/*! 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 43 E ssm Suppose a neutrino is crea... [FREE SOLUTION] | 91影视

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Chapter 32: Problem 43

E ssm Suppose a neutrino is created and has an energy of 35 MeV. (a) Assuming the neutrino, like the photon, has no mass and travels at the speed of light, find the momentum of the neutrino. (b) Determine the de Broglie wavelength of the neutrino.

Short Answer

Expert verified
(a) Momentum \( p = 1.87 \times 10^{-22} \text{ kg m/s} \). (b) Wavelength \( \lambda = 3.54 \times 10^{-14} \text{ m} \).

Step by step solution

01

Identify the Given Values

We know the energy, \( E = 35 \text{ MeV} \). We convert this energy into joules: \( 1 \text{ MeV} = 1.602 \times 10^{-13} \text{ J} \). So, \( E = 35 \times 1.602 \times 10^{-13} \text{ J} \).
02

Use Energy-Momentum Relation

For particles traveling at the speed of light, like neutrinos, the energy \( E \) is related to the momentum \( p \) by \( E = pc \). Since \( c = 3 \times 10^8 \text{ m/s} \), we can rearrange the formula to find the momentum: \( p = \frac{E}{c} \).
03

Calculate the Momentum

Substitute the values into the equation: \[ p = \frac{35 \times 1.602 \times 10^{-13} \text{ J}}{3 \times 10^8 \text{ m/s}} \]Calculate to find the momentum, \( p \).
04

Apply de Broglie Wavelength Formula

The de Broglie wavelength \( \lambda \) is given by \( \lambda = \frac{h}{p} \), where \( h = 6.626 \times 10^{-34} \text{ m}^2 \text{ kg/s} \). Use the momentum \( p \) from Step 3 to find \( \lambda \).
05

Calculate the de Broglie Wavelength

Substitute the momentum into the de Broglie formula:\[ \lambda = \frac{6.626 \times 10^{-34} \text{ m}^2 \text{ kg/s}}{p} \]Compute to obtain \( \lambda \).

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Key Concepts

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

Neutrino Energy
Neutrinos, despite being one of the most elusive particles, possess energy much like any other particle. When you come across a problem where a neutrino has an energy of 35 MeV, it鈥檚 crucial to understand the significance of this measurement. Energy for subatomic particles is often given in electron volts (eV). Here, we have 35 MeV, which is a mega electron volt (1 MeV = \(1.602 \times 10^{-13}\) joules). This conversion is key when you are dealing with formulas that require standard international units.

Understanding the energy of a neutrino helps us make calculations about other properties, such as its momentum. You'll typically use its energy in equations related to its speed and its negligible mass. As neutrinos travel at the speed of light, they have energy similar to that of photons, even though they are fundamental particles with no electric charge and only tiny interactions with other matter.
Momentum of a Neutrino
Momentum is a fundamental concept in physics and is crucial for understanding the behavior of particles like neutrinos. For particles that travel at the speed of light, such as neutrinos in this textbook problem, their momentum can be calculated directly from their energy. This is because for massless particles or those assumed to have negligible mass, the energy is directly proportional to their momentum.

To find the momentum of a neutrino with 35 MeV energy, we use the formula:
  • \( E = pc \), where \( E \) is energy, \( p \) is momentum, and \( c \) is the speed of light.
This relation simplifies to \( p = \frac{E}{c} \). By substituting the given energy (converted to joules) and the speed of light, you can calculate the neutrino's momentum in the appropriate standard unit of kilograms meter per second (kg路m/s).

This understanding of neutrino momentum is essential for delving into interactions these particles might have in particle physics experiments.
Energy-Momentum Relation
The energy-momentum relation is a critical aspect of relativistic physics, especially for particles like neutrinos. What makes this relation intriguing is its application to particles that travel at or near the speed of light. For such particles, the energy is directly related to momentum without the typical rest mass term you find in classical physics.

