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In a Young's double-slit experiment that uses electrons, the angle that locates the first-order bright fringes is \(\theta_{\mathrm{A}}=1.6 \times 10^{-4}\) degrees when the magnitude of the electron momentum is \(p_{\mathrm{A}}=1.2 \times 10^{-22} \mathrm{kg} \cdot \mathrm{m} / \mathrm{s} .\) With the same double slit, what momentum magnitude \(p_{\mathrm{B}}\) is necessary so that an angle of \(\theta_{\mathrm{B}}=4.0 \times 10^{-4}\) degrees locates the first-order bright fringes?

Short Answer

Expert verified
The required momentum magnitude \(p_{B}\) is \(4.8 \times 10^{-23} \text{ kg} \cdot \text{m/s}\).

Step by step solution

01

Understand the relationships

In the double-slit experiment, we can use the formula for the angle of interference fringes: \(\sin \theta = \frac{m\lambda}{d}\), where \(\lambda\) is the wavelength, \(\theta\) is the angle of the fringe, \(m\) is the fringe order, and \(d\) is the distance between the slits. Since \(\theta\) is very small, \(\sin\theta \approx \theta\) in radians. The relationship between wavelength and momentum is \(\lambda = \frac{h}{p}\), where \(h\) is Planck's constant.
02

Convert angles to radians

Convert \(\theta_{A}\) and \(\theta_{B}\) from degrees to radians. Use the conversion: \(1\) degree = \(\frac{\pi}{180}\) radians. Thus, \(\theta_{A} = 1.6 \times 10^{-4} \times \frac{\pi}{180}\) radians and \(\theta_{B} = 4.0 \times 10^{-4} \times \frac{\pi}{180}\) radians.
03

Use the experimental setup information

For the first-order bright fringe (\(m=1\)), we establish that \(\theta_{A}\) corresponds to \(p_{A}\). Thus, \(p_{A} = \frac{h}{d\theta_{A}}\). Rearranging, \(d = \frac{h}{p_{A}\theta_{A}}\). We can equate this expression for the slit separation \(d\) to find \(p_{B}\) using \(p_{B} = \frac{h}{d\theta_{B}}\).
04

Express \(p_{B}\) in terms of known quantities

Substituting for \(d\) from earlier, \(p_{B} = \frac{h\theta_{A}}{p_{A}\theta_{B}}\). This expression allows us to directly solve for \(p_{B}\) using known quantities \(\theta_{A}\), \(\theta_{B}\), and \(p_{A}\).
05

Substitute and calculate \(p_{B}\)

Let's substitute the values: Planck's constant \(h = 6.626 \times 10^{-34} \text{ J} \cdot \text{s}\), \(\theta_{A} = 1.6 \times 10^{-4} \times \frac{\pi}{180}\), \(\theta_{B} = 4.0 \times 10^{-4} \times \frac{\pi}{180}\), and \(p_{A} = 1.2 \times 10^{-22} \text{ kg} \cdot \text{m/s}\). Solving, we find \(p_{B} = 4.8 \times 10^{-23} \text{ kg} \cdot \text{m/s}\).

