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Chapter 4: Problem 9

A beam of neutrons with kinetic energy \(0.1 \mathrm{eV}\) and intensity \(10^{6} \mathrm{~s}^{-1}\) is incident normally on a thin foil of \({ }_{92}^{235} \mathrm{U}\) of effective density \(10^{-1} \mathrm{~kg} \mathrm{~m}^{-2}\). The beam can undergo (i) isotropic elastic scattering, with a cross-section \(\sigma_{\text {el }}=3 \times 10^{-2} \mathrm{~b}\), (ii) radiative capture, with a cross-section \(\sigma_{\text {cap }}=10^{2} \mathrm{~b}\), or (iii) it can fission a \({ }_{92}^{235} \mathrm{U}\) nucleus, with a cross-section \(\sigma_{\text {fission }}=3 \times 10^{2} b\). Calculate the attenuation of the beam and the flux of elastically scattered particles \(5 \mathrm{~m}\) from the foil.

Short Answer

Expert verified
The beam is attenuated to approximately zero if attenuation is very large. Elastically scattered flux at 5m is insignificant due to enormous attenuation.

Step by step solution

01

Convert Cross-Sections from Barns to m虏

Given cross-sections are in barns where 1 barn = \(10^{-28}\) m虏. Convert the cross-sections as follows: - \(\sigma_{\text{el}} = 3 \times 10^{-2} \text{ b} = 3 \times 10^{-2} \times 10^{-28}\, \text{m}^2\)- \(\sigma_{\text{cap}} = 10^2 \text{ b} = 10^2 \times 10^{-28} \,\text{m}^2\)- \(\sigma_{\text{fission}} = 3 \times 10^2 \text{ b} = 3 \times 10^2 \times 10^{-28} \,\text{m}^2\)
02

Calculate the Total Cross-Section

The total cross-section \(\sigma_{\text{total}}\) is the sum of elastic scattering, radiative capture, and fission cross-sections:\[\sigma_{\text{total}} = \sigma_{\text{el}} + \sigma_{\text{cap}} + \sigma_{\text{fission}} = (3 \times 10^{-2} + 10^2 + 3 \times 10^2) \times 10^{-28} \,\text{m}^2\]
03

Determine the Attenuation of the Beam

The attenuation of the beam is given by the exponential decay formula:\[I = I_0 e^{-n\sigma_{\text{total}}d}\]where \(I_0 = 10^6 \,\text{s}^{-1}\) and \(d = 10^{-1} \,\text{kg/m}^2\). We need to calculate the number density \(n\) using \(n = \frac{\rho}{M} \frac{N_A}{A}\), where \(\rho = 10^{-1} \,\text{kg/m}^2, M = 235 \,\text{g/mol}, N_A = 6.022 \times 10^{23} \,\text{mol}^{-1}\).Convert \(M\) to kg using \(M = 235 \times 10^{-3} \,\text{kg/mol}\).
04

Calculate Number Density and Attenuation

Calculate \(n\):\[n = \frac{10^{-1} \,\text{kg/m}^2 }{235 \times 10^{-3} \,\text{kg/mol}} \times 6.022 \times 10^{23} \,\text{mol}^{-1} = 2.57 \times 10^{21} \,\text{m}^{-2}\]Calculate attenuation:\[I = 10^6 e^{-(2.57 \times 10^{21})(3.1 \times 10^{-26})(10^{-1})}\]where \(\sigma_{\text{total}} = 3.1 \times 10^{-26} \,\text{m}^2\).Evaluate the exponent and calculate \(I\).
05

Calculate Flux of Elastically Scattered Particles at 5m

The flux of the elastically scattered particles is given by:\[\phi = n_0 \sigma_{\text{el}} J_{el} \]where \(J_{el} = \frac{I}{4\pi r^2}\) is the intensity at a distance \(r = 5 \,\text{m}\).Substitute values \(n_0 = I_0\), \(\sigma_{\text{el}} = 3 \times 10^{-30} \,\text{m}^2\), and evaluate the flux.

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Key Concepts

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

Attenuation of Neutron Beam
When a neutron beam passes through a material, it does not remain unchanged. Instead, its intensity reduces or attenuates as it interacts with the material. This is known as the attenuation of a neutron beam. Attenuation happens due to the various possible interactions that neutrons can have with the atoms in the material, such as scattering, capture, or causing fission.

The level of attenuation is characterized by an exponential decay law, given by the formula \( I = I_0 e^{-n \sigma_{\text{total}} d} \) where \( I_0 \) is the initial intensity of the neutron beam, \( I \) is the intensity after passing through the material, \( n \) is the number density of target atoms, \( \sigma_{\text{total}} \) is the total cross-section of all interactions, and \( d \) is the thickness or density of the material. This formula helps us predict how much of the neutron beam will be absorbed or scattered when it exits the material.

Understanding attenuation is crucial in nuclear physics and applications like medical imaging or radiotherapy, where precise control over the neutron dose is necessary.
Neutron Cross-Section
The neutron cross-section is a measure of the probability of a neutron interacting with a nucleus. It is a crucial parameter in nuclear physics, dictating how likely a neutron is to undergo certain reactions, such as elastic scattering, radiative capture, or fission, when it comes into contact with a nuclear target.

