/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 A neutral pion of (rest) mass \(... [FREE SOLUTION] | 91Ó°ÊÓ

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A neutral pion of (rest) mass \(m\) and (relativistic) momentum \(p=\) \(\frac{3}{4} m_{c}\) decays into two photons. One of the photons is emitted in the same direction as the original pion, and the other in the opposite direction. Find the (relativistic) energy of each photon.

Short Answer

Expert verified
Photon energies are \( mc^2 \) and \( \frac{1}{2} mc^2 \).

Step by step solution

01

Understanding Conservation of Energy

In a decay process, the total energy before and after must be conserved. Initially, the energy of the pion is a combination of its rest energy and kinetic energy due to its momentum. Use the formula for total relativistic energy: \[ E = \sqrt{(pc)^2 + (m_0 c^2)^2} \]where \( p \) is the momentum and\( m_0 \) is the rest mass of the pion.
02

Calculating the Initial Energy of the Pion

Substitute the given momentum \( p = \frac{3}{4} mc \) into the energy equation to find the total energy of the pion before decay:\[ E = \sqrt{\left(\frac{3}{4} m c^2\right)^2 + (m c^2)^2} \]\[ E = \sqrt{\frac{9}{16} m^2 c^4 + m^2 c^4} \]\[ E = m c^2 \sqrt{\frac{25}{16}} = \frac{5}{4} m c^2 \]
03

Applying Conservation of Momentum

The momentum before and after the decay must also be conserved. Before decay, the momentum is \( \frac{3}{4} m c \). After decay, because the photons are emitted in opposite directions, the total momentum remains \( \frac{3}{4} m c \), with one photon carrying positive momentum and the other carrying negative momentum.
04

Understanding Photon Energies

Each photon's energy is related to its momentum by \( E = pc \). Let \( E_1 \) and \( E_2 \) be the energies of the two photons moving in opposite directions. By conservation of momentum, we have:\[ E_1 - E_2 = \frac{3}{4} m c^2 \]
05

Applying Conservation of Energy for Photons

The total energy of the photons must add up to the pion's total energy before decay:\[ E_1 + E_2 = \frac{5}{4} m c^2 \]
06

Solving for Energies of Photons

Now solve the system of equations from Steps 4 and 5:1. \( E_1 - E_2 = \frac{3}{4} m c^2 \)2. \( E_1 + E_2 = \frac{5}{4} m c^2 \)Adding these equations gives:\[ 2E_1 = \frac{8}{4} m c^2 = 2 m c^2 \]\[ E_1 = m c^2 \]Subtracting the first equation from the second gives:\[ 2E_2 = \frac{2}{4} m c^2 = \frac{1}{2} m c^2 \]\[ E_2 = \frac{1}{2} m c^2 \]
07

Conclusion

We find that the energy of the photon moving in the same direction as the original pion is \( E_1 = m c^2 \), and the energy of the photon moving in the opposite direction is \( E_2 = \frac{1}{2} m c^2 \).

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Key Concepts

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

Conservation of Energy
In the world of physics, conservation of energy is a fundamental principle. It states that energy can neither be created nor destroyed, but only transformed from one form to another. This principle is particularly important in the analysis of particle decay.

For a neutral pion with a given rest mass and relativistic momentum, the energy prior to decay must equal the total energy of the resulting particles—here, two photons. The energy of the pion is a combination of its rest energy, given by \(E_0 = m_0 c^2\), and its kinetic energy due to momentum \(p\).

The total relativistic energy formula helps calculate this:
  • \(E = \sqrt{(pc)^2 + (m_0 c^2)^2}\): accounts for both kinetic and rest energy.
In our problem, the initial energy computing processes yielded a pion energy equivalent to \(\frac{5}{4}m c^2\). This remains consistent with conservation, ensuring all subsequent energy calculations reflect this total.
Photon Energy
When discussing photon energy, it’s important to understand that photons are unique particles of light, possessing energy directly related to their momentum. In physics, energy of a photon \(E\) can be expressed using the formula:
  • \(E = pc\): where \(p\) is the momentum and \(c\) is the speed of light.
In the decay of a neutral pion into two photons, determining each photon's energy involves leveraging this relationship.

Combining the conservation of energy and the photon energy concept, the equation setup involves each photon moving in opposite directions. With known initial total energy at \(\frac{5}{4}m c^2\), energy division reveals
  • The forward photon energy: \(m c^2\)
  • The backward photon energy: \(\frac{1}{2}m c^2\)
These values satisfy both the energy and momentum conservation laws—a calculated distribution once the system of equations is resolved.
Momentum Conservation
Momentum conservation is another strong pillar in physics, essential in analyzing systems where particles interact or transform. Momentum, a vector quantity involving both magnitude and direction, must be equal before and after any particle decay event.

Applying this principle to the decay of a neutral pion, initially moving with momentum \(\frac{3}{4}m c\), leads to:
  • Post-decay, momentum remains the same overall as photons compositionally preserve system momentum.
  • The emitted photons, traveling in opposite directions, balance each other's momentum: one carries positive and the other negative momentum.
When calculating, the equations for momentum on both sides ensure neutrality:
  • Forward photon: \(E_1 = m c^2\)
  • Reverse photon: \(E_2 = \frac{1}{2}m c^2\)
These solutions epitomize momentum conservation, a crucial consideration especially as it accounts for directionality in these dynamic interactions.

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Most popular questions from this chapter

A particle of mass \(m\) whose total energy is twice its rest energy collides with an identical particle at rest. If they stick together, what is the mass of the resulting composite particle? What is its velocity?

Inertial system \(\overrightarrow{\mathcal{S}}\) moves at constant velocity \(\mathbf{v}=\beta c(\cos \phi \hat{\mathbf{x}}+\) \(\sin \phi \hat{\mathbf{y}})\) with respect to \(\mathcal{S}\). Their axes are parallel to one another, and their origins coincide at \(t=\bar{t}=0\), as usual. Find the Lorentz transformation matrix \(\Lambda\) (Eq. 12.25). $$ \left[\begin{array}{cccc} \gamma & -\gamma \beta \cos \phi & -\gamma \beta \sin \phi & 0 \\ \text { Answer: } & \left(\begin{array}{ccc} \gamma & \left(\gamma \cos ^{2} \phi+\sin ^{2} \phi\right) & (\gamma-1) \sin \phi \cos \phi & 0 \\ -\gamma \beta \cos \phi & (\gamma-1) \sin \phi \cos \phi & \left(\gamma \sin ^{2} \phi+\cos ^{2} \phi\right) & 0 \\ -\gamma \beta \sin \phi & (\gamma-1) & 0 & 1 \end{array}\right) \\ 0 & 0 & 0 & 0 \end{array}\right] $$

In a laboratory experiment, a muon is observed to travel \(800 \mathrm{~m}\) before disintegrating. A graduate student looks up the lifetime of a muon \(\left(2 \times 10^{-6} \mathrm{~s}\right)\) and concludes that its speed was $$ v=\frac{800 \mathrm{~m}}{2 \times 10^{-6} \mathrm{~s}}=4 \times 10^{8} \mathrm{~m} / \mathrm{s} $$ Faster than light! Identify the student's error, and find the actual speed of this muon.

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A Lincoln Continental is twice as long as a VW Beetle, when they are at rest. As the Continental overtakes the VW, going through a speed trap, a (stationary) policeman observes that they both have the same length. The VW is going at half the speed of light. How fast is the Lincoln going? (Leave your answer as a multiple of c.)

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