Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The neutral pion \(\pi^{0}\) is an unstable particle (mass \(m=135\ \mathrm{MeV} / c^{2}\) ) that can decay into two photons, \(\pi^{0} \rightarrow \gamma+\gamma .\) (a) If the pion is at rest, what is the energy of each photon? (b) Suppose instead that the pion is traveling along the \(x\) axis and that the photons are observed also traveling along the \(x\) axis, one forward and one backward. If the first photon has three times the energy of the second, what was the pion's original speed \(v ?\)

Short Answer

Expert verified
(a) 67.5 MeV per photon. (b) Pion's speed was \( \frac{c}{2} \).

Step by step solution

01

Analyze the Decay Process at Rest

For part (a), when the neutral pion \( \pi^0 \) is at rest and decays into two photons, the total energy of the pion is its rest mass energy \( E = m c^2 \), where \( m = 135 \text{ MeV}/c^2 \). Each photon carries half of this energy since the initial system is symmetric and both photons share equal energy.
02

Calculate Photon Energy at Rest

The energy of each photon is given by \( E_{\gamma} = \frac{E}{2} = \frac{m c^2}{2} \). Substituting the given mass, we find \( E_{\gamma} = \frac{135 \text{ MeV}}{2} = 67.5 \text{ MeV} \). Thus, each photon has an energy of 67.5 MeV.
03

Analyze the Moving Pion Scenario

For part (b), when the pion is moving along the x-axis, let's consider the conservation of energy and momentum. If one photon moves in the forward direction with energy \( E_{1} \) and the other in the opposite direction with energy \( E_{2} \), and \( E_{1} = 3E_{2} \), then the initial energy \( E_{\pi} \) of the pion equals \( E_{1} + E_{2} \).
04

Set Up Equations for Conservation of Momentum

Similarly, conservation of momentum dictates that \( p_{1} - p_{2} = p_{\pi} \), where \( p_{1} \) and \( p_{2} \) are the magnitudes of momentum for the photons. Since energy \( E = pc \) for photons, we have \( p_{1} - p_{2} = \frac{E_{\pi}}{c} \).
05

Substitute Energies and Solve for Velocity

Given that \( E_{1} = 3E_{2} \), substitute \( E_{1} = 3E_{2} \) into the energy equation \( E_{\pi} = E_{1} + E_{2} = 4E_{2} \) and momentum equation \( \frac{E_{\pi}}{c} = \frac{3E_{2}}{c} - \frac{E_{2}}{c} = 2\frac{E_{2}}{c} \). Solving these equations, we find \( E_{\pi} = 4E_{2} \) and \( p_{\pi}c = 2E_{2} \), which gives \( v = \frac{p_{\pi}c^2}{E_{\pi}} = \frac{2E_{2}c^2 / c}{4E_{2}} = \frac{1}{2}c \). Thus, the original velocity of the pion was \( v = \frac{c}{2} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Photon Energy
When a neutral pion (\(\pi^0\)) decays while at rest, it transforms into two photons. The energy of each photon can be found using the principle of energy conservation. Initially, the pion has an energy equal to its rest mass energy, given by \(E = mc^2\). For a pion with mass \(m = 135 \mathrm{MeV}/c^2 \), this energy is 135 \mathrm{MeV}.Since the pion is at rest and decays evenly into two photons, the energy is shared equally between the two photons. Thus, each photon obtains half of the original energy, calculated as:
  • \(E_{\gamma} = \frac{mc^2}{2} = \frac{135\ \mathrm{MeV}}{2} = 67.5\ \mathrm{MeV} \)
Consequently, each photon has an energy of 67.5 MeV. This simple yet vital calculation demonstrates the symmetry and balanced energy sharing in particle decay when at rest. Understanding energy distribution is crucial in particle physics, especially in scenarios involving conservation laws.
Conservation of Momentum
In scenarios where the pion (\(\pi^0\)) is moving, it's key to consider both energy and momentum conservation to understand how the particles interact post-decay.Momentum conservation plays a critical role. For instance, when the pion is originally moving along the x-axis, it decays into two photons, one moving forward and the other backward. The energies of these photons are different, where one photon has three times the energy of the other. Let these energies be \(E_1 = 3E_2\) and \(E_2\). The total original energy \(E_{\pi}\) of the pion is thus:
  • \( E_{\pi} = E_1 + E_2 = 4E_2 \)
For momentum conservation, since photons have energy \(E = pc\), the relation is:
  • \(p_1 - p_2 = \frac{E_{\pi}}{c} = p_{\pi}\)
These equations capture the entirety of energy and momentum before and after the decay, maintaining the conservation laws integral in physics that hold true in all inertial frames. Recognizing these concepts allows for deeper analysis and understanding of particle interactions and transformations.
Relativistic Velocity Calculation
To find the original velocity of a moving pion (\(\pi^0\)) before it decays, we delve into relativistic physics.Given the scenario where the energy of the first photon is three times the energy of the second (\(E_1 = 3E_2\)), we need to use the relationship between energy and momentum. The equations drawn from energy and momentum conservation provide the foundation for calculating the velocity.The equations are:
  • \(E_{\pi} = 4E_2 \)
  • \(\frac{E_{\pi}}{c} = 2\frac{E_{2}}{c} = p_{\pi} \)
By solving for velocity, we equate momentum :
  • \(v = \frac{p_{\pi}c^2}{E_{\pi}} = \frac{2E_2c^2 / c}{4E_2} = \frac{1}{2}c \)
This reveals that the pion was originally moving at half the speed of light \(v = \frac{c}{2}\). Relativistic calculations are essential when dealing with high-speed particles, ensuring our theoretical predictions match observed behaviors.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider two events that occur at positions \(\mathbf{r}_{1}\) and \(\mathbf{r}_{2}\) and times \(t_{1}\) and \(t_{2} .\) Let \(\Delta \mathbf{r}=\mathbf{r}_{2}-\mathbf{r}_{1}\) and \(\Delta t=t_{2}-t_{1} .\) Write down the Lorentz transformation for \(\mathbf{r}_{1}\) and \(t_{1}\), and likewise for \(\mathbf{r}_{2}\) and \(t_{2},\) and deduce the transformation for \(\Delta \mathbf{r}\) and \(\Delta t\). Notice that differences \(\Delta \mathbf{r}\) and \(\Delta t\) transform in exactly the same way as \(\mathbf{r}\) and \(t\). This important property follows from the linearity of the Lorentz transformation.

