/*! 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 38 A beam of 40 -V electrons travel... [FREE SOLUTION] | 91Ó°ÊÓ

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A beam of 40 -V electrons traveling in the \(+x\) direction passes through a slit that is parallel to the \(y\) -axis and 5.0\(\mu \mathrm{m}\) wide. The diffraction pattern is recorded on a screen 2.5 \(\mathrm{m}\) from the slit. (a) What is the de Broglie wavelength of the electrons? (b) How much time does it take the electrons to travel from the slit to the screen? (c) Use the width of the central diffraction pattern to calculate the uncertainty in the \(y\) -component of momentum of an electron just after it has passed through the slit. (d) Use the result of part (c) and the Heisenberg uncertainty principle (Eq, 39.11 for \(y\) ) to estimate the minimum uncertainty in the \(y\) -coordinate of an electron just after it has passed through the slit. Compare your result to the width of the slit.

Short Answer

Expert verified
(a) 6.02 x 10^-11 m, (b) 2.11 x 10^-7 s, (c) 1.301 x 10^-28 kg m/s, (d) 4.05 x 10^-7 m; uncertainty is less than slit width.

Step by step solution

01

Calculate Electron Speed

Use the electron's kinetic energy to determine its speed. The kinetic energy is given by \( KE = \frac{1}{2}mv^2 = 40\, \text{eV} \). Convert this energy into joules by using \( 1\, \text{eV} = 1.602 \times 10^{-19}\, \text{J} \): \[ KE = 40 \times 1.602 \times 10^{-19} \approx 6.408 \times 10^{-18} \text{ J} \]Now solve for speed \( v \):\[ v = \sqrt{\frac{2 \cdot KE}{m}} \]Substitute electron mass \( m = 9.11 \times 10^{-31}\, \text{kg} \):\[ v = \sqrt{\frac{2 \cdot 6.408 \times 10^{-18}}{9.11 \times 10^{-31}}} \approx 1.187 \times 10^7\, \text{m/s} \]
02

Calculate de Broglie Wavelength

The de Broglie wavelength \( \lambda \) can be found using the formula \( \lambda = \frac{h}{mv} \), where \( h \) is Planck's constant, \( 6.626 \times 10^{-34}\, \text{J s} \). Substitute the values for \( m \) and \( v \):\[ \lambda = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 1.187 \times 10^7} \approx 6.02 \times 10^{-11}\, \text{m} \]
03

Calculate Travel Time to Screen

Determine the travel time from the slit to the screen using the distance and speed: \[ t = \frac{d}{v} = \frac{2