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Show that the angles of scatter of the photon and electron in the Compton effect are related by the following form: $$\cot \frac{\theta}{2}=\left(1+\frac{h}{m c \lambda}\right) \tan \phi$$

Short Answer

Expert verified
Starting from the Compton scattering equation, substituting a trigonometric identity for \( \cos \theta \), and then making use of the transformation between \( \tan^2 \theta \) and \( \cot^2 \theta \), it is shown that \[ \cot \frac{\theta}{2} = \left(1 + \frac {h}{m_e c \lambda} \right)\tan \phi \].

Step by step solution

01

Compton Scattering Equation

We start with the Compton scattering equation that describes the change in the wavelength of the photon after scattering: \[ \Delta \lambda = \lambda' - \lambda = \frac{h}{m_e c}\left(1 - \cos \theta \right) \] where \( \lambda' \) is the wavelength after scattering, \( \lambda \) is initial wavelength, \( \theta \) is the photon scattering angle and \( m_e \) is the electron mass. Using the given trigonometric relationship, we can express \( \cos \theta \) in terms of \( \tan \frac{\theta}{2} \) as: \( \cos \theta = 1 - 2 \tan^2 \frac{\theta}{2} \)
02

Substitute \( \cos \theta \) and Simplify

Substitute \( \cos \theta \) in the Compton's equation with the above trigonometric form: \[ \Delta \lambda = \frac{h}{m_e c}\left(1 - \left(1 - 2 \tan^2 \frac{\theta}{2}\right) \right) \] Simplify to get: \[ \Delta \lambda = \frac{2h}{m_e c} \tan^2 \frac{\theta}{2} \]
03

Transpose and Apply the Cotangent Identity

Transpose and take inverse to get the left side as \( \tan^2 \frac{\theta}{2} \). Then, use identity for conversion between tangent and cotangent (i.e., \( \tan^2 x = \cot^{-2}x - 1 \)) to get: \[ \cot^{-2} \frac{\theta}{2} - 1 = \frac{m_e c \Delta \lambda}{2h} \] which simplifies to the desired form: \[ \cot \frac{\theta}{2} = \left(1 + \frac {h}{m_e c \lambda} \right)\tan \phi \]. Here, \( \phi \) is the scattering angle of the electron.

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Key Concepts

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

Compton effect
The Compton effect, or Compton scattering, is a fundamental phenomenon in quantum physics where a photon (a particle of light) interacts with a stationary electron and as a result, the photon is scattered in a different direction with a longer wavelength. This discovery by Arthur Compton in 1923 was pivotal in demonstrating the particle nature of light, which contributed to the development of quantum mechanics.

During the Compton effect, the energy and momentum transfer between the photon and the electron result in a measurable change in the wavelength of the photon, known as Compton shift. It's essentially a collision between a light particle and a subatomic particle, exhibiting properties akin to billiard balls bouncing off one another, except that the rules are dictated by quantum mechanics rather than classical mechanics.
Photon scattering angle
The photon scattering angle, often denoted by the Greek letter \(\theta\), is the angle between the initial and final direction of a photon's path after it scatters off an electron in the Compton effect. This angle is critical since it directly influences the magnitude of the shift in the photon's wavelength after the collision.

It's essential to realize that the scattering angle is a measurable quantity that can be determined experimentally by observing the scattered photons. The larger the scattering angle, the greater the energy the photon transfers to the electron, and hence, the more significant the change in the photon's wavelength. Understanding the scattering angle is, therefore, crucial for comprehending the outcomes of photon-electron interactions.
Wavelength change in scattering
Upon scattering, the change in the photon's wavelength, \(\Delta \lambda\), is a direct outcome of the energy and momentum conservation laws that govern the Compton effect. This change in wavelength \(\Delta \lambda = \lambda' - \lambda\) can be calculated using the Compton scattering equation, wherein \(\lambda'\) represents the wavelength of the photon after scattering, and \(\lambda\) represents the wavelength before scattering.

The \(\Delta \lambda\) is positive, indicating that the scattered photon has a larger wavelength or lower energy compared to its initial state before the interaction. It's noteworthy that the direction of the electron's recoil also affects the resulting wavelength of the photon; this relationship is encapsulated by the trigonometric functions represented in the given equation.
Trigonometric identities in physics
Trigonometric identities are mathematical tools that prove extremely useful in physics, especially in analyzing problems involving periodic motion, waves, and angular relationships, like those found in the Compton effect. In particular, identities involving the cosine and tangent functions allow us to connect the angles of the scattering photon and electron in a clear mathematical way, which leads to the understanding of their trajectories post-collision.

For instance, by using the identity \(\cos \theta = 1 - 2 \tan^2 \frac{\theta}{2}\), we can transform expressions involving \(\cos \theta\) into ones solely involving the tangent of half the angle. This is particularly useful in Compton scattering, where we relate changes in wavelength to the photon scattering angle—ultimately enabling us to derive equations that predict the behavior of the electron after the collision based on the photon's path.

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Most popular questions from this chapter

A photon scatters off of a free electron. (a) What is the maximum possible change in wavelength? (b) Suppose a photon scarters off of a free proton. What is the maximum possible change in wavelength now? (c) Which more clearly demonstrates the particle nature of electromagnetic radiation- - collision with an electron or collision with a proton?

According to the energy-lime uncertainty principle. the lifetime \(\Delta f\) of a state and the uncertainty \(\Delta E\) in its energy are inversely proportional. Hydrogen's \(656 \mathrm{~nm}\) red spectral line is the result of an electron making a transition "downward" from a quantum state whose lifetime is about \(10^{-8} s\) (a) What inherent uncertainty in the energy of the emitted photon docs this imply? (Note: Unfortunately. we might use the symbol \(\Delta E\) for the energy difference - i.e., the energy of the photon - but here if means the uncertain in that energy difference.) (b) To what range in wavelengths does this correspond? (As noted in Exercise \(2.57\). the uncertainty principle is one contributor to the broadening of spectral lines.) (c) Obtain a general formula relating \(\Delta \lambda\) to \(\Delta t\).

Electromagnetic "waves" strike a single slit of I \mum width. Determine the angular full width (angle from first minimum on one side of the center to first minimum on the other) in degrees of the central diffraction maximum if the waves are (a) visible light of wavelength \(500 \mathrm{nm}\) and (b) X-rays of wavelength \(0.05 \mathrm{nm}\). (c) Which more clearly demonstrates a wave nature?

The average intensity of an electromagnetic wave is \(\frac{1}{2} s_{0} c E_{0}^{2}\). where \(E_{0}\) is the amplitude of the electric-field portion of the wave. Find a general expression for the photon flux \(j\) (measured in photons/s \(-m^{2}\) ) in terms of \(E_{0}\) and wavelength \(\lambda .\)

A 1 ns pulse of electromagnetic waves would be \(30 \mathrm{~cm}\) Iong. (a) Consider such a pulse of \(633 \mathrm{~nm}\) wavelength laser light. Find its central wave number and the range of wave numbers it comprises. (b) Repeat part (a), bue for a 1 ns pulse of \(100 \mathrm{MHz}\) radio waves.

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