(a) What is a particle's speed if its kinetic energy \(T\) is equal to its rest energy? ( \(\mathbf{b}\) ) What if its energy \(E\) is equal to \(n\) times its rest energy?

One way to create exotic heavy particles is to arrange a collision between two lighter particles $$a+b \rightarrow d+e+\cdots+g$$ where \(d\) is the heavy particle of interest and \(e, \cdots, g\) are other possible particles produced in the reaction. (A good example of such a process is the production of the \(\psi\) particle in the process \(\left.e^{+}+e^{-} \rightarrow \psi, \text { in which there are no other particles } e, \cdots, g .\right)\) (a) Assuming that \(m_{d}\) is much heavier that any of the other particles, show that the minimum (or threshold) energy to produce this reaction in the CM frame is \(E_{\mathrm{cm}} \approx m_{d} c^{2} .\) (b) Show that the threshold energy to produce the same reaction in the lab frame, where the particle \(b\) is initially at rest, is \(E_{\mathrm{lab}} \approx m_{d}^{2} c^{2} / 2 m_{b} .\) (c) Calculate these two energies for the process \(e^{+}+e^{-} \rightarrow \psi,\) with \(m_{e} \approx 0.5 \mathrm{MeV} / c^{2}\) and \(m_{\psi} \approx 3100 \mathrm{MeV} / c^{2} .\) Your answers should explain why particle physicists go to the trouble and expense of building colliding-beam experiments.

A rocket traveling at speed \(\frac{1}{2} c\) relative to frame \(\mathcal{S}\) shoots forward bullets traveling at speed \(\frac{3}{4} c\) relative to the rocket. What is the speed of the bullets relative to \(\mathcal{S} ?\)

A mad physicist claims to have observed the decay of a particle of mass \(M\) into two identical particles of mass \(m,\) with \(M<2 m .\) In response to the objections that this violates conservation of energy, he replies that if \(M\) was traveling fast enough it could easily have energy greater than \(2 m c^{2}\) and hence could decay into the two particles of mass \(m\). Show that he is wrong. [He has forgotten that both energy and momentum are conserved. You can analyse this problem in terms of these two conservation laws, but it is much simpler to go to the rest frame of \(M .]\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free