An alpha particle and an antiproton are released from rest a great distance apart. They are oppositely charged so each accelerates toward the other. What are the speeds of the two particles when they are 2.5 nm apart?


alpha particle has a charge of +2e and is four times as massive as a proton

antiproton has a charge of -e but has the same mass as a proton

Because momentum is conserved, and total momentum is zero, the alpha particle will have 1/4 the velocity of the antiproton.

Total energy remains zero. That means the two particles acquire kinetic energy equal to the potential energy LOSS, which is

k*2*e^2/R

where e is the electron charge, k is the Coulomb constant and R is the final separation. The number "2" is there because the alpha has charge +2e, and the antiproton has charge -e. It is the product than matters.

Set that equal to the final kinetic energy and solve for the velocity of either particle. The second particle will have to be expressed in terms of the first. For example,
V(alpha) = (1/4)*V(antiproton)

To determine the speeds of the alpha particle and the antiproton when they are 2.5 nm apart, we can use the principles of conservation of energy and conservation of momentum.

First, let's consider the conservation of energy. We can assume that the potential energy between the two particles is zero when they are infinitely far apart. As they get closer, they will gain kinetic energy at the expense of potential energy.

At any distance r between the particles, the potential energy U is given by the equation:
U = k * (q1 * q2) / r

Where k is the electrostatic constant (approximated to be 8.99 x 10^9 N*m²/C²), q1 and q2 are the charges of the two particles, and r is the distance between them.

Since the alpha particle has a charge of +2e and the antiproton has a charge of -e, the potential energy can be written as:
U = k * (2e * (-e)) / r = -2k * e² / r

At any distance r, the total mechanical energy E of the system is the sum of the kinetic energy and the potential energy:
E = K + U

Since the particles start from rest, their initial kinetic energy is zero. Thus, E = U.

Let's assume the final distance between the particles is d = 2.5 nm = 2.5 x 10^(-9) m.

At this final distance, the mechanical energy is:
E = -2k * e² / d

Now, let's consider the conservation of momentum. As the particles accelerate toward each other, their momentum changes. However, the total momentum of the system remains constant.

The momentum of an object is given by the equation:
p = m * v

Where p is the momentum, m is the mass of the object, and v is the velocity.

Given that the alpha particle is four times as massive as a proton, its mass will be 4 * m_p, where m_p is the mass of a proton.

At any distance r, the total momentum of the system is the sum of the momenta of the alpha particle and the antiproton:
p = m_alpha * v_alpha + m_antiproton * v_antiproton

Since the particles start from rest, their initial momentum is zero. Thus, the total momentum at the final distance is also zero:
m_alpha * v_alpha + m_antiproton * v_antiproton = 0

Now, we have two equations: one for the energy and one for the momentum. We can solve the system of equations to find the velocities of the particles at a distance of 2.5 nm.

Substituting the expressions for the potential energy and the momentum into the energy equation, we have:
-2k * e² / d = (1/2) * m_alpha * v_alpha² + (1/2) * m_antiproton * v_antiproton²

Let's plug in the given values:
-2 * (8.99 x 10^9 N*m²/C²) * (1.6 x 10^(-19) C)² / (2.5 x 10^(-9) m) = (1/2) * (4 * m_p) * v_alpha² + (1/2) * m_p * v_antiproton²

To solve for the velocities, we need to know the values of the charges (e) and the masses (m_p). However, the given question states that the antiproton and the proton have the same mass, so we can substitute m_alpha = 4m_p and m_antiproton = m_p into the equation:

-2 * (8.99 x 10^9 N*m²/C²) * (1.6 x 10^(-19) C)² / (2.5 x 10^(-9) m) = (1/2) * (4 * m_p) * v_alpha² + (1/2) * m_p * v_antiproton²

Solving this equation will give us the values of v_alpha (velocity of the alpha particle) and v_antiproton (velocity of the antiproton) at a distance of 2.5 nm apart.