/*! 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 69 \(\bullet\) The neutral \(\pi^{\... [FREE SOLUTION] | 91Ó°ÊÓ

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\(\bullet\) The neutral \(\pi^{\circ}\) meson is an unstable particle produced in high-energy particle collisions. Its mass is about 264 times that of the electron, and it exists for an average lifetime of \(8.4 \times 10^{-17}\) s before decaying into two gamma-ray photons. Assuming that the mass and energy of the particle are related by the Einstein relation \(E=m c^{2},\) find the uncertainty in the mass of the particle and express it as a fraction of the particle's mass.

Short Answer

Expert verified
The uncertainty in mass is approximately 0.947% of the particle's mass.

Step by step solution

01

Understand the Problem

We need to use the average lifetime of the particle to determine the uncertainty in its energy using the energy-time uncertainty principle and then convert that energy uncertainty to mass uncertainty.
02

Recall the Energy-Time Uncertainty Principle

According to the Heisenberg uncertainty principle, the uncertainty in energy \( \Delta E \) and the uncertainty in time \( \Delta t \) are related by: \\[ \Delta E \Delta t \geq \frac{\hbar}{2} \]where \( \hbar \) is the reduced Planck's constant, approximately \( 1.055 \times 10^{-34} \, \text{Js} \).
03

Calculate Uncertainty in Energy

Given \( \Delta t = 8.4 \times 10^{-17} \, \text{s} \), we find \( \Delta E \) as follows:\[ \Delta E \approx \frac{\hbar}{2 \Delta t} = \frac{1.055 \times 10^{-34} \, \text{Js}}{2 \times 8.4 \times 10^{-17} \, \text{s}} \]Calculate \( \Delta E \) from this expression.
04

Convert Energy Uncertainty to Mass Uncertainty

From the relationship \( E = mc^2 \), the uncertainty in the mass \( \Delta m \) can be expressed as:\[ \Delta m = \frac{\Delta E}{c^2} \] where \( c = 3 \times 10^8 \, \text{m/s} \) is the speed of light. Substitute \( \Delta E \) to find \( \Delta m \).
05

Calculate the Fraction of the Particle's Mass

The mass of the particle is given as approximately 264 times the mass of the electron, where the mass of an electron \( m_e \) is \( 9.11 \times 10^{-31} \, \text{kg} \). Thus, the mass of the meson is:\[ m_{\pi} = 264 \times 9.11 \times 10^{-31} \, \text{kg} \]The fraction of the uncertainty to the mass is:\[ \frac{\Delta m}{m_{\pi}} \]Compute this final fraction.

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Key Concepts

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

Neutral Pi Meson
The neutral pi meson, often denoted as the \(\pi^0\) meson, is a fascinating particle found in the realm of particle physics.It is part of the meson family, which are particles made up of quark-antiquark pairs.Uniquely, the neutral pi meson has no electric charge, distinguishing it from its charged counterparts.

