Problem 56
Question
The mass of a \(\pi^{0}\) can be measured by observing the reaction \(\pi^{-}+\mathrm{p} \rightarrow \pi^{0}+\mathrm{n}\) at very low incident \(\pi^{-}\) kinetic energy (assume it is zero). The neutron is observed to be emitted with a kinetic energy of \(0.60 \mathrm{MeV}\). Use conservation of energy and momentum to determine the \(\pi^{0}\) mass.
Step-by-Step Solution
Verified Answer
The mass of \( \pi^{0} \) is approximately 137.67 MeV/c².
1Step 1: Understand the Reaction
We have a reaction \( \pi^{-} + p \rightarrow \pi^{0} + n \). This means a \( \pi^{-} \) particle (pion) and a proton (p) form a \( \pi^{0} \) particle and a neutron (n). We need to determine the mass of the \( \pi^{0} \) particle. The incident \( \pi^{-} \) has essentially no kinetic energy.
2Step 2: Apply Conservation of Energy
According to the conservation of energy, the total energy before the reaction must equal the total energy after the reaction. Initially, we have the rest energies: \[ E_{initial} = m_{\pi^{-}}c^2 + m_{p}c^2 \]After the reaction, we have:\[ E_{final} = m_{\pi^{0}}c^2 + m_{n}c^2 + KE_{n} \]where \( KE_{n} = 0.60 \text{ MeV} \).
3Step 3: Utilize Known Masses
The mass of the \( \pi^{-} \) is \( 139.57 \text{ MeV/c}^2 \) , the proton mass is \( 938.27 \text{ MeV/c}^2 \), and the neutron mass is \( 939.57 \text{ MeV/c}^2 \). Using conservation of energy, we equate the initial and final energies and solve for \( m_{\pi^{0}}c^2 \): \[ m_{\pi^{-}}c^2 + m_{p}c^2 = m_{\pi^{0}}c^2 + m_{n}c^2 + 0.60 \]
4Step 4: Rearrange to Find \( m_{\pi^{0}} \)
Rearrange the energy conservation equation to solve for \( m_{\pi^{0}} \) \[ m_{\pi^{0}}c^2 = m_{\pi^{-}}c^2 + m_{p}c^2 - m_{n}c^2 - 0.60 \]
5Step 5: Calculate \( m_{\pi^{0}} \)
Substitute in the known values:\[ m_{\pi^{0}}c^2 = 139.57 \text{ MeV} + 938.27 \text{ MeV} - 939.57 \text{ MeV} - 0.60 \text{ MeV} \]Compute the result:\[ m_{\pi^{0}}c^2 = 137.67 \text{ MeV} \]Thus, the mass of \( \pi^{0} \) is \( 137.67 \text{ MeV/c}^2 \).
Key Concepts
Conservation of MomentumParticle PhysicsPion Mass Calculation
Conservation of Momentum
In any physical reaction, the total momentum before the interaction must equal the total momentum afterward. This principle is known as the conservation of momentum. It is crucial in particle physics because it allows us to predict the motion and behavior of particles after collisions or reactions.
When dealing with subatomic particles like pions and protons, understanding momentum becomes especially important. For a reaction involving minimal kinetic energy, like the one described, we assume the particles are initially at rest. This simplifies calculations because the initial momentum is effectively zero.
When dealing with subatomic particles like pions and protons, understanding momentum becomes especially important. For a reaction involving minimal kinetic energy, like the one described, we assume the particles are initially at rest. This simplifies calculations because the initial momentum is effectively zero.
- Before the reaction: The total momentum is zero because the particles are assumed to have negligible motion relative to each other.
- After the reaction: The momentum of the resulting particles (\(\pi^{0}\) and neutron) must also sum to zero, considering other forces are minimal.
Particle Physics
Particle physics is the branch of physics that studies the smallest constituents of matter and the forces that govern them. Understanding particle reactions involves considering fundamental particles like protons, neutrons, and pions. These particles participate in reactions that often conserve both energy and momentum.
- Pions (\(\pi\) particles): These are mesons predominantly responsible for mediating the strong nuclear force between nucleons, like protons and neutrons.
- Proton and Neutron: Part of the baryon family in particle physics. The proton has a slightly lower mass than the neutron, influencing reaction outcomes like energy release and particle speeds.
Pion Mass Calculation
Determining the mass of a particle such as the \(\pi^{0}\) pion involves careful use of conservation laws. These measurements are not only theoretical but often crucial to experiments in particle physics. Let's break down the steps one might take to measure the pion's mass in practice:
Initial total energy: \( m_{\pi^{-}}c^2 + m_{p}c^2 \) equals
Final total energy: \( m_{\pi^{0}}c^2 + m_{n}c^2 + KE_{n} \).
By rearranging the terms, we solve for \( m_{\pi^{0}} \), using known values for other particle masses.
This computation reveals the rest mass energy of the \(\pi^{0}\) particle, enhancing our understanding of particle masses and interactions. It's a prime example of how theoretical concepts apply in experimental physics.
- Setup: Conduct reactions at controlled, low kinetic energy conditions to minimize unaccounted variables.
- Observation: Utilizing detectors, measure emitted particle energies accurately, as done with the neutron's kinetic energy.
- Calculation: Apply conservation laws, initially equating total energy (rest mass energy plus kinetic energy) before and after the reaction:
Initial total energy: \( m_{\pi^{-}}c^2 + m_{p}c^2 \) equals
Final total energy: \( m_{\pi^{0}}c^2 + m_{n}c^2 + KE_{n} \).
By rearranging the terms, we solve for \( m_{\pi^{0}} \), using known values for other particle masses.
This computation reveals the rest mass energy of the \(\pi^{0}\) particle, enhancing our understanding of particle masses and interactions. It's a prime example of how theoretical concepts apply in experimental physics.
Other exercises in this chapter
Problem 53
Symmetry breaking occurs in the electroweak theory at about \(10^{-18} \mathrm{~m}\). Show that this corresponds to an energy that is on the order of the mass o
View solution Problem 55
How many fundamental fermions are there in a water molecule?
View solution Problem 57
(a) Show that the so-called unification distance of \(10^{-31} \mathrm{~m}\) in grand unified theory is equivalent to an energy of about \(10^{16} \mathrm{GeV}\
View solution Problem 60
For the reaction \(p+p \rightarrow 3 p+\bar{p},\) where one of the initial protons is at rest, use relativistic formulas to show that the threshold energy is \(
View solution