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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.)

Short Answer

Expert verified
The law of reflection (\( \theta_i = \theta_r \)) follows from conservation of energy and angular momentum in hard-sphere scattering.

Step by step solution

01

Define the System

Consider a projectile particle with initial velocity \( v_i \) approaching a hard sphere with radius \( R \). This particle is scattered off the sphere's surface, with an incidence angle \( \theta_i \) and a reflection angle \( \theta_r \). Our goal is to prove that \( \theta_i = \theta_r \).
02

Use Conservation of Energy

Initially, the kinetic energy of the particle is given by \( KE_i = \frac{1}{2}mv_i^2 \), where \( m \) is the mass of the particle. For a hard-sphere scattering, the kinetic energy after the collision (\( KE_f \)) remains the same, so \[ KE_i = KE_f = \frac{1}{2}mv_f^2 \].This implies that the speed of the particle does not change during scattering.
03

Utilize Conservation of Angular Momentum

If the sphere is centered at the origin and the particle's initial position is \( (x_i, y_i) \), then the initial angular momentum \( L_i \) relative to the center of the sphere is given by \[ L_i = m(v_{i,x}y - v_{i,y}x) \].After reflection, the angular momentum \( L_f \) is \[ L_f = m(v_{f,x}y - v_{f,y}x) \],and since angular momentum is conserved, \[ L_i = L_f. \]
04

Analyze Angular Momentum and Energy Relations

The equality \( L_i = L_f \) implies that the magnitudes of the components of velocity perpendicular to the radius vector \( \mathbf{r} \) at the point of contact remain unchanged. Since the speed remains the same (from conservation of energy), the direction must remain symmetric about the normal to the surface. This symmetry in directional change results in equal angles of incidence and reflection. Thus, \( \theta_i = \theta_r \).
05

Conclude with the Law of Reflection

By conserving energy and angular momentum, we confirm that the direction of the particle changes symmetrically about the normal at the surface contact point. This symmetry derives the law of reflection where \( \theta_i = \theta_r \).

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Key Concepts

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

Conservation of Energy
When we talk about the conservation of energy, it means that the energy in a closed system remains constant over time. For hard sphere scattering, this concept plays a crucial role. Before hitting the sphere, a particle has kinetic energy given by \( KE_i = \frac{1}{2}mv_i^2 \), where \( m \) is the particle's mass and \( v_i \) its velocity.
After the particle bounces off the hard sphere, its kinetic energy does not change. This unchanging kinetic energy read as \( KE_f = \frac{1}{2}mv_f^2 \) implies that the speed of the particle before and after the collision is the same.
Conservation of energy ensures that the scattering does not absorb or dissipate kinetic energy. This aspect of hard sphere scattering means focusing on velocity changes purely in direction, not in magnitude.
Conservation of Angular Momentum
Angular momentum refers to the rotation of an object around a point. In our scenario, it concerns the motion of the particle around the hard sphere. Angular momentum is calculated using \( L = mv_x y - mv_y x \), where \( x \) and \( y \) are the coordinates of the particle's position relative to the sphere's center.
  • The principle of conservation of angular momentum tells us that \( L_i = L_f \).
    This equality suggests that the particle's path curvature around the sphere does not change the perpendicular speed components.
  • Since the force during scattering is symmetric (equally pulling from all directions), the particle's angular momentum stays constant—another affirmation that only the direction changes, not the speed.
This conservation is pivotal for proving that the law of reflection holds: the incidence and reflection angles remain equal.
Law of Reflection
The law of reflection states that the angle at which a particle, or any wave, hits a surface equals the angle at which it bounces away. For hard spheres, the incidence angle \( \theta_i \) matches the reflection angle \( \theta_r \).
  • This law often applies to physical optics and general physics, ensuring that scenes in mirrors look realistic.
  • The surface normal, a line at 90 degrees to the sphere's touching point during collision, is the symmetry axis.
  • The conservation laws of energy and angular momentum guarantee that the change in direction after bouncing is symmetrical.
Therefore, the particle's path after striking the sphere maintains this symmetry, confirming \( \theta_i = \theta_r \) and respecting the law of reflection, core to hard sphere scattering.

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Most popular questions from this chapter

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.

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 \(0.1 \mathrm{nm}\). \(\left(1 \mathrm{nm}=10^{-9} \mathrm{m} .\right)\)

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.)

In their famous experiment, Rutherford's assistants, Geiger and Marsden, detected the scattered alpha particles using a zinc sulphide screen, which produced a tiny flash of light when struck by an alpha particle. If their screen had area \(1 \mathrm{mm}^{2}\) and was \(1 \mathrm{cm}\) from the target, what solid angle did it subtend?

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