Skip to main content
Ch 36: Special Relativity
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 36: Special RelativityProblem 75
Chapter 36, Problem 75

Some particle accelerators allow protons (p⁺) and antiprotons (p⁻) to circulate at equal speeds in opposite directions in a device called a storage ring. The particle beams cross each other at various points to cause p⁺ + p⁻ collisions. In one collision, the outcome is p⁺ + p⁻ → e⁺ + e⁻ + γ + γ, where γ represents a high-energy gamma-ray photon. The electron and positron are ejected from the collision at 0.9999995c and the gamma-ray photon wavelengths are found to be 1.0 x 10-6 nm. What were the proton and antiproton speeds, as a fraction of c, prior to the collision?

Verified step by step guidance
1
Understand the problem: The collision involves protons and antiprotons moving at equal speeds in opposite directions. The goal is to determine their speeds as a fraction of the speed of light (c) before the collision. The key concepts here are relativistic energy and momentum conservation.
Step 1: Write down the total energy of the system before the collision. The total energy of each proton or antiproton is given by the relativistic energy formula: E = γmc², where γ = 1 / √(1 - v²/c²) is the Lorentz factor, m is the rest mass of the proton, and v is the speed of the proton as a fraction of c. Since the protons and antiprotons are moving at equal speeds, their total energy is 2γmc².
Step 2: Write down the total energy of the system after the collision. The products of the collision are an electron (e⁻), a positron (e⁺), and two gamma-ray photons (γ). The total energy after the collision is the sum of the relativistic energies of the electron and positron, plus the energy of the two gamma-ray photons. The energy of each photon is given by E_γ = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon.
Step 3: Apply conservation of energy. The total energy before the collision (2γmc²) must equal the total energy after the collision. Set up the equation: 2γmc² = 2γ_e m_e c² + 2E_γ, where γ_e is the Lorentz factor for the electron and positron, m_e is the rest mass of the electron, and E_γ is the energy of each gamma-ray photon. Substitute the given values for the photon wavelength and the electron speed (0.9999995c) to calculate γ_e and E_γ.
Step 4: Solve for the proton speed. Rearrange the energy conservation equation to isolate γ for the proton. Use the relationship γ = 1 / √(1 - v²/c²) to solve for v, the speed of the proton as a fraction of c. This will give you the proton and antiproton speeds prior to the collision.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Relativistic Speeds

In physics, relativistic speeds refer to velocities that are a significant fraction of the speed of light (c). At these speeds, classical mechanics no longer accurately describes motion, and relativistic effects such as time dilation and length contraction become significant. Understanding how objects behave at relativistic speeds is crucial for analyzing high-energy particle collisions, as it affects the energy and momentum calculations involved.
Recommended video:
Guided course
07:59
Speed Distribution & Special Speeds of Ideal Gases

Conservation of Energy and Momentum

The conservation of energy and momentum is a fundamental principle in physics stating that in an isolated system, the total energy and total momentum remain constant before and after a collision. In particle physics, this principle is used to analyze the outcomes of collisions, such as the creation of new particles. For the given collision, one must account for the initial energies and momenta of the protons and antiprotons to determine their speeds.
Recommended video:
Guided course
05:58
Conservation Of Momentum

Photon Wavelength and Energy Relation

The energy of a photon is inversely related to its wavelength, described by the equation E = hc/λ, where E is energy, h is Planck's constant, c is the speed of light, and λ is the wavelength. In high-energy collisions, the emitted gamma-ray photons can provide insights into the energy released during the interaction. Understanding this relationship is essential for calculating the energy involved in the collision and the resulting particle speeds.
Recommended video:
Guided course
05:42
Unknown Wavelength of Laser through Double Slit
Related Practice
Textbook Question

The sun radiates energy at the rate 3.8 x 1026 W. The source of this energy is fusion, a nuclear reaction in which mass is transformed into energy. The mass of the sun is 2.0 x 1030 kg. Fusion takes place in the core of a star, where the temperature and pressure are highest. A star like the sun can sustain fusion until it has transformed about 0.10% of its total mass into energy, then fusion ceases and the star slowly dies. Estimate the sun's lifetime, giving your answer in billions of years.

1
views
Textbook Question

An electron moving to the right at 0.90c collides with a positron moving to the left at 0.90c. The two particles annihilate and produce two gamma-ray photons. What is the wavelength of the photons?

1
views
Textbook Question

The nuclear reaction that powers the sun is the fusion of four protons into a helium nucleus. The process involves several steps, but the net reaction is simply 4p → 4He + energy. The mass of a proton, to four significant figures, is 1.673 x 10-27 kg, and the mass of a helium nucleus is known to be 6.644 x 10-27 kg. What fraction of the initial rest mass energy is this energy?

1
views