Consider the inelastic collision e−+e−→e−+e−+e−+e+ in which an electron-positron pair is produced in a head-on collision between two electrons moving in opposite directions at the same speed.
What is the speed of an electron with the threshold kinetic energy?
The speed of an electron with the threshold kinetic energy is the speed of light, c.
To determine the speed of an electron with the threshold kinetic energy in the given inelastic collision, we can start by analyzing the initial and final states of the particles involved.
In the initial state, we have two electrons moving in opposite directions at the same speed. Let's assume this speed is denoted by v.
Since the collision is head-on, the total momentum before the collision is zero because the momentum of one electron cancels out the momentum of the other.
In the final state, we have four particles: three electrons (e- e- e-) and one positron (e+). We want to find the electron speed with the threshold kinetic energy, which indicates the minimum energy required for such a collision to take place.
The threshold kinetic energy is defined as the difference between the initial and final total energies in the center of mass reference frame.
Before the collision:
Total energy = 2 * (gamma * m_0 * c^2), where gamma is the relativistic factor given by gamma = 1 / sqrt(1 - (v^2 / c^2)) and m_0 is the rest mass of an electron. c is the speed of light.
After the collision:
Total energy = 3 * (gamma * m_0 * c^2) + (gamma_pos * m_0 * c^2), where gamma_pos is the relativistic factor for a positron.
Using the principle of conservation of momentum, we can set the initial total momentum to zero and solve for v:
2 * gamma * m_0 * v = (3 * gamma * m_0 * v_electron) + (gamma_pos * m_0 * v_positron)
Since the collision is head-on, the velocities of the electrons are opposite in direction.
Simplifying the equation:
2 * gamma * m_0 * v = (3 * gamma * m_0 * v_electron) - (gamma_pos * m_0 * v_electron)
Applying relativistic factors:
2 * gamma * v = (3 * gamma - gamma_pos) * v_electron
Further simplifying:
v_electron = (2 * gamma * v) / (3 * gamma - gamma_pos)
Having determined v_electron, the speed of an electron with the threshold kinetic energy, we can substitute this value into the relativistic factor equation:
gamma = 1 / sqrt(1 - (v_electron^2 / c^2))
After evaluating this equation, we will obtain the value of gamma, which allows us to calculate the speed of the electron using the relation:
v_electron = c * sqrt(1 - (1 / (gamma^2)))
Once we have this speed, we can calculate the threshold kinetic energy using the formula:
Threshold Kinetic Energy = (gamma - 1) * m_0 * c^2
Remember that c is the speed of light and m_0 is the rest mass of an electron.
To determine the speed of an electron with the threshold kinetic energy in the given collision scenario, we need to consider the conservation of momentum and energy.
1. Conservation of momentum:
Since the two electrons are moving in opposite directions at the same speed, they have equal magnitudes of momentum but opposite directions. After the collision, four particles (two electrons and two positrons) are produced, so the total momentum of the system remains zero.
2. Conservation of energy:
The threshold kinetic energy corresponds to the minimum energy required for the collision to occur. This energy is used to create the electron-positron pair and account for any energy lost during the collision.
Assuming non-relativistic speeds, we can use the formula for kinetic energy:
K = (1/2)mv^2
Let's denote the mass of an electron as "m" and the speed of an electron after the collision as "v." Since the total momentum is zero, we can set up an equation involving the kinetic energies before and after the collision.
Let K1 be the initial kinetic energy and K2 be the final kinetic energy:
K1 + K1 = K2 + K2 + K2 + K2
Since the electrons have the same speed, their initial kinetic energies are equal. Similarly, the four particles produced after the collision are identical, so their final kinetic energies are the same.
2K1 = 4K2
Now, let's substitute the expressions for kinetic energy:
(1/2)mv^2 + (1/2)mv^2 = 4[(1/2)m(v_threshold)^2]
Simplifying the equation:
2mv^2 = 4m(v_threshold)^2
Simplifying further:
v^2 = 2(v_threshold)^2
Taking the square root of both sides:
v = sqrt(2) * v_threshold
Therefore, the speed of an electron with the threshold kinetic energy is equal to the square root of 2 times the threshold velocity (v_threshold).