Question

A proton and an antiproton annihilate, producing two photons. Find the energy, frequency, and wavelength of each photon (a) if the p and \(\overline{p}\) are initially at rest and (b) if the p and \(\overline{p}\) collide head-on, each with an initial kinetic energy of 620 MeV.

Short answer

(a) 938.27 MeV; frequency \(1.48 \times 10^{23}\, \text{Hz}\); wavelength \(2.02 \times 10^{-15}\, \text{m}\). (b) 1558.27 MeV; frequency \(2.47 \times 10^{23}\, \text{Hz}\); wavelength \(1.21 \times 10^{-15}\, \text{m}\).

Step-by-step solution

Analyze Rest Annihilation Energy

When a proton (p) and an antiproton (\(\overline{p}\)) annihilate at rest, their combined rest mass energy is converted into photon energy. The rest mass energy of each particle is given by \(E = mc^2\), where \(m_{\text{p}} = 938.27\, \text{MeV}/c^2\). Thus, \(E = 2 \times 938.27\, \text{MeV} = 1876.54\, \text{MeV}\) total for two photons, so each photon has \(E = 938.27\, \text{MeV}\).

Calculate Energy of Photon (at rest)

Each photon's energy is half the rest mass energy of the combined system at annihilation, hence \(E_\gamma = 938.27\, \text{MeV}\).

Calculate Frequency of Photon at Rest

Frequency \(f\) is calculated using the formula \(E = hf\), where \(h\) is Planck's constant \(6.626 \times 10^{-34}\, \text{Js}\). Therefore, \(f = \frac{E}{h} = \frac{938.27\times 10^6 \times 1.602 \times 10^{-19}\, \text{J}}{6.626 \times 10^{-34}\, \text{Js}}\).

Calculate Wavelength of Photon at Rest

The wavelength \(\lambda\) can be found using \(\lambda = \frac{c}{f}\), where \(c\) is the speed of light \(3 \times 10^8\, \text{m/s}\). Use the frequency found in Step 3 to find \(\lambda\).

Analyze Kinetic Energy Contribution (Head-on Collision)

For a head-on collision with kinetic energy, the total energy before annihilation becomes the sum of the rest mass energy and kinetic energy of each particle: \(E = 2 \times (938.27\, \text{MeV} + 620\, \text{MeV})\).

Calculate Each Photon's Energy (head-on)

Each photon's energy is \(E_\gamma = \frac{2 \times 1558.27\, \text{MeV}}{2} = 1558.27\, \text{MeV}\) after the kinetic energy is included.

Calculate Frequency of Photon (head-on)

The frequency \(f\) is calculated from \(E_\gamma = hf\) as \(f = \frac{1558.27 \times 10^6 \times 1.602 \times 10^{-19}\, \text{J}}{6.626 \times 10^{-34}\, \text{Js}}\).

Calculate Wavelength of Photon (head-on)

Using \(\lambda = \frac{c}{f}\) with the new frequency from Step 7, calculate \(\lambda\) for the head-on collision scenario.

Key concepts

Photon Energy

Photon energy plays a crucial role in the annihilation of a proton and an antiproton. When these particles annihilate, they transform entirely into energy in the form of photons. This transformation uses the principle given by Einstein's equation, \(E = mc^2\). In this context, \(m\) is the mass of the proton or antiproton, and \(c\) is the speed of light.

• For a proton or antiproton at rest, each has a rest mass energy of 938.27 MeV.
• Upon annihilation, both rest mass energies add up, giving a total of 1876.54 MeV, which is shared between the two photons produced.
• Consequently, each photon in this scenario has an energy of 938.27 MeV.

However, if the particles collide head-on, they possess additional kinetic energy before annihilating, affecting the photons' energy:

• In a head-on collision with 620 MeV kinetic energy per particle, the total energy for photon creation becomes \(2 \times (938.27 \text{MeV} + 620 \text{MeV}) = 3116.54 \text{MeV}\).
• Thus, each photon now receives an increased energy of 1558.27 MeV.

Frequency Calculation

Calculating the frequency of a photon involves understanding the relationship between energy and frequency through Planck's equation: \(E = hf\). Here, \(E\) denotes the energy of the photon, \(h\) is Planck's constant \((6.626 \times 10^{-34} \text{ Js})\), and \(f\) is the frequency we wish to determine.

For photons resulting from proton-antiproton annihilation:

• At rest, a photon with 938.27 MeV will have its energy converted to joules by multiplying by \(1.602 \times 10^{-19}\) J/MeV.
• The frequency \(f\) is calculated as \(f = \frac{E}{h}\), where \(E\) is the energy converted to joules.

In a head-on collision scenario, the frequency increases:

• With each photon's energy at 1558.27 MeV, convert it to joules similarly.
• Once again, apply \(f = \frac{E}{h}\) to find the higher frequency, as the energy is greater here.

Wavelength Determination

The wavelength of a photon is closely related to its frequency through the equation \(\lambda = \frac{c}{f}\), where \(c\) is the speed of light \((3 \times 10^8 \text{ m/s})\). Understanding the wavelength helps in visualizing the electromagnetic properties of photons resulting from annihilation.

• For photons with rest energy, once the frequency is computed, \(\lambda\) can be derived to show how the photon's wavelength fits into the electromagnetic spectrum.
• Longer wavelengths equate to lower frequencies, reflecting lower-energy photons, while shorter wavelengths indicate higher energy.

If there is a kinetic energy influence:

• The increased frequency from head-on collisions results in shorter wavelengths for those photons. This means that these high-energy photons are more to the 'high-frequency' end of the spectrum.
• Knowing the wavelength is pivotal in applications, like understanding the type of radiation produced.