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Our definition of the scattering cross section, \(N_{\mathrm{sc}}=N_{\text {inc }} n_{\text {tar }} \sigma,\) applies to an experiment using a narrow beam of projectiles all of which pass through a wide target assembly. Experimenters sometimes use a wide incident beam, which completely engulfs a small target assembly (the beam of photons from a car's headlamp, directed at a small piece of plastic, for example). Show that in this case \(N_{\mathrm{sc}}=n_{\text {inc }} N_{\mathrm{tar}} \sigma\) where \(n_{\text {inc }}\) is the density (number/area) of the incident beam, viewed head-on, and \(N_{\text {tar }}\) is the total number of targets in the target assembly.

Short Answer

Expert verified
The scattering formula changes to \( N_{\mathrm{sc}} = n_{\text{inc}} N_{\mathrm{tar}} \sigma \) for a wide incident beam and small target.

Step by step solution

01

Understand the Initial Formula

The initial equation given is \( N_{\mathrm{sc}}=N_{\text{inc}} n_{\text{tar}} \sigma \), where \( N_{\mathrm{sc}} \) is the number of scattered particles, \( N_{\text{inc}} \) is the total number of incident beam particles, \( n_{\text{tar}} \) is the target density (number/volume), and \( \sigma \) is the scattering cross-section.
02

Adapt for Wide Beam Situation

In the problem's scenario, we have a wide beam and a small target assembly. So we redefine our variables: \( n_{\text{inc}} \) becomes the density of the incident beam (number/area) and \( N_{\text{tar}} \) becomes the total count of targets in the assembly.
03

Derive the New Formula

The number of incident particles hitting the small target is related to the total beam density. For each particle of the target, the incident particle's density is \( n_{\text{inc}} \), so we multiply this by the total number of targets \( N_{\text{tar}} \) and the scattering cross-section \( \sigma \), resulting in \( N_{\mathrm{sc}} = n_{\text{inc}} N_{\mathrm{tar}} \sigma \).

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Key Concepts

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

Target Assembly
A target assembly in a scattering experiment refers to the collection of particles or objects being hit by an incoming stream of particles, also known as the incident beam. This assembly acts like a target for the particles in the beam to interact with.
In the context of the problem, when experimenters use a small target assembly engulfed by a wide incident beam, the assembly must be small compared to the dimensions of the beam. This setup allows researchers to examine interactions across a broad area.
By understanding the properties of the target assembly, one can predict how many particles from the incident beam will potentially scatter upon contact with it. This relationship is crucial for calculating the scattering cross-section, which tells us the likelihood of scattering events occurring within a certain target size.
Incident Beam
The incident beam is the stream of particles directed towards the target assembly in a scattering experiment. These particles might be electrons, protons, or photons, depending on the experimental setup.
A wide incident beam can cover a large area, making it optimal for experiments involving small target assemblies. This kind of beam ensures that all potential scattering points in the target assembly are illuminated or reached by the incoming particles.
Understanding the nature of the incident beam, especially its density and spread, helps in assessing how it would interact with the target and how many particles are expected to scatter. The density of the incident beam can be expressed as the number of particles per unit area when viewed head-on.
Particle Density
Particle density is a measure used to describe how concentrated particles are in a given space. In our problem, we consider both the target density and the incident beam density.
* **Target Density**: Refers to how many particles are present in a specified volume of the target assembly. It influences how many particles from the incident beam will collide with and scatter from the assembly. * **Incident Beam Density**: When viewed head-on, it is measured as the number of particles per unit area. This density determines how many particles will engage the target assembly in the case where a wide beam is used.
High particle density in either the target or the beam increases the likelihood of particle interactions, thereby raising the number of scattering events detected.
Scattering Particles
Scattering particles are the particles from the incident beam that interact with the target assembly and are deflected in various directions as a result of this interaction.
When the incident beam strikes the target assembly, some particles will pass through, some will be absorbed, and others will be scattered. The scattered particles are what experimenters are typically interested in measuring, as they provide insights into the properties and characteristics of the target they interact with.
The scattering process is quantified using the scattering cross-section, a crucial factor in determining how likely particles in the beam will scatter upon hitting the target. The cross-section, combined with particle densities, helps in calculating the number of expected scattering events. This number is vital in various research fields, including nuclear physics and material science, where understanding microscopic interactions is key.

