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[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.

Short Answer

Expert verified
The scattering angle \( \theta = 2\left(\arcsin \frac{b}{R} - \arcsin \frac{\zeta b}{R}\right) \) with \( \zeta = 0.5 \). The differential cross-section varies with \( \theta \). Total cross-section captures the interaction scale.

Step by step solution

01

Understanding the Spherical Well Potential

The potential energy in the spherical well depends on the distance \( r \) from the origin. Within the well \((0 \leq r \leq R)\), the potential energy \( U(r) \) is \(-U_0\). Outside the well \((r > R)\), \( U(r) = 0 \). The projectile receives an impulse at \( r = R \), affecting its kinetic energy.
02

Sketching the Orbit with Impact Parameter \( b < R \)

The projectile approaches the spherical well and enters the region \( r < R \) where it receives an inward impulse, increasing its kinetic energy by \( U_0 \). Inside, it moves freely and reverses its path upon reaching the closest distance of approach \( d \). Exiting the well, kinetic energy decreases by \( U_0 \). Sketch shows symmetrical entry and exit angles, indicating a change in travel path when crossing \( r = R \).
03

Conservation of Energy inside the Well

Energy conservation gives kinetic energy change at \( r = R \). Initial kinetic energy outside \( E = \frac{p_o^2}{2m} \), becomes \( E + U_0 \) inside due to potential change. This energy gives an inside momentum \( p = \sqrt{p_o^2 + 2mU_0} \). Solving gives \( \zeta = \frac{p_o}{p} = \sqrt{\frac{p_o^2}{p_o^2 + 2mU_0}} \).
04

Closest Distance by Conservation of Angular Momentum

Conservation of angular momentum states \( p_o b = p d \). Solving gives \( d = \frac{p_o b}{p} = \zeta b \), showing closest approach distance \( d \) in terms of momenta and impact parameter.
05

Scattering Angle and Its Derivation

Utilize geometry of path inside the well from the sketch; total angle \( \theta = 2(\sin^{-1} \frac{b}{R} - \sin^{-1} \frac{\zeta b}{R}) \). Input given momentum ratio \( \zeta = 0.5 \) to derive \( \theta \) as function of \( b \).
06

Plotting \( \theta \) versus \( b \)

Substitute \( \zeta = 0.5 \) in \( \theta(b) \). Plot \( \theta \) from \( b = 0 \) to \( R \), showing changes in scattering angle relative to impact parameter. Observe trend in \( \theta \) as \( b \) changes.
07

Calculating Differential Cross Section \( \frac{d\sigma}{d\Omega} \)

Differentiate \( \theta(b) \) with respect to \( b \) using Leibniz's rule to find \( \frac{d\theta}{db} \), then transform relation to find \( \frac{d\sigma}{d\Omega} = \frac{b}{\sin\theta} \left| \frac{d\theta}{db} \right|^{-1} \).
08

Parametric Plot of \( \frac{d\sigma}{d\Omega} \) versus \( \theta \)

Plot \((\theta, \frac{d\sigma}{d\Omega})\) parametrically from \( b=0 \) to \( r=R \), showing dependency on scattering angle. Comment on differential characteristic of cross-section.
09

Total Cross Section via Integration

Integrate \( \frac{d\sigma}{d\Omega} \) over all solid angles to derive total cross-section \( \sigma \). Evaluate and explain differences in captures and deflected trajectories by spherical well interaction.

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Key Concepts

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

Conservation of Energy
In physics, the conservation of energy principle tells us that the total energy in an isolated system remains constant over time. This includes both kinetic energy and potential energy. In the context of a projectile moving in a spherical well potential, this principle helps us determine how the projectile's energy changes as it crosses different radial distances.

Initially, outside the well, the projectile has kinetic energy given by the equation \( E = \frac{p_o^2}{2m} \), where \( p_o \) is the momentum and \( m \) is the mass. When the projectile enters the well (at radius \( r = R \)), it experiences an energy change. Its kinetic energy increases due to the well's potential depth \( U_0 \). The new kinetic energy inside the well becomes \( E + U_0 \), leading to a change in its momentum to \( p = \sqrt{p_o^2 + 2mU_0} \).

