/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 A 200 -MeV photon is scattered a... [FREE SOLUTION] | 91Ó°ÊÓ

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A 200 -MeV photon is scattered at \(40.0^{\circ}\) by a free proton that is initially at rest. Find the energy (in MeV) of the scattered photon.

Short Answer

Expert verified
The energy of the scattered photon is \(180.6\) MeV.

Step by step solution

01

Identify all the variables

The initial energy of the photon \(E_{{\text{{i}}}}\) = 200 MeV, the scattering angle \(\theta = 40^{\circ}\), and the proton mass \(m_p = 938\) MeV/c².
02

Convert the angle from degrees to radians

This conversion is made because the trigonometric functions in the formula work with the radian measurement. The conversion formula is radians = degrees × \(\pi / 180\). Calculating this gives: \(\theta = 40^{\circ} * \pi / 180 = 0.698\) rad.
03

Plug all the values into the Compton scattering formula

Substituting these values into the formula gives \(E_f = \frac{{200\, \text{{MeV}}}}{{1 + \frac{{200\, \text{{MeV}}}}{{938\, \text{{MeV/c}}^{2}}}(1 - \cos(0.698))}}\).
04

Compute \(\cos(\theta)\) and simplify the equation

The cosine of 0.698 rad is 0.766. Substituting this value and simplifying the formula gives: \(E_f = \frac{{200\, \text{{MeV}}}}{{1 + \frac{{200\, \text{{MeV}}}}{{938\, \text{{MeV/c}}^{2}}}(1 - 0.766)}}\). Further simplification gives: \(E_f = 180.6\) MeV

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Key Concepts

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

Photon Energy Calculation
Understanding how to calculate the energy of a photon is fundamental in physics, especially when delving into topics like Compton scattering. The energy of a photon can be determined using the formula:
\[ E = h u \]
where \(h\) is Planck's constant, and \(u\) (nu) is the frequency of the photon. In the Compton effect, which involves the scattering of a photon by a particle such as an electron or proton, the energy and momentum of the photon are conserved. The change in energy of the photon is called the 'Compton Shift' and is observed as a shift in the wavelength of the scattered photon.

This shift occurs due to the transfer of energy to the particle, which often ends up being a less energetic, lower frequency photon. When the energy of a photon is given in electron-volts (eV) or mega-electron-volts (MeV), as commonly seen in particle physics, it represents the photon's ability to interact with matter. Calculating photon energy becomes a crucial step in understanding phenomena like the one described in the exercise, where a 200 MeV photon collides with a proton.
Compton Scattering Formula
Compton scattering is a quantum mechanical phenomenon where a photon scatters off a target particle, such as an electron or a proton. The formula that describes this scattering event is central to understanding how the energy of a photon changes as a result of the interaction. The formula is derived from principles of conservation of energy and momentum and is given by:
\[ E_f = \frac{E_i}{1 + \frac{E_i}{m_pc^2}(1 - \cos \theta)} \]
where \(E_f\) is the energy of the scattered photon, \(E_i\) is the initial energy of the incoming photon, \(m_p\) is the mass of the target particle, \(c\) is the speed of light, and \(\theta\) is the scattering angle in radians.

This formula allows us to calculate the scattered photon's energy after the interaction. As you can see in the exercise, the scattering angle's cosine plays a significant role in determining how much energy is lost by the photon. By plugging in the known values such as the initial photon energy, particle mass, and the scattering angle, one can compute the final energy of the photon using the Compton scattering formula.
Special Relativity
The theory of special relativity, formulated by Albert Einstein, revolutionized the way we think about space, time, and energy. Two of its most famous aspects are the concepts that the speed of light in a vacuum is the same for all observers, no matter their relative motion, and the equivalence of mass and energy, represented by the iconic equation
\[ E = mc^2 \]
In the context of Compton scattering, special relativity plays a crucial role. Photons, as packets of electromagnetic energy, are directly affected by relativistic principles. One aspect of special relativity particularly relevant to Compton scattering is the relativistic energy-momentum relationship, which ensures conservation laws hold true even at high velocities close to the speed of light.

This relationship is particularly significant when dealing with particles such as electrons and protons in high-energy photon interactions, as their relativistic mass changes due to the energy transfer from the photon. In the practice of these calculations, mass is often converted into energy units, as seen with the proton's mass expressed in MeV/c² in the textbook example, so it can be easily used within the formulas derived from relativistic principles.

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Most popular questions from this chapter

Calculate the minimum-wavelength \(\mathrm{x}\) -ray that can be produced when a target is struck by an electron that has been accelerated through a potential difference of (a) \(15.0 \mathrm{kV}\) and (b) \(1.00 \times 10^{2} \mathrm{kV}\). (c) What happens to the minimum wavelength as the potential difference increases?

X-rays of wavelength \(0.140 \mathrm{~nm}\) are reflected from a certain crystal, and the first-order maximum occurs at an angle of \(14.4^{\circ} .\) What value does this give for the interplanar spacing of the crystal?

In the ground state of hydrogen, the uncertainty in the position of the electron is roughly \(0.10 \mathrm{~nm}\). If the speed of the electron is approximately the same as the uncertainty in its speed, about how fast is it moving?

A particle of mass \(m\) and charge \(q\) is accelerated from rest through a potential difference \(\Delta V\). (a) Use conservation of energy to find a symbolic expression for the momentum of the particle in terms of \(m, q\), and \(\Delta V\). Assume the particle's speed isn't relativistic. (b) Write a symbolic expression for the de Broglie wavelength using the result of part (a). (c) If an electron and proton go through the same potential difference but in opposite directions, which particle will have the shorter wavelength?

X-rays with an energy of \(300 \mathrm{keV}\) undergo Compton scattering from a target. If the scattered rays are deflected at \(37.0^{\circ}\) relative to the direction of the incident rays, find (a) the Compton shift at this angle, (b) the energy of the scattered x-ray, and (c) the kinetic energy of the recoiling electron.

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