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Key Concepts

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

Electron Wave-Particle Duality
In physics, the concept known as the electron wave-particle duality is fundamental to understanding how particles like electrons behave in experiments such as Young's double-slit experiment. This concept suggests that particles, which we traditionally think of as discrete, solid objects, can also exhibit properties of waves. It's a captivating dual nature.
  • When electrons pass through a double-slit, they create an interference pattern typical of waves, even though they are individual particles.
  • This duality can be puzzling but illustrates the quantum nature of matter.
In this experimental setup, electrons are not merely tiny balls flying through space. Instead, they demonstrate wavelike behavior, producing patterns that are influenced by their respective wavelengths. Understanding this aspect is crucial because it helps explain why we see interference fringes—bands of light and dark—when electrons pass through the slits. Their wave nature is at play, interfering constructively and destructively just like light waves would.
Momentum and Wavelength Relationship
The relationship between momentum and wavelength is central to many quantum experiments, including the double-slit experiment described in the exercise. In classical terms, momentum is simply mass times velocity. However, in quantum mechanics, any particle with momentum also has an associated wavelength, given by the de Broglie relationship.
  • The formula is \[ \lambda = \frac{h}{p} \]where \(\lambda\) is the wavelength, \(h\) is Planck's constant, and \(p\) is the momentum.
  • This equation shows that as momentum increases, the wavelength decreases.
In Young's experiment, adjusting the electron's momentum changes the observed interference pattern. When the electron has a small momentum, its wavelength is larger, affecting the spacing of these interference fringes. By carefully controlling the momentum, scientists can manipulate and study these fascinating patterns.
Interference Fringes
Interference fringes are an essential part of Young's double-slit experiment. These fringes occur due to the wave-like properties of particles such as electrons. When we shoot particles like electrons through two slits, they act as waves and create a pattern of light and dark bands, known as interference fringes.
  • ((Constructive interference)) occurs when the crests of two waves overlap, resulting in a bright fringe.
  • ((Destructive interference)) happens when a crest meets a trough, leading to a dark fringe.
The angle at which these fringes appear can be calculated with the formula \[ \sin \theta = \frac{m \lambda}{d} \]where \(\theta\) is the angle, \(m\) is the fringe order (1 for first-order), \(\lambda\) is the wavelength, and \(d\) is the distance between the slits. The formula shows that the angle depends on the wavelength, meaning by measuring these angles, we can infer the wavelength of the electrons, linking directly to their wave-particle nature.
Planck's Constant
Planck's constant \((h)\) is a crucial value in physics that relates the energy of photons to their frequency. It plays a key role in quantum mechanics and is integral to understanding phenomena like the electron momentum-wavelength relationship.
  • Planck's constant is approximately \[ 6.626 \times 10^{-34} ext{ J} imes ext{s} \]
  • In the context of the double-slit experiment, \(h\) is used in the de Broglie equation to relate the momentum of electrons to their wavelength.
By using Planck’s constant, we can bridge the gap between classical and quantum physics. In Young's double-slit experiment, it's the key figure that allows us to calculate the wavelength from momentum, thereby predicting the resulting interference pattern. Understanding \(h\) allows us to quantify and predict how particles behave at a quantum level, underlining the strange but beautiful nature of the quantum world.

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Most popular questions from this chapter

An electron and a proton have the same kinetic energy and are moving at speeds much less than the speed of light. Concepts: (i) How is the de Broglie wavelength \(\lambda\) related to the magnitude \(p\) of the momentum? (ii) How is the magnitude of the momentum related to the kinetic energy of a particle of mass \(m\) that is moving at a speed that is much less than the speed of light? (iii) Which has the greater de Broglie wavelength, the electron or the proton? Calculations: Determine the ratio of the de Broglie wavelength of the electron to that of the proton.

An AM radio station broadcasts an electromagnetic wave with a frequency of 665 kHz, whereas an FM station broadcasts an electromagnetic wave with a frequency of 91.9 MHz. How many AM photons are needed to have a total energy equal to that of one FM photon?

The minimum uncertainty \(\Delta y\) in the position \(y\) of a particle is equal to its de Broglie wavelength. Determine the minimum uncertainty in the speed of the particle, where this minimum uncertainty \(\Delta v_{y}\) is expressed as a percentage of the particle's speed \(v_{y}\) (Percentage \(=\frac{\Delta v_{y}}{v_{y}} \times 100 \%\)) Assume that relativistic effects can be ignored.

Sunlight, whose visible wavelengths range from 380 to \(750 \mathrm{nm}\), is incident on a sodium surface. The work function for sodium is \(W_{0}=2.28 \mathrm{eV}\). Concepts: (i) Will electrons with a greater value of \(\mathrm{KE}_{\text {max }}\) be emitted when the incident photons have a relatively greater or relatively smaller amount of energy? (ii) In the range of visible wavelengths, which wavelength corresponds to incident photons that carry the greatest energy? (iii) What is the smallest value of \(\mathrm{KE}_{\max }\) with which an electron can be ejected from the sodium? Calculations: Find the maximum kinetic energy \(\mathrm{KE}_{\max }\) (in joules) of the photoelectrons emitted from the surface and the range of wavelengths that will cause photoelectrons to be emitted.

An electron and a proton have the same speed. Ignore relativistic effects and determine the ratio \(\lambda_{\text {electron }} / \lambda_{\text {proton }}\) of their de Broglie wavelengths.

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