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Consider the case where an electron \(\mathrm{e}^{-}\) and a positron \(\mathrm{e}^{+}\) annihilate each other and produce photons. Assume that these two particles collide head-on with equal, but small, speeds. (a) Show that it is not possible for only one photon to be produced. (Hint: Consider the conservation law that must be true in any collision.) (b) Show that if only two photons are produced, they must travel in opposite directions and have equal energy. (c) Calculate the wavelength of each of the photons in part (b). In what part of the electromagnetic spectrum do they lie?

Short Answer

Expert verified
(a) A single photon violates momentum conservation. (b) Two photons travel opposite with equal energy. (c) Wavelength: About 2.43 pm; it's gamma radiation.

Step by step solution

01

Identify the Conservation Laws

In particle interactions like annihilation, two key conservation laws apply: conservation of energy and conservation of momentum. For part (a), we'll test whether a single photon can satisfy these laws.
02

Consider the Conservation of Momentum for One Photon

Initially, the electron and positron have equal but opposite momenta, so their total momentum is zero. If only one photon is produced, it must carry away zero momentum for total momentum to be conserved, which contradicts the principle that a photon must have momentum.
03

Conclusion for Part (a)

Since a photon cannot have zero momentum (as it would imply zero energy and thus nonexistence), it is impossible for a single photon to be produced in this annihilation, as it cannot satisfy momentum conservation.
04

Analyze Two Photons Conservation of Momentum

For part (b), we assume two photons are produced. Since the initial momentum is zero, the photons' total momentum must also be zero. This requires them to travel in opposite directions with equal magnitude of momentum.
05

Analyze Energy Conservation for Two Photons

The total energy before annihilation is the sum of the rest energy of the electron and positron \[ E = 2m_e c^2 \].Since energy is conserved, each photon carries an energy equal to \( m_e c^2 \).
06

Use Energy to Find Photon Wavelength

The energy of a photon is given by \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant and \( c \) is the speed of light. Solving for \( \lambda \), we have:\[\lambda = \frac{hc}{m_e c^2} = \frac{h}{m_e c}.\]Substitute known values \( h = 6.626 \times 10^{-34} \, \text{Js}, \)\(m_e = 9.109 \times 10^{-31} \, \text{kg}\), \( c = 3 \times 10^8 \, \text{m/s} \) to find \( \lambda. \)
07

Calculate the Wavelength

Plug in the values:\[\lambda = \frac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 3 \times 10^8} \approx 2.43 \times 10^{-12} \, \text{m}\]This wavelength lies in the gamma-ray region of the electromagnetic spectrum.

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Key Concepts

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

Electron-Positron Annihilation
In the realm of particle physics, electron-positron annihilation is a fascinating phenomenon. When an electron \(e^{-}\) meets its antimatter counterpart, the positron \(e^{+}\), they can annihilate each other. This collision converts their rest mass into energy, typically resulting in the production of photons. This process vividly illustrates the famous equation by Einstein, \(E=mc^2\), where mass is converted directly into energy.
In this type of collision, the conservation laws of physics play a pivotal role. The annihilation must obey both the conservation of energy and the conservation of momentum. Thus, direct conversion into other particles and energy forms should respect these laws, making electron-positron annihilation a relevant example in studying fundamental physics processes.
This collision usually happens at a microscopic scale, often requiring the particles to collide head-on at nearly equal speeds. Understanding this annihilation gives insightful perspectives into particle interactions and their implications in the universe.
Conservation of Momentum
When discussing particle interactions like electron-positron annihilation, the conservation of momentum is a fundamental principle. Initially, in the system of an electron and a positron colliding head-on, their momenta are equal and opposite, resulting in a total system momentum of zero. In any collision, this total momentum must remain constant, according to the conservation of momentum.
If one were to consider the possibility of producing a single photon during annihilation, it becomes clear why this scenario doesn't work. A photon cannot have zero momentum—if it did, its energy would also be zero, implying its nonexistence. Thus, a single photon would violate the conservation of momentum, confirming that two photons must be created in reality.
  • Both photons travel in opposite directions to conserve overall momentum.
  • They share equal momentum magnitude but are directionally opposite.
Through this example, conservation of momentum is highlighted significantly and demonstrates its pivotal role in governing particle physics interactions.
Photon Wavelength
During electron-positron annihilation, the energy of the resultant photons is key to determining their wavelength. Each photon's energy is equivalent to the rest energy of the original particles, as governed by the conservation of energy.
Using the formula \(E = \frac{hc}{\lambda}\), where \(E\) is the energy, \(h\) is Planck's constant, and \(c\) is the speed of light, you can solve for \(\lambda\) (the wavelength). The energy of each photon is half the total rest energy of the pair \(e^{-}+e^{+}\). Thus, the photon's wavelength can be calculated as follows:
  • Substitute known constants into the equation: \(E = m_e c^2\).
  • Rearrange to solve for \(\lambda\).
  • \

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