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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 = \(1 - A\)

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