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An electron accelerates through a potential difference of \(50.0 \mathrm{kV}\) in an \(\mathrm{x}\) -ray tube. When the electron strikes the target, \(70.0 \%\) of its kinetic energy is imparted to a single photon. Find the photon's frequency, wavelength, and magnitude of momentum.

Short Answer

Expert verified
The photon's frequency is approximately \( 8.44 \times 10^{18} \) Hz, its wavelength is about \( 3.55 \times 10^{-11} \) m, and the magnitude of its momentum is roughly \( 1.867 \times 10^{-23} \) kgm/s.

Step by step solution

01

Calculate Initial Kinetic Energy

The initial kinetic energy (KE) of the electron can be calculated using the formula KE = eV, where e is the charge of the electron (\(1.6 \times 10^{-19}\) C) and V is the potential difference provided in the problem (\(50.0 \times 10^3\) V). This yields a kinetic energy of \( KE = 1.6 \times 10^{-19} \times 50.0 \times 10^3 = 8.0 \times 10^{-15} \) Joules.
02

Calculate the Photon's Energy

When the electron strikes the target, 70% of its kinetic energy is imparted to a single photon. Therefore, we can calculate the energy of the photon E using the formula: \(E = 0.70 \times KE = 0.70 \times 8.0 \times 10^{-15} = 5.6 \times 10^{-15}\) Joules.
03

Compute the Photon's Frequency

The photon's frequency can be calculated using Planck's equation \(E = hf\), where h is Planck's constant \((6.63 \times 10^{-34}\) Js), E is the energy of the photon and f is the frequency. Rearranging the Planck's equation, we get \( f = \frac{E}{h} = \frac{5.6 \times 10^{-15}}{6.63 \times 10^{-34}} = 8.44 \times 10^{18}\) Hz.
04

Calculate the Wavelength of the Photon

The wavelength (\(\lambda\)) of the photon can be found using the relationship of wave speed \(c = f \lambda\), where c is the speed of light \(3.0 \times 10^8\) ms\(^{-1}\). Rearranging the equation for \(\lambda\), we get \( \lambda = \frac{c}{f} = \frac{3.0 \times 10^8}{8.44 \times 10^{18}} = 3.55 \times 10^{-11}\) m.
05

Determine the Momentum of the Photon

The magnitude of the photon's momentum can be computed with the equation \( p = \frac{E}{c} = \frac{5.6 \times 10^{-15}}{3.0 \times 10^8} = 1.867 \times 10^{-23} \) kgm/s.

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Key Concepts

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

Planck's equation
Imagine an invisible particle carrying a tiny packet of light energy zooming through space; that's a photon, and the amount of energy it holds can be understood using Planck's equation. This fascinating equation, named after physicist Max Planck, is a fundamental piece in the puzzle of quantum mechanics. It's written as \( E = hf \), where \( E \) represents the energy of the photon, \( h \) is Planck's constant (a fundamental constant in physics with a value of \( 6.63 \times 10^{-34} \) joule-second), and \( f \) stands for the frequency of the photon.

Using Planck's equation, we can calculate the energy of the photons produced in various scenarios, such as an electron striking a target in an X-ray tube, like in our original exercise. Importantly, this energy is directly proportional to frequency: higher frequency means higher energy. This is crucial for understanding not just how objects at the atomic and subatomic level behave, but also for applications ranging from medical imaging to the study of the stars.
Electron kinetic energy
Let's now zoom into an electron speeding up through an electric field. The energy that this electron gains as it speeds up is what we call its kinetic energy. But how do we calculate this energy for electrons? It's quite simple: when an electron accelerates through a potential difference, its kinetic energy increases by an amount equal to the charge of the electron multiplied by the potential difference (voltage). We denote this by the equation \( KE = eV \), with \( KE \) being the kinetic energy, \( e \) the charge of the electron (a constant \(1.6 \times 10^{-19}\) Coulombs), and \( V \) the potential difference in volts.