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Most popular questions from this chapter

A blueberry pancake has diameter 15 cm and contains 6 large blueberries, each of diameter 1 cm. Find the cross section \(\sigma\) of a blueberry and the "target" density \(n_{\text {tar }}\) (number/area) of berries in the pancake, as seen from above. What is the probability that a skewer, jabbed at random into the pancake, will hit a berry (in terms of \(\sigma\) and \(n_{\text {tar }}\) and then numerically)?

In deriving the cross section for scattering by a hard sphere, we used the "law of reflection," that the angles of incidence and reflection of a particle bouncing off a hard sphere are equal, as in Figure 14.10. Use conservation of energy and angular momentum to prove this law. (The definition of "hard-sphere scattering" is that a projectile bounces with its kinetic energy unchanged. That the force is spherically symmetric implies, as usual, that angular momentum about the sphere's center is conserved.)

A beam of particles is directed through a tank of liquid hydrogen. If the tank's length is \(50 \mathrm{cm}\) and the liquid density is 0.07 gram/cm \(^{3}\), what is the target density (number/area) of hydrogen atoms seen by the incident particles?

The cross section for scattering a certain nuclear particle by a copper nucleus is 2.0 barns. If \(10^{9}\) of these particles are fired through a copper foil of thickness \(10 \mu \mathrm{m}\), how many particles are scattered? (Copper's density is 8.9 gram/cm \(^{3}\) and its atomic mass is \(63.5 .\) The scattering by any atomic electrons is completely negligible.)

[Computer] Consider a point projectile moving in a fixed, spherical force whose potential energy is $$U(r)=\left\\{\begin{array}{ll} -U_{0} & (0 \leq r \leq R) \\ 0 & (R < r) \end{array}\right.\quad\quad\quad\quad\quad\quad(14.60)$$ where \(U_{\mathrm{o}}\) is a positive constant. This so-called spherical well represents a projectile which moves freely in either of the regions \(r < R\) and \(R < r_{1}\) but, when it crosses the boundary \(r=R_{1}\) receives a radially inward impulse that changes its kinetic energy by \(\pm U_{\mathrm{o}}\left(+U_{\mathrm{o}}\right.\) going inward, \(-U_{\circ}\) going outward). (a) Sketch the orbit of a projectile that approaches the well with momentum \(p_{\mathrm{o}}\) and impact parameter \(b < R .\) (b) Use conservation of energy to find the momentum \(p\) of the projectile inside the well \((r < R)\). Let \(\zeta\) denote the momentum ratio \(\zeta=p_{\mathrm{o}} / p\) and let \(d\) denote the projectile's distance of closest approach to the origin. Use conservation of angular momentum to show that \(d=\zeta b\). (c) Use your sketch to prove that the scattering angle \(\theta\) is $$\theta=2\left(\arcsin \frac{b}{R}-\arcsin \frac{\zeta b}{R}\right).\quad\quad\quad\quad\quad\quad(14.60)$$ This gives \(\theta\) as a function of \(b\), which is what you need to get the cross section. The relation depends on the momentum ratio \(\zeta,\) which in turn depends on the incoming momentum \(p_{\mathrm{o}}\) and the well depth \(U_{\mathrm{o}} .\) Plot \(\theta\) as a function of \(b\) for the case that \(\zeta=0.5 .\) ( \(\mathbf{d}\) ) By differentiating \(\theta\) with respect to \(b\), find an expression for the differential cross section as a function of \(b\), and make a plot of \(d \sigma / d \Omega\) against \(\theta\) for the case that \(\zeta=0.5 .\) Comment. [Hint: To plot as a function of \(\theta\) you don't need to solve for \(b\) in terms of \(\theta ;\) instead, you can make a parametric plot of the point \((\theta, d \sigma / d \Omega)\) as a function of the parameter b running from 0 to \(R\).] (e) By integrating \(d \sigma / d \Omega\) over all directions, find the total cross section.

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