The relation is given by the equation:
  • \( E^2 = (pc)^2 + (m_0c^2)^2 \)
For massless particles, or those effectively treated as massless (\( m_0 = 0 \)), such as neutrinos, the equation simplifies to \( E = pc \). This simplification allows for straightforward calculations as we no longer deal with the mass term. This encapsulates the behavior of photons as well and highlights the universality of the energy-momentum relationship in high-energy physics.

Being comfortable with this concept unlocks a deeper understanding of how particles carry energy, move through space, and interact with their environment.

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Most popular questions from this chapter

E V-HINT A neutral pion \(\pi^{0}\) (rest energy \(=135.0\) MeV ) produced in a high-energy particle experiment moves at a speed of \(0.780 \mathrm{c} .\) After a very short time, it decays into two \(\gamma\) -ray photons. One of the \(\gamma\) -ray photons has an energy of 192 MeV. What is the energy (in MeV) of the second \(\gamma\) -ray photon? Take relativistic effects into account.

M Deuterium \(\left({ }_{1}^{2} \mathrm{H}\right)\) is an attractive fuel for fusion reactions because it is abundant in the oceans, where about \(0.015 \%\) of the hydrogen atoms in the water \(\left(\mathrm{H}_{2} \mathrm{O}\right)\) are deuterium atoms. (a) How many deuterium atoms are there in one kilogram of water? (b) If each deuterium nucleus produces about 7.2 MeV in a fusion reaction, how many kilograms of water would be needed to supply the energy needs of the United States for one year, estimated to be \(1.1 \times 10^{20} \mathrm{J} ?\)

When considering the biological effects of ionizing radiation, the concept of biologically equivalent dose is especially important. Its importance lies in the fact that the biologically equivalent dose incorporates both the amount of energy per unit mass that is absorbed and the effectiveness of a particular type of radiation in producing a certain biological effect. Problem 55 examines this concept and also reviews the notions of power (Section 6.7 ) and intensity (Section 16.7\()\) of a wave. Problem 56 illustrates the decay of a particle into two photons, and provides a review of the principles of conservation of energy and conservation of momentum. M CHALK SSM A patient is being given a chest X-ray. The X-ray beam is turned on for \(0.20 \mathrm{s},\) and its intensity is \(0.40 \mathrm{W} / \mathrm{m}^{2} .\) The area of the chest being exposed is \(0.072 \mathrm{m}^{2},\) and the radiation is absorbed by \(3.6 \mathrm{kg}\) of tissue. The relative biological effectiveness (RBE) of the X-ray for this tissue is \(1.1 .\) Concepts: (i) How is the power of the beam related to the beam intensity? (ii) How is the energy absorbed by the tissue related to the power of the beam? (iii) What is the absorbed dose? (iv) How is the biologically equivalent dose related to the absorbed dose? Calculation: Calculate the biologically equivalent dose received by the patient.

E GO Someone stands near a radioactive source and receives doses of the following types of radiation: \(\gamma\) rays \((20 \mathrm{mrad}, \mathrm{RBE}=1),\) electrons \((30 \mathrm{mrad}, \mathrm{RBE}=1),\) protons \((5 \mathrm{mrad}, \mathrm{RBE}=10),\) and slow neutrons \((5 \mathrm{mrad}\) \(\mathrm{RBE}=2\) ). Rank the types of radiation, highest first, as to which produces the largest biologically equivalent dose.

\(\mathrm{M}\) A Rough Measure of Exposure. You and your team are designing a crude device to estimate radiation exposure. The device consists of a set of parallel plates with a large voltage across them. The circular plates have a radius of \(3.50 \mathrm{cm}\) and are separated by distance of \(1.00 \mathrm{cm} .\) The region between the plates is filled with dry air at standard temperature and pressure (STP: \(T=0^{\circ}, P=1\) ). A cubic meter of dry air at STP has a mass of \(1.29 \mathrm{kg}\) When radiation ionizes an air molecule, the stripped electron is accelerated by the electric field between the plates and is registered as a current. When the device is located near a particularly strong source, it registers a current of \(1.30 \times 10^{-9}\) A. What is the exposure between the plates after \(5.00 \mathrm{s}\) (in roentgens)?
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