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An electron and a positron have masses of \(9.11 \times 10^{-31} \mathrm{~kg} .\) They collide and both vanish, with only electromagnetic radiation appearing after the collision. If each particle is moving at a speed of \(0.20 c\) relative to the laboratory before the collision, determine the energy of the electromagnetic radiation.

Short Answer

Expert verified
The energy of the electromagnetic radiation is \(1.642 \times 10^{-13} \text{ J}\).

Step by step solution

01

Identify the Problem

We must find the energy of the electromagnetic radiation produced when an electron and a positron collide and annihilate.
02

Understand Mass-Energy Equivalence

Using Einstein's mass-energy equivalence formula, we know that the energy associated with the rest mass of a particle is given by \(E = mc^2\).
03

Calculate the Rest Energy of Each Particle

The rest energy of each particle (electron or positron) is \(E_0 = mc^2\), where \(m = 9.11 \times 10^{-31} \text{ kg}\) and \(c = 3 \times 10^8 \text{ m/s}\). Calculate: \[E_0 = (9.11 \times 10^{-31} \text{ kg})(3 \times 10^8 \text{ m/s})^2\]\[E_0 = 8.19 \times 10^{-14} \text{ J}\].
04

Calculate Kinetic Energy of Each Particle

The kinetic energy \(KE\) is given by \(KE = \frac{1}{2} mv^2\), but at relativistic speeds, we use \(KE = (\gamma - 1)mc^2\), where \(\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}\) and \(v = 0.20c\).Calculate \(\gamma\):\[\gamma = \frac{1}{\sqrt{1 - (0.20)^2}} = 1.0206\]Calculate \(KE\):\[KE = (1.0206 - 1) \times 9.11 \times 10^{-31} \text{ kg} \times (3 \times 10^8 \text{ m/s})^2\]\[KE = 1.68 \times 10^{-16} \text{ J}\] per particle.
05

Calculate Total Energy Before Collision

Total energy per particle before the collision includes rest energy and kinetic energy:\[\text{Total Energy per Particle} = E_0 + KE = 8.19 \times 10^{-14} \text{ J} + 1.68 \times 10^{-16} \text{ J}\]\[\text{Total Energy per Particle} \approx 8.21 \times 10^{-14} \text{ J}\].
06

Calculate Total Energy of System

The total energy of the system before the collision is the sum of the energies of the electron and positron:\[\text{Total Energy} = 2 \times 8.21 \times 10^{-14} \text{ J}\]\[\text{Total Energy} = 1.642 \times 10^{-13} \text{ J}\].
07

Conclude Energy of Electromagnetic Radiation

Since energy is conserved and both particles annihilate, the energy of the electromagnetic radiation is equal to the total energy of the system:\[E_{\text{radiation}} = 1.642 \times 10^{-13} \text{ J}\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electron-Positron Annihilation
Electron-positron annihilation is a fascinating process that illustrates the mass-energy equivalence in a vivid way. When an electron, a negatively charged particle, meets its antimatter counterpart, the positron, which has a positive charge, they collide and effectively cancel each other out. This collision results in their mass being converted into energy. This energy is released in the form of electromagnetic radiation, typically gamma rays.

The incredible aspect of this process is that the entire rest mass of these particles disappears, with only energy left as a product. According to the mass-energy equivalence principle, articulated by Einstein, the rest mass energy of a particle is defined by the equation: \[ E = mc^2 \]where \( E \) is the energy, \( m \) is the rest mass, and \( c \) is the speed of light. This formula demonstrates that mass can be converted into a large amount of energy, even if the mass is quite small, due to the enormous value of \( c^2 \).

In electron-positron annihilation, the conservation of energy dictates that the energy available before the annihilation, including kinetic energy and rest mass energy, must equal the energy of the photons emitted, which is the essence of mass-energy equivalence.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. For everyday speeds, it's often calculated using the classical formula:\[ KE = \frac{1}{2}mv^2 \] where \( KE \) is kinetic energy, \( m \) is mass, and \( v \) is velocity. However, when particles like electrons and positrons approach relativistic speeds, meaning speeds close to the speed of light, their kinetic energy calculations require modifications to account for relativistic effects.

Under these circumstances, we use the relativistic expression for kinetic energy:\[ KE = (\gamma - 1)mc^2 \] where \( \gamma \) (gamma) is the Lorentz factor, given by:\[ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \]This factor accounts for the relativistic changes experienced due to high velocities.

It ensures that as an object's velocity increases, particularly as it nears the speed of light, its mass effectively becomes infinitely heavy, and thus, infinite energy would be required to reach the speed of light, which is why it's impossible.

In the context of the electron-positron annihilation exercise, both particles have significant kinetic energy, contributing to the total energy that will be converted into electromagnetic radiation upon annihilation.
Relativistic Effects
Relativistic effects become significant when particles travel at speeds approaching the speed of light. These effects include changes in mass, time, and energy as perceived by an observer.

One of the key differences within relativistic physics, as opposed to classical physics, is how velocity affects an object's mass and energy. As an object's speed increases, its relativistic mass increases, which is an indicator of how much energy it has. Consequently, it takes more force to accelerate the object than would be predicted by Newtonian physics alone.

This is explained by the previously mentioned Lorentz factor, \( \gamma \), which becomes significantly larger as the speed of an object nears the speed of light, leading to notable relativistic effects. These adjustments ensure that fundamental laws of physics, like conservation of energy and momentum, hold true in all frames of reference.

In scenarios involving particles like electrons and positrons moving at relativistic speeds before annihilation, these effects ensure accurate calculations of kinetic and total energies. They factor importantly into the total energy, predicted accurately using the equation: \[ E = \gamma mc^2 \]. This encompasses both rest mass energy and kinetic energy in systems where relativity cannot be ignored. Thus, understanding relativistic effects is imperative when dealing with high-speed particles in both theoretical physics and practical applications.

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Most popular questions from this chapter

A Klingon spacecraft has a speed of \(0.75 c\) with respect to the earth. The Klingons measure \(37.0 \mathrm{~h}\) for the time interval be tween two events on the earth. What value for the time interval would they measure if their ship had a speed of \(0.94 c\) with respect to the earth?

A rectangle has the dimensions of \(3.0 \mathrm{~m} \times 2.0 \mathrm{~m}\) when viewed by someone at rest with respect to it. When you move past the rectangle along one of its sides, the rectangle looks like a square. What dimensions do you observe when you move at the same speed along the adjacent side of the rectangle?

A tourist is walking at a speed of \(1.3 \mathrm{~m} / \mathrm{s}\) along a \(9.0-\mathrm{km}\) path that follows an old canal. If the speed of light in a vacuum were \(3.0 \mathrm{~m} / \mathrm{s}\), how long would the path be, according to the tourist?

Two spaceships \(A\) and \(B\) are exploring a new planet. Relative to this planet, spaceship A has a speed of \(0.60 c,\) and space ship \(B\) has a speed of \(0.80 c\). What is the ratio \(D_{\mathrm{A}} / D_{\mathrm{B}}\) of the values for the planet's diameter that each spaceship measures in a direction that is parallel to its motion?

A woman and a man are on separate rockets, which are flying parallel to each other and have a relative speed of \(0.940 c\). The woman measures the same value for the length of her own rocket and for the length of the man's rocket. What is the ratio of the value that the man measures for the length of his own rocket to the value he measures for the length of the woman's rocket?

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