This change illustrates the conservation of energy: as the projectile passes the boundary, it gains potential energy that transforms into additional kinetic energy. By comparing the initial and new momenta through the ratio \( \zeta = \frac{p_o}{p} \), we understand how energy is conserved and redistributed within the system.
Angular Momentum
Angular momentum is a fundamental concept in physics that is conserved in systems with rotational symmetry. It is akin to the conservation of linear momentum but for rotating systems. In a spherical well potential, the conservation of angular momentum helps us determine the projectile's closest distance of approach, denoted as \( d \).

Angular momentum \( L \) for a particle is calculated as \( L = p_o b \), where \( b \) is the impact parameter. As the projectile enters the well, conservation of angular momentum states \( p_o b = p d \). This equality helps us solve for the closest distance \( d \) using the momentum inside the well \( p \), leading to the relationship \( d = \zeta b \).

This equation showcases that the distance of closest approach depends on both the initial conditions of the projectile's path and how its momentum changes due to the spherical potential well.
Scattering Angle
The scattering angle is a measure of how much a projectile's path is deflected due to interaction with a potential, such as the spherical well. To derive this angle geometrically, we refer to the angles the projectile's path makes as it travels through and exits the well.

The angle \( \theta \) can be calculated using the formula \( \theta = 2(\arcsin \frac{b}{R} - \arcsin \frac{\zeta b}{R}) \), where \( b \) is the impact parameter and \( R \) is the radius of the spherical well boundary. Here, \( \zeta \) represents the momentum ratio, affecting the dispersion of the projectile.

Plotting \( \theta \) against \( b \), especially for a given \( \zeta \, (e.g., \, 0.5) \), shows us how the scattering angle varies with an impact parameter. This plot provides insights into how the well influences projectile deflection, providing a direct view of the geometrical impact of the potential on the particle's trajectory.
Cross Section Analysis
Cross section analysis in scattering theory is crucial for understanding how likely a projectile will scatter at a particular angle. It describes the spatial distribution of scattered particles, given by the function \( \frac{d\sigma}{d\Omega} \), known as the differential cross section.

To find this, differentiate the scattering angle \( \theta(b) \) with respect to \( b \) and apply transformations, arriving at \( \frac{d\sigma}{d\Omega} = \frac{b}{\sin\theta} \left| \frac{d\theta}{db} \right|^{-1} \). This function is essential because it provides a detailed picture of how the directionality of scattering varies as a function of \( \theta \).

By plotting \( \frac{d\sigma}{d\Omega} \) against \( \theta \) parametrically, one can intuitively visualize how scattering behavior changes with different impact parameters. Integrating this differential quantity over all solid angles yields the total cross section \( \sigma \), summing up the entire scattering interaction dynamics facilitated by the spherical well potential. Such analysis helps explain variations in scattering efficiency depending on incoming conditions.

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Most popular questions from this chapter

Calculate the solid angles subtended by the moon and by the sun, both as seen from the earth. Comment on your answers. (The radii of the moon and sun are \(R_{\mathrm{m}}=1.74 \times 10^{6} \mathrm{m}\) and \(R_{\mathrm{s}}=6.96 \times 10^{8}\) m. Their distances from earth are \(\left.d_{\mathrm{m}}=3.84 \times 10^{8} \mathrm{m} \text { and } d_{\mathrm{s}}=1.50 \times 10^{11} \mathrm{m} .\right)\)

The differential cross section for scattering 6.5-MeV alpha particles at 120^ off a silver nucleus is about 0.5 barns/sr. If a total of \(10^{10}\) alphas impinge on a silver foil of thickness \(1 \mu \mathrm{m}\) and if we detect the scattered particles using a counter of area \(0.1 \mathrm{mm}^{2}\) at \(120^{\circ}\) and \(1 \mathrm{cm}\) from the target, about how many scattered alphas should we expect to count? (Silver has a specific gravity of \(10.5,\) and atomic mass of 108.)

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)?

(a) A certain nucleus has radius \(5 \mathrm{fm}\). ( \(1 \mathrm{fm}=10^{-15} \mathrm{m}\).) Find its cross section \(\sigma\) in barns. (1 barn \(=10^{-28} \mathrm{m}^{2} .\) ) (b) Do the same for an atom of radius \(0.1 \mathrm{nm}\). \(\left(1 \mathrm{nm}=10^{-9} \mathrm{m} .\right)\)

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.)

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