In our exercise, the electron's kinetic energy is transformed when it strikes a target, and a significant portion of that energy becomes a photon's energy. Mastering the concept of kinetic energy is essential for anyone wanting to delve into the behavior of particles in fields, be it electric or magnetic, and is a cornerstone of physics that leads into various advanced topics, including the deeper quantum realm.
Wave-particle duality
If you've ever been amazed by the concept that light behaves both as a wave and as a particle, then you're already introduced to the wave-particle duality. This duality is a fundamental principle of quantum mechanics, showing us that particles like photons and electrons exhibit properties of both waves and particles. It can be a bit mind-bending, but to put it into perspective:

When we consider the wave nature, we talk about wavelengths and frequencies. When we consider the particle nature, we think about individual quanta of energy, like photons in our exercise. Photons, for instance, carry energy as described by Planck's equation and exhibit momentum, reinforcing their particle characteristics. But they also have a wavelength, as shown in the exercise, highlighting their wave-like nature.

Understanding wave-particle duality allows us to grasp why certain phenomena, like interference patterns and diffraction, occur. It is a cornerstone of modern physics, explaining everything from the radiation of black bodies to the uncertainty principle, and it deeply influences our understanding of the very fabric of reality.

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Most popular questions from this chapter

Higher-energy photons might be desirable for the treatment of certain tumors. Which of these actions would generate higher-energy photons in this linear accelerator? (a) Increasing the number of electrons that hit the tungsten target; (b) accelerating the electrons through a higher potential difference; (c) both (a) and (b); (d) none of these.

A photon with wavelength \(0.1100 \mathrm{nm}\) collides with a free electron that is initially at rest. After the collision the wavelength is \(0.1132 \mathrm{nm}\). (a) What is the kinetic energy of the electron after the collision? What is its speed? (b) If the electron is suddenly stopped (for example, in a solid target), all of its kinetic energy is used to create a photon. What is the wavelength of this photon?

When ultraviolet light with a wavelength of \(400.0 \mathrm{nm}\) falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be \(1.10 \mathrm{eV}\). What is the maximum kinetic energy of the photoelectrons when light of wavelength \(300.0 \mathrm{nm}\) falls on the same surface?

A photon with wavelength \(\lambda\) is incident on a stationary particle with mass \(M,\) as shown in Fig. \(\mathbf{P 3 8 . 3 9 .}\) The photon is annihilated while an electron-positron pair is produced. The target particle moves off in the original direction of the photon with speed \(V_{M}\). The electron travels with speed \(v\) at angle \(\phi\) with respect to that direction. Owing to momentum conservation, the positron has the same speed as the electron. (a) Using the notation \(\gamma=\left(1-v^{2} / c^{2}\right)^{-1 / 2}\) and \(\gamma_{M}=\left(1-V_{M}^{2} / c^{2}\right)^{-1 / 2},\) write a relativistic expression for energy conservation in this process. Use the symbol \(m\) for the electron mass. (b) Write an analogous expression for momentum conservation. (c) Eliminate the ratio \(h / \lambda\) between your energy and momentum equations to derive a relationship between \(v / c, V_{M} / c,\) and \(\phi\) in terms of the ratio \(m / M\), keeping \(\gamma\) and \(\gamma_{M}\) as a useful shorthand notation. (d) Consider the case where \(V_{M} \ll c\), and use a binomial expansion to derive an expression for \(\gamma_{M}\) to the first order in \(V_{M}\). Use that result to rewrite your previous result as an expression for \(V_{M}\) in terms of \(c, v, m / M,\) and \(\phi .\) (e) Are there choices of \(v\) and \(\phi\) for which \(V_{M}=0 ?\) (f) Suppose the target particle is a proton. If the electron and positron remain stationary, so that \(v=0\), then with what speed does the proton move, in \(\mathrm{km} / \mathrm{s} ?\) (g) If the electron and positron each have total energy \(5.00 \mathrm{MeV}\) and move with \(\phi=60^{\circ}\), then what is the speed of the proton? (Hint: First solve for \(\gamma\) and \(v .)\) (h) What is the energy of the incident photon in this case?

A horizontal beam of laser light of wavelength \(585 \mathrm{nm}\) passes through a narrow slit that has width \(0.0620 \mathrm{~mm}\). The intensity of the light is measured on a vertical screen that is \(2.00 \mathrm{~m}\) from the slit. (a) What is the minimum uncertainty in the vertical component of the momentum of each photon in the beam after the photon has passed through the slit? (b) Use the result of part (a) to estimate the width of the central diffraction maximum that is observed on the screen.

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