A proton and an antiproton annihilate, producing two photons. Find the energy, frequency, and wavelength of each photon if the ppp and p‾\overline{p}p collide head-on, each with an initial kinetic energy of 620620 MeV.

Ch 44: Particle Physics and Cosmology

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 44: Particle Physics and Cosmology

Problem 4b

Chapter 44, Problem 4b

A proton and an antiproton annihilate, producing two photons. Find the energy, frequency, and wavelength of each photon if the $p$ and $\overline{p}$ collide head-on, each with an initial kinetic energy of $620$ MeV.

Verified step by step guidance

Understand the problem: When a proton and an antiproton annihilate, their total energy (rest energy + kinetic energy) is converted into the energy of the two photons. The energy of each photon can be determined using the principle of conservation of energy.

Step 1: Calculate the total energy of the proton and antiproton. The rest energy of a proton (or antiproton) is given by Einstein's equation: $ E_{rest} = m c^2 $, where $ m $ is the mass of the proton ($ 1.67 \times 10^{-27} \; \text{kg} $) and $ c $ is the speed of light ($ 3.00 \times 10^8 \; \text{m/s} $). Add the given kinetic energy (620 MeV) to the rest energy to find the total energy of each particle.

Step 2: Use the conservation of energy principle. The total energy of the system before annihilation is the sum of the total energies of the proton and antiproton. Since the two photons share this energy equally, divide the total energy by 2 to find the energy of each photon.

Step 3: Relate the energy of each photon to its frequency using the Planck-Einstein relation: $ E = h f $, where $ h $ is Planck's constant ($ 6.63 \times 10^{-34} \; \text{J·s} $) and $ f $ is the frequency. Rearrange the equation to solve for $ f $: $ f = \frac{E}{h} $.

Step 4: Relate the frequency of each photon to its wavelength using the wave equation: $ c = \lambda f $, where $ \lambda $ is the wavelength and $ c $ is the speed of light. Rearrange the equation to solve for $ \lambda $: $ \lambda = \frac{c}{f} $.

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Key Concepts

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

Mass-Energy Equivalence

Mass-energy equivalence, expressed by Einstein's equation E=mc², states that mass can be converted into energy and vice versa. In particle physics, this principle is crucial when considering the annihilation of particles, such as a proton and an antiproton, where their mass is converted into energy in the form of photons.

Photon Energy

The energy of a photon is directly related to its frequency and inversely related to its wavelength, described by the equation E=hf, where E is energy, h is Planck's constant, and f is frequency. This relationship allows us to calculate the energy of the photons produced in the annihilation process based on the total energy available from the colliding particles.

Kinetic Energy in Particle Collisions

In particle collisions, the total energy is the sum of the rest mass energy and the kinetic energy of the particles involved. For the proton and antiproton with 620 MeV of kinetic energy each, this total energy must be considered when calculating the resulting energy of the photons produced during their annihilation, as it contributes significantly to the energy available for conversion into photon energy.

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A proton and an antiproton annihilate, producing two photons. Find the energy, frequency, and wavelength of each photon if the ppp and p‾\(\overline{p}\)p collide head-on, each with an initial kinetic energy of 620620 MeV.

Verified video answer for a similar problem:

Key Concepts

Mass-Energy Equivalence

Photon Energy

Kinetic Energy in Particle Collisions

A proton and an antiproton annihilate, producing two photons. Find the energy, frequency, and wavelength of each photon if the $p$ and $\overline{p}$ collide head-on, each with an initial kinetic energy of $620$ MeV.