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Magazine Chapter 27: Problem 35
(II) What is the minimum photon energy needed to produce a \(\mu^+\mu^-\) pair? The mass of each \(\mu\) (muon) is 207 times the mass of an electron. What is the wavelength of such a photon?
Short Answer
Expert verified
Minimum photon energy is 211.454 MeV; wavelength is \(5.87 \times 10^{-15}\) m.
Step by step solution
01
Understand the Problem
We need to find the minimum photon energy required to produce a muon-antimuon pair, denoted as \(\mu^+\mu^-\). To do this, we first need to understand that the photon energy must equal twice the rest energy of a muon, because a photon will simultaneously create a muon and an antimuon.
02
Calculate Muon Rest Energy
The rest mass of an electron is approximately \(0.511 \text{ MeV}/c^2\). Since a muon has a mass 207 times that of an electron, the rest mass of a muon is \(207 \times 0.511 \text{ MeV}/c^2 = 105.727 \text{ MeV}/c^2\). The rest energy of one muon is therefore \(105.727 \text{ MeV}\).
03
Find Total Energy for Muon Pair Production
To produce both a \(\mu^+\) and a \(\mu^-\) particle, we require a total energy of twice the rest energy of a single muon, which is \(2 \times 105.727 \text{ MeV} = 211.454 \text{ MeV}\). This is the minimum energy required for the photon.
04
Use Energy-Wavelength Relation
According to the energy-wavelength relationship for photons, \(E = \frac{hc}{\lambda}\), where \(E\) is the energy, \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength. Rearranging gives \(\lambda = \frac{hc}{E}\).
05
Plug in Constants to Find Wavelength
Using \(h = 4.1357 \times 10^{-15} \text{ eV}\cdot\text{s}\) and \(c = 3 \times 10^8 \text{ m/s}\), we substitute these into the rearranged formula along with the energy \(E = 211.454 \text{ MeV} = 211454 \text{ eV}\):\[\lambda = \frac{4.1357 \times 10^{-15} \times 3 \times 10^8}{211454} \approx 5.87 \times 10^{-15} \text{ m}.\]This is the wavelength of the photon with the minimum energy required for \(\mu^+\mu^-\) pair production.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Muon
Muons are elementary particles similar to electrons, but with a greater mass. They are part of the second generation of leptons in the Standard Model of particle physics. Muons carry an electric charge of \(e\) and a spin of \(\frac{1}{2}\), classifying them as fermions. They have a mass approximately 207 times that of an electron. This makes them more massive than electrons and capable of generating different interactions and reactions in high-energy physics experiments.
- Muons have both positive (\(\mu^+\)) and negative (\(\mu^-\)) charge variants, known as muons and antimuons, respectively.
- Unlike electrons, muons are unstable and decay into electrons, neutrinos, and antineutrinos, with a mean lifetime of about 2.2 microseconds.
- Muons are commonly produced in cosmic ray interactions in the Earth's atmosphere or artificially in particle accelerators.
Pair Production
Pair production is a fascinating quantum phenomenon where energy is directly transformed into matter. It occurs when a photon's energy is sufficient to produce a particle-antiparticle pair. In this context, pair production refers to forming a muon (\(\mu^+\)) and its antiparticle, an antimuon (\(\mu^-\)). This process only occurs when the photon's energy exceeds the combined rest mass energy of the created particles.
- The minimum energy required to produce a \(\mu^+\mu^-\) pair is thus twice the rest energy of a single muon, allowing the simultaneous creation of both particles.
- This threshold energy ensures conservation of energy and momentum, key principles of physics guiding such interactions.
- Pair production is instrumental in studying fundamental forces and the properties of particles since it illustrates the interchangeability of energy and mass, a cornerstone of Einstein's theory of relativity.
Wavelength Calculation
Calculating the wavelength of a photon given its energy involves using the simple yet powerful energy-wavelength relationship. The formula \(E = \frac{hc}{\lambda}\) connects the energy \(E\) of a photon to its wavelength \(\lambda\), where \(h\) is Planck's constant and \(c\) is the speed of light. By rearranging the equation, we find the wavelength:
\[ \lambda = \frac{hc}{E} \]
\[ \lambda = \frac{hc}{E} \]
- Using the values \(h = 4.1357 \times 10^{-15} \) eVs and \(c = 3 \times 10^8 \) m/s, plugging in the energy value \(E = 211.454 \) MeV becomes straightforward.
- After converting \(E\) into electronvolts and inserting these constants, one can compute the resulting wavelength.
- This calculated wavelength represents the smallest wavelength鈥攁nd thus the maximum energy鈥攐f a photon capable of \(\mu^+\mu^-\) pair production.
Rest Energy
Rest energy signifies the energy inherent to a particle due to its mass, unassociated with any kinetic or potential energy it might possess. Einstein's iconic equation \(E = mc^2\) defines rest energy, illustrating the intrinsic link between mass and energy. For a muon, the rest mass energy is vastly greater than that of an electron, due to its larger mass.
- The rest energy of a muon is calculated using its rest mass and equals approximately \(105.727 \) MeV.
- In pair production, the photon's energy must be at least equivalent to the combined rest energy of the two muons involved, highlighting energy conservation.
- This concept is foundational in particle physics, explaining how particles acquire mass and interact with other forces.