This particle is quite unstable, with a very short average lifetime of\(8.4 \times 10^{-17}\) seconds.It plays a crucial role in high-energy particle physics due to its process of decay into two gamma-ray photons.Understanding the \(\pi^0\) meson involves examining its mass relative to more familiar subatomic particles.For instance, its mass is about 264 times that of an electron.This relative mass helps physicists study the behaviors and interactions of fundamental particles in atomic nuclei.
Mass-Energy Equivalence
Mass-energy equivalence is a vital concept in physics, brought into the realm of common knowledge by Albert Einstein.It is encapsulated in the famous equation \(E=mc^{2}\), which reveals that mass (\(m\)) and energy (\(E\)) are interchangeable.
  • \(E\) - Energy of the particle
  • \(m\) - Mass of the particle
  • \(c\) - Speed of light in a vacuum, approximately \(3 \times 10^{8}\) m/s
Through this principle, even a small amount of mass can correspond to a substantial amount of energy.In our context, the energy of the neutral pi meson as it decays plays a significant role.By applying this principle, the energy uncertainty derived from the particle's limited lifetime allows us to calculate the mass uncertainty as well.This powerful relationship is at the heart of particle physics and cosmology.
Average Lifetime
The concept of average lifetime is crucial when dealing with unstable particles like the neutral pi meson.The average lifetime of a particle tells us how long it typically exists before decaying into other particles.
In our scenario, the \(\pi^0\) meson has an extremely short average lifetime of \(8.4 \times 10^{-17}\) seconds.This fleeting lifespan signifies that the meson rapidly transforms after being produced in a high-energy collision, resulting in its decay into gamma-ray photons.Measurement of such short lifetimes requires precision.This precision allows us to use the Heisenberg uncertainty principle to analyze particle behaviors and uncertainties in their properties like energy and mass.By knowing the average lifetime, we gain insight into the particle's stability and its subsequent interactions in the universe.
Gamma-Ray Photons
At the instance of decay, the neutral pi meson emits two gamma-ray photons.Gamma rays are a form of electromagnetic radiation, much like visible light but with much higher energy.These photons are incredibly energetic, making them essential in various fields of science and technology, from medical imaging to nuclear physics.
When the \(\pi^0\) meson decays, the transformation is an example of how energy is conserved through different forms.The mass of the meson is converted into the energy of the gamma-ray photons, showcasing mass-energy equivalence.Gamma-ray photons are uncharged and have high penetrating power.This makes them useful for studying fundamental physics processes where other means might fail due to interference.Using these photons, scientists explore the building blocks of the universe in both natural environments and specialized laboratory settings.

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Most popular questions from this chapter

\bullet A triply ionized beryllium ion, \(\mathrm{Be}^{3+}\) (a beryllium atom with three electrons removed), behaves very much like a hydrogen atom, except that the nuclear charge is four times as great. (a) What is the ground-level energy of \(\mathrm{Be}^{3+} ?\) How does this compare with the ground-level energy of the hydrogen atom? (b) What is the ionization energy of Be \(^{3+} ?\) How does this compare with the ionization energy of the hydrogen atom? (c) For the hydrogen atom, the wavelength of the photon emitted in the transition \(n=2\) to \(n=1\) is 122 nm. (See Example 28.6 . What is the wavelength of the photon emitted when a \(\mathrm{Be}^{3+}\) ion undergoes this transition? (d) For a given value of \(n,\) how does the radius of an orbit in \(\mathrm{Be}^{3+}\) compare with that for hydrogen?

An incident \(x\) -ray photon is scattered from a free electron that is initially at rest. The photon is scattered straight back at an angle of \(180^{\circ}\) from its initial direction. The wavelength of the scattered photon is 0.0830 \(\mathrm{nm.}\) (a) What is the wavelength of the incident photon? (b) What is the magnitude of the momentum of the electron after the collision? (c) What is the kinetic energy of the electron after the collision?

\(\bullet\) A specimen of the microorganism Gastropus hyptopus measures 0.0020 \(\mathrm{cm}\) in length and can swim at a speed of 2.9 times its body length per second. The tiny animal has a mass of roughly \(8.0 \times 10^{-12} \mathrm{kg}\) . (a) Calculate the de Broglie wave- length of this organism when it is swimming at top speed. (b) Calculate the kinetic energy of the organism (in eV) when it is swimming at top speed.

\(\bullet\) (a) What accelerating potential is needed to produce electrons of wavelength 5.00 \(\mathrm{nm}\) ? (b) What would be the energy of photons having the same wavelength as these electrons? (c) What would be the wavelength of photons having the same energy as the electrons in part (a)?

\(\bullet\) In a parallel universe, the value of Planck's constant is 0.0663 \(\mathrm{J} \cdot\) s. Assume that the physical laws and all other physical constants are the same as in our universe. In this other universe, two physics students are playing catch with a baseball. They are 50 \(\mathrm{m}\) apart, and one throws a 0.10 \(\mathrm{kg}\) ball with a speed of 5.0 \(\mathrm{m} / \mathrm{s}\) (a) What is the uncertainty in the ball's horizontal momentum in a direction perpendicular to that in which it is being thrown if the student throwing the ball knows that is located within a cube with volume 1000 \(\mathrm{cm}^{3}\) at the time she throws it? (b) By what horizontal distance could the ball miss the second student?

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