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A photon scatters in the backward direction (\(\phi = 180^\circ\)) from a free proton that is initially at rest. What must the wavelength of the incident photon be if it is to undergo a 10.0\(\%\) change in wavelength as a result of the scattering?

Short Answer

Expert verified
The initial wavelength is approximately \(1.32 \times 10^{-14} \text{ m}\).

Step by step solution

01

Understand the Problem

We need to calculate the initial wavelength of a photon that undergoes Compton scattering, resulting in a 10% increase in wavelength when scattering backward from a proton at rest.
02

Compton Wavelength Change Formula

The change in wavelength \(\Delta \lambda\) for Compton scattering is given by the formula:\[\Delta \lambda = \frac{h}{{m_pc}}(1 - \cos \phi)\]where \(h\) is Planck's constant, \(m_p\) is the proton mass, \(c\) is the speed of light, and \(\phi\) is the scattering angle.
03

Substitute Known Values

Since the photon scatters backward, \(\phi = 180^\circ\), hence \(\cos 180^\circ = -1\). Substitute this into the equation:\[\Delta \lambda = \frac{h}{{m_pc}}(1 - (-1)) = \frac{2h}{{m_pc}}\]
04

Express Wavelength Change as Percentage

We know that the change in wavelength is 10% of the initial wavelength, \(\lambda\). So, \(\Delta \lambda = 0.1 \lambda\). Thus,\[0.1 \lambda = \frac{2h}{{m_pc}}\]
05

Solve for Initial Wavelength

Rearrange the equation to solve for \(\lambda\):\[\lambda = \frac{20h}{{m_pc}}\]
06

Substitute Constants and Compute

Substitute numerical values \(h = 6.626 \times 10^{-34} \text{J} \cdot \text{s}\), \(m_p = 1.672 \times 10^{-27} \text{kg}\), and \(c = 3.00 \times 10^8 \text{m/s}\) into the equation and calculate \(\lambda\):\[\lambda = \frac{20(6.626 \times 10^{-34})}{{1.672 \times 10^{-27} \times 3.00 \times 10^8}} \approx 1.32 \times 10^{-14} \text{m}\]

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Key Concepts

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

Photon Wavelength
The wavelength of a photon is a crucial factor in the phenomenon of Compton scattering. In this process, a photon collides with a particle, like a proton, leading to a change in its wavelength. This change is a direct result of the transfer of energy and momentum between the photon and the particle it interacts with.

A photon's initial wavelength before scattering is denoted as \( \lambda \), and after the scattering event, the change in wavelength is represented by \( \Delta \lambda \). The incident wavelength can be crucial in determining properties such as energy and frequency of a photon. In our exercise, understanding how much the wavelength changes during scattering helps us find the initial wavelength. Knowing that the wavelength increases by 10% aids in solving for the original wavelength using the Compton wavelength change formula.

Overall, the wavelength is an essential parameter that connects directly with the energy level of the photon and is a telling factor in interactions involving photons.
Scattering Angle
The scattering angle, denoted as \( \phi \), plays a significant role in Compton scattering. This angle determines the direction in which the photon moves after interacting with a particle. It is a pivotal parameter in calculating the change in the photon's properties, especially the change in wavelength.

When a photon is scattered at an angle \( \phi \), the greater the angle (up to 180 degrees), the more significant the change in wavelength. In our exercise, the scattering angle is specified to be \( 180^\circ \), implying a full reversal of direction in the scattered photon. This specific angle leads to the maximum possible wavelength change in Compton scattering because \( \cos 180^\circ = -1 \). As such, the equation for wavelength change, \( \Delta \lambda = \frac{h}{{m_pc}}(1 - \cos \phi) \), shows a doubling of Planck's constant over the particle's rest mass and light speed product, resulting in maximum wavelength elongation.

Understanding the concept of scattering angles is crucial in physics problems because it influences the calculation and outcome of photon interactions, directly affecting the energy and direction of scattered photons.
Planck's Constant
Planck's constant, denoted by \( h \), is a fundamental constant in physics with the value \( 6.626 \times 10^{-34} \text{J} \cdot \text{s} \). It is a cornerstone of quantum mechanics, relating to the quantization of various physical properties. It appears prominently in Planck's law, Einstein's photoelectric equation, and, relevant to our topic, the Compton scattering formula.

In Compton scattering, Planck's constant helps determine the amount of wavelength change a photon experiences. The equation \( \Delta \lambda = \frac{h}{{m_pc}}(1 - \cos \phi) \) encapsulates the relationship between the scattering angle, the mass of the scattering particle, and the speed of light. Planck's constant is pivotal, demonstrating how tiny units of energy, or quanta, lead to measurable changes in a particle's wavelength when scattering occurs.

The presence of Planck's constant in these calculations emphasizes the deep connections between wave and particle characteristics, forming the basis of quantum mechanical principles that govern behavior at microscopic scales.

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Most popular questions from this chapter

A photon of green light has a wavelength of 520 nm. Find the photon's frequency, magnitude of momentum, and energy. Express the energy in both joules and electron volts.

In developing night-vision equipment, you need to measure the work function for a metal surface, so you perform a photoelectric-effect experiment. You measure the stopping potential \(V_0\) as a function of the wavelength \(\lambda\) of the light that is incident on the surface. You get the results in the table. In your analysis, you use \(c\) = 2.998 \(\times\) 10\(^8\) m/s and \(e\) = 1.602 \(\times\) 10\(^{-19}\) C, which are values obtained in other experiments. (a) Select a way to plot your results so that the data points fall close to a straight line. Using that plot, find the slope and y-intercept of the best-fit straight line to the data. (b) Use the results of part (a) to calculate Planck's constant \(h\) (as a test of your data) and the work function (in eV) of the surface. (c) What is the longest wavelength of light that will produce photoelectrons from this surface? (d) What wavelength of light is required to produce photoelectrons with kinetic energy 10.0 eV?

X rays with initial wavelength 0.0665 nm undergo Compton scattering. What is the longest wavelength found in the scattered x rays? At which scattering angle is this wavelength observed

Nuclear fusion reactions at the center of the sun produce gamma-ray photons with energies of about 1 MeV (10\(^6\) eV). By contrast, what we see emanating from the sun's surface are visiblelight photons with wavelengths of about 500 nm. A simple model that explains this difference in wavelength is that a photon undergoes Compton scattering many times-in fact, about 10\(^{26}\) times, as suggested by models of the solar interior-as it travels from the center of the sun to its surface. (a) Estimate the increase in wavelength of a photon in an average Compton-scattering event. (b) Find the angle in degrees through which the photon is scattered in the scattering event described in part (a). (\(Hint\): A useful approximation is cos \(\phi \approx 1 - \phi^2 /2\), which is valid for \(\phi \ll\) 1. Note that \(\phi\) is in radians in this expression.) (c) It is estimated that a photon takes about 10\(^6\) years to travel from the core to the surface of the sun. Find the average distance that light can travel within the interior of the sun without being scattered. (This distance is roughly equivalent to how far you could see if you were inside the sun and could survive the extreme temperatures there. As your answer shows, the interior of the sun is \(very\) opaque.)

(a) What is the minimum potential difference between the filament and the target of an x-ray tube if the tube is to produce x rays with a wavelength of 0.150 nm? (b) What is the shortest wavelength produced in an x-ray tube operated at 30.0 kV?

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