the kinetic energy is at its maximum.

Conservation of Momentum

The conservation of momentum is integral in analyzing the motion of charged particles. Initially, both the proton and the alpha particle are at rest, implying that the total momentum of the system is zero.

• Initial momentum: \( p_i = 0 \)

During the interaction, even though the forces exchanged are internal to the system of particles, the total momentum remains constant due to the absence of external forces.

• Conservation equation: \( m_1 v_1 + m_2 v_2 = 0 \)

Given that the alpha particle's mass is four times that of the proton, any velocity imparted to the alpha particle results in the proton moving four times faster in the opposite direction:

• \( v_1 = -4v_2 \)

This relationship helps in expressing the velocities in terms of each other when solving energy conservation equations.

Kinetic Energy

Kinetic energy is the energy associated with the motion of objects, defined by the formula:

• \( KE = \frac{1}{2}mv^2 \)

In this scenario, kinetic energy becomes significant as the proton and alpha particle are released and begin to move apart due to their mutual electric forces. Initially, both particles have zero kinetic energy because they start from rest.

As the system evolves and the particles accelerate away from each other, their potential energy gradually converts into kinetic energy. At maximum separation, their velocities are maximal, and thus their kinetic energies reach their peak values.

• Max kinetic energy is achieved when potential energy is totally converted.

Understanding how energy is transformed into kinetic energy is essential for predicting the particles' behavior and velocities over time.

Electric Potential Energy

The concept of electric potential energy is central to understanding the forces between charged particles. It is the energy stored due to the relative positions of charged particles:

• Formula: \( U = \frac{K_e q_1 q_2}{r} \)

Where \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between them. For a proton and an alpha particle, the initial separation provides a significant electric potential energy due to their opposite charges.

This initial potential energy is wholly converted into kinetic energy as particles move infinitely apart under the action of their electric forces. At the starting point, the particles' potential energy determines the system's total capacity to perform work by facilitating the particles' motion.

• The closer the charges, the higher the potential energy.
• As distance increases, potential energy decreases but kinetic energy increases.

This transformation is a critical aspect of many electrostatic applications and experiments.