Each reaction has its own cross-section, denoted as \( \sigma_{\text{el}} \), \( \sigma_{\text{cap}} \), and \( \sigma_{\text{fission}} \) for elastic scattering, capture, and fission, respectively. The units of these cross-sections are barns (b), where 1 barn is \( 10^{-28} \) m虏, representing a very small area on the atomic scale. Together, these cross-sections combine to form the total cross-section \( \sigma_{\text{total}} \), which is essential in calculating the attenuation of a neutron beam.

Understanding neutron cross-sections allows researchers and engineers to design materials and processes for applications such as nuclear reactors, radiation shielding, and even medical therapies where nuclear reactions play a critical role.
Isotropic Elastic Scattering
Isotropic elastic scattering occurs when a neutron collides with a nucleus and bounces off without any energy loss but changes direction. The term "isotropic" suggests that the neutrons scatter uniformly in all directions. This type of interaction is vital because it affects how neutron beams distribute as they pass through a material.

Elastic scattering is described by the elastic scattering cross-section \( \sigma_{\text{el}} \), which tells us the probability of this scattering occurring. When the beam undergoes isotropic elastic scattering, it maintains its kinetic energy but spreads across an angle, which can influence how neutrons penetrate or traverse materials.

In practical applications, isotropic elastic scattering is considered when determining the spread of neutrons in materials, which is important for reactor design and safety analysis, ensuring that neutrons are not concentrated in any one area but distributed evenly throughout a system.

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Most popular questions from this chapter

Estimate the thickness of iron through which a beam of neutrinos with energy \(300 \mathrm{GeV}\) must travel if 1 in \(10^{9}\) of them is to interact. Assume that at high energies the neutrino-nucleon total cross-section is given approximately by \(\sigma_{\nu} \approx 10^{-38} E_{\nu} \mathrm{cm}^{2}\), where \(E_{\nu}\) is given in GeV. The density of iron is \(\rho=7.9 \mathrm{~g} \mathrm{~cm}^{-3}\).

Alpha particles are accelerated in a cyclotron operating with a magnetic field of magnitude \(B=0.8 \mathrm{~T}\). If the extracted beam has an energy of \(12 \mathrm{MeV}\), calculate the extraction radius and the orbital frequency of the beam (the so-called cyclotron frequency).

A liquid hydrogen target of volume \(125 \mathrm{~cm}^{3}\) and density \(\rho=0.071 \mathrm{gcm}^{-3}\) (to two significant figures) is bombarded with a mono-energetic beam of negative pions with a flux \(2 \times 10^{7} \mathrm{~m}^{-2} \mathrm{~s}^{-1}\) and the reaction \(\pi^{-}+p \rightarrow \pi^{0}+n\) observed by detecting the photons from the decay of the \(\pi^{0}\). Calculate the rate of photons emitted from the target per second if the cross-section is \(40 \mathrm{mb}\).

At the HERA collider (which was operational until 2007) at the DESY Laboratory in Hamburg, a \(20 \mathrm{GeV}\) electron beam collided with a \(300 \mathrm{GeV}\) proton beam at a crossing angle of \(10^{\circ}\). Evaluate the total centre-of-mass energy and calculate what beam energy would be required in an electron accelerator with a fixed-target proton to achieve the same total centre-ofmass energy.

The reaction \(e^{+} e^{-} \rightarrow \tau^{+} \tau^{-}\)is studied using a collider with equal beam energies of \(5 \mathrm{GeV}\). The differential cross- section is given by $$ \frac{\mathrm{d} \sigma}{\mathrm{d} \Omega}=\frac{\alpha^{2} \hbar^{2} c^{2}}{4 E_{\mathrm{CM}}^{2}}\left(1+\cos ^{2} \theta\right) $$ where \(E_{\mathrm{CM}}\) is the total centre-of-mass energy and \(\theta\) is the angle between the incoming \(e^{-}\)and the outgoing \(\tau^{-}\). If the detector can only record an event if the \(\tau^{+} \tau^{-}\)pair makes an angle of at least \(30^{\circ}\) relative to the beam line, what fraction of events will be recorded? What is the total crosssection for this reaction in nanobarns? If the reaction is recorded for \(10^{7} \mathrm{~s}\) at a luminosity of \(L=10^{31} \mathrm{~cm}^{-2} \mathrm{~s}^{-1}\), how many events are expected? Suppose the detector is of cylindrical construction and at increasing radii from the beam line there is a drift chamber, an electromagnetic calorimeter, a hadronic calorimeter, and finally muon chambers. If in a particular event the tau decays are $$ \tau^{-} \rightarrow \mu^{-}+\bar{\nu}_{\mu}+\nu_{\tau} \text { and } \tau^{+} \rightarrow e^{+}+\bar{\nu}_{\tau}+\nu_{e}, $$ what signals would be observed in the various parts of the detector?
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