Problem 6 - University Physics with Modern Physics

Question

A proton and an antiproton annihilate, producing two photons. Find the energy, frequency, and wavelength of each photon emitted in the center-of-momentum reference frame (a) if the initial kinetic energies of the proton and antiproton are negligible and (b) if each particle has an initial kinetic energy of 830 \(\mathrm{MeV}\) .

Step-by-Step Solution

Answer

(a) Each photon has energy 938 MeV, frequency \(2.27 \times 10^{23} \, \text{Hz}\), and wavelength \(1.32 \times 10^{-15} \, \text{m}\). (b) Each photon has energy 1763 MeV, frequency \(4.26 \times 10^{23} \, \text{Hz}\), and wavelength \(7.04 \times 10^{-16} \, \text{m}\).

Step 1: Understanding the Problem

We have a proton and antiproton annihilating each other, resulting in the creation of two photons. We need to find the energy, frequency, and wavelength of each photon in two scenarios: (a) when the initial kinetic energy of both particles is negligible and (b) when the initial kinetic energy for each particle is 830 MeV.

Step 1: Proton Rest Mass Energy

The rest mass energy of a proton (or antiproton) can be calculated using the equation: \[ E = m c^2 \]where \( m \) is the mass of the proton, approximately \( 938 \, \text{MeV}/c^2 \). Therefore, the rest mass energy is \( 938 \, \text{MeV} \) for each particle.

Step 2: Total Energy Calculation for Part (a)

In this case, since the initial kinetic energies are negligible, the total energy involved in the annihilation is just the rest mass energies of the proton and antiproton combined:\[ E_{ ext{total(a)}} = 938 \, \text{MeV} + 938 \, \text{MeV} = 1876 \, \text{MeV} \] This energy is shared equally between the two photons, so each photon has energy\[ E_{ ext{photon(a)}} = \frac{1876 \, \text{MeV}}{2} = 938 \, \text{MeV} \]

Step 3: Calculate Frequency of Photons (Part a)

The energy of a photon is also given by \[ E = h f \]where \( h \) is Planck's constant (\( 4.135667696 \times 10^{-15} \, \text{eV}\cdot\text{s} \)) and \( f \) is the frequency. Rearranging gives:\[ f = \frac{E}{h} \]Substitute \( E = 938 \, \text{MeV} = 938 \times 10^6 \, \text{eV} \): \[ f_{\text{photon(a)}} = \frac{938 \times 10^6 \, \text{eV}}{4.135667696 \times 10^{-15} \, \text{eV}\cdot\text{s}} \approx 2.27 \times 10^{23} \, \text{Hz} \]

Step 4: Calculate Wavelength of Photons (Part a)

The relationship between frequency \( f \), wavelength \( \lambda \), and the speed of light \( c \) is given by:\[ \lambda = \frac{c}{f} \]Substitute \( c = 3 \times 10^8 \, \text{m/s} \):\[ \lambda_{\text{photon(a)}} = \frac{3 \times 10^8 \, \text{m/s}}{2.27 \times 10^{23} \, \text{Hz}} \approx 1.32 \times 10^{-15} \, \text{m} \]

Step 5: Total Energy Calculation for Part (b)

Each particle has an initial kinetic energy of 830 MeV, in addition to their rest mass energy. Therefore, the total energy in part (b) is:\[ E_{ ext{total(b)}} = (938 \, \text{MeV} + 830 \, \text{MeV}) \times 2 = 3526 \, \text{MeV} \]Each photon thus has:\[ E_{\text{photon(b)}} = \frac{3526 \, \text{MeV}}{2} = 1763 \, \text{MeV} \]

Step 6: Calculate Frequency of Photons (Part b)

Using the same process, find the frequency for each photon:\[ f_{\text{photon(b)}} = \frac{1763 \times 10^6 \, \text{eV}}{4.135667696 \times 10^{-15} \, \text{eV}\cdot\text{s}} \approx 4.26 \times 10^{23} \, \text{Hz} \]

Step 7: Calculate Wavelength of Photons (Part b)

Substitute these values back into our wavelength equation:\[ \lambda_{\text{photon(b)}} = \frac{3 \times 10^8 \, \text{m/s}}{4.26 \times 10^{23} \, \text{Hz}} \approx 7.04 \times 10^{-16} \, \text{m} \]

Key Concepts

Proton-Antiproton Annihilation

When a proton and its antimatter counterpart, the antiproton, meet, they undergo a fascinating process called annihilation. In this process, the mass of the proton and antiproton is converted entirely into energy. This energy is typically in the form of photons, which are light particles. Their combined rest mass energies produce two high-energy photons.

In the exercise scenario where the kinetic energy of the particles is negligible, each proton and antiproton contributes its rest mass energy, totaling 1876 MeV combined. This entire energy transforms into two photons of equal energy, 938 MeV each.

In the second scenario where each particle has 830 MeV of kinetic energy added, the total energy involved rises to 3526 MeV. Thus, each photon formed has more energy, 1763 MeV, compared to before. Annihilation showcases the impressive transformation of matter into energy, illustrating Einstein's mass-energy equivalence, encapsulated in the equation \( E=mc^2 \).

Photon Frequency

Frequency is a fundamental property of photons, representing the number of wave crests passing a particular point per second. It’s directly related to the energy of the photon through Planck's equation \( E = hf \), where \( E \) is the energy, \( h \) is Planck’s constant (\( 4.135667696 \times 10^{-15} \, \text{eV}\cdot\text{s} \)), and \( f \) is the frequency.

In the exercise, two distinct scenarios are considered. For abolition in part (a), each photon emerges with a frequency of approximately \( 2.27 \times 10^{23} \, \text{Hz} \) since they carry an energy of 938 MeV each. The higher the energy of a photon, the greater its frequency. Moving to part (b) with additional kinetic energy, the photons each have a frequency boosted to \( 4.26 \times 10^{23} \, \text{Hz} \), due to their increased energy of 1763 MeV.

Understanding photon frequency is crucial, as it influences the behavior of photons, including how they interact with matter and transmit information through electromagnetic waves.

Photon Wavelength

Wavelength is another critical characteristic of photons. It is the distance between successive peaks of the wave and is inversely related to frequency. The relationship between wavelength \( \lambda \), frequency \( f \), and the speed of light \( c \) is given by the formula \( \lambda = \frac{c}{f} \).

In the exercise, the photons in scenario (a) have a wavelength calculated to be approximately \( 1.32 \times 10^{-15} \, \text{m} \). This short wavelength is derived from their high frequency. For scenario (b), the increased photon energy results in a shorter wavelength of \( 7.04 \times 10^{-16} \, \text{m} \), reflecting the increase in frequency and energy.

Shorter wavelengths indicate higher energies and frequencies. In practical terms, this means that high-energy photons, like those produced in proton-antiproton annihilation, are forms of gamma rays, which possess the ability to penetrate materials and trigger significant physical or chemical changes.