Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 39: Problem 58

If a neutrino beam of \(300 .\) GeV energy is passed through a 1.00 -cm-thick sheet of aluminum, how much of the beam is attenuated, if the cross section of interactions is given by \(\sigma(E)=\left(1.00 \cdot 10^{-38} \mathrm{~cm}^{2} \mathrm{GeV}^{-1}\right) E ?\)

Short Answer

Expert verified
Answer: To find the amount of beam attenuated, first calculate the cross-section \(\sigma(300)\) using the given formula. Then, find the number density \(N\) of aluminum by using its density, atomic mass, and Avogadro's number. Finally, calculate the attenuation factor \(A\) using the Beer-Lambert Law formula along with the calculated values of \(\sigma(300)\) and \(N\). The amount of beam attenuated is given by the difference \(1 - A\).

Step by step solution

01

Definition of the problem

Given the energy of the neutrino beam (300 GeV) and the thickness of the aluminum sheet (1.00 cm), we want to find the fraction of the beam attenuated after passing through the sheet. The attenuation depends on the cross-section, which is a function of energy.
02

Calculate the cross-section at the given energy

First, we need to calculate the cross-section \(\sigma\) at the given neutrino energy of 300 GeV. Using the provided formula, we can plug in the energy to obtain the cross-section: \(\sigma(E) = (1.00 \cdot 10^{-38}~\mathrm{cm}^2\mathrm{GeV}^{-1}) E\) \(\sigma(300) = (1.00 \cdot 10^{-38}~\mathrm{cm}^2\mathrm{GeV}^{-1}) (300~\mathrm{GeV})\)
03

Calculate the beam attenuation factor

To find the amount of beam attenuated (the attenuation factor \(A\)), we will use the Beer-Lambert Law, which is given as: \(A = e^{-\sigma N l}\) Where \(N\) is the number density of the target material (aluminum in this case), and \(l\) is the thickness of the target material. To find the value of \(N\), we'll need to consider the properties of aluminum: - Atomic mass: \(m_a=26.98~\mathrm{g/mol}\) - Density: \(\rho=2.70~\mathrm{g/cm}^3\) - Avogadro's number: \(N_A=6.02 \times 10^{23} \, \mathrm{/mol}\) The number density \(N\) can be calculated as: \(N= \frac{ (\rho N_A) }{ m_a }\)
04

Calculate the number density of Aluminum

Now, let's plug in the values for aluminum and calculate the number density: \(N = \frac{ (2.70~\mathrm{g/cm^3})(6.02 \times 10^{23} \, \mathrm{/mol}) }{ 26.98~\mathrm{g/mol} }\)
05

Calculate the attenuation factor for the neutrino beam

Now that we have the values for the cross-section \(\sigma(300)\) and number density \(N\), let's plug them into the Beer-Lambert Law formula, using the thickness of the aluminum sheet (1.00 cm) for \(l\): \(A = e^{-\sigma(300) N (1.00~\mathrm{cm})}\)
06

Compute the amount of beam attenuated

Finally, calculate the value of \(A\), the attenuation factor, using the above formula. This value will represent the fraction of the beam attenuated after passing through the aluminum sheet: Amount of beam attenuated = \(1 - A\)

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.

Start your free trial

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

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

A Geiger-Marsden experiment, where \(\alpha\) particles are scattered off of a thin gold film, yields an intensity of particles of \(I\left(90^{\circ}\right)=100 .\) counts/s at a scattering angle of \(90^{\circ} \pm 1^{\circ} .\) What will be the intensity of particles at a scattering angle of \(60^{\circ} \pm 1^{\circ}\) if the scattering obeys the Rutherford formula?

Draw a quark-level Feynman diagram for the decay of a neutral kaon into two charged pions, \(K^{0} \rightarrow \pi^{+}+\pi^{-}\).

If the energy of the virtual photon mediating an electronproton scattering, \(e^{-}+p \rightarrow e^{-}+p\), is given by \(E\), what is the range of this electromagnetic interaction in terms of \(E ?\)

Determine the approximate probing distance of a photon with an energy of \(2.0 \mathrm{keV}\).

Within three years after it begins operation, the proton beam at the Large Hadron Collider at CERN is expected to reach a luminosity of \(10^{34} \mathrm{~cm}^{-2} \mathrm{~s}^{-1}\) (this means that in a \(1-\mathrm{cm}^{2}\) area, \(10^{34}\) protons encounter each other every second). The cross section for collisions, which could lead to direct evidence of the Higgs boson, is approximately \(1 \mathrm{pb}\) (picobarn). [These numbers were obtained from "Introduction to LHC physics," by G. Polesello, Journal of Physics: Conference Series \(53(2006), 107-116 .]\) If the accelerator runs without interruption, approximately how many of these Higgs events can one expect in one year at the LHC?
See all solutions

Recommended explanations on Physics Textbooks

Magnetism and Electromagnetic Induction

Read Explanation

Fields in Physics

Read Explanation

Classical Mechanics

Read Explanation

Magnetism

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Electromagnetism

Read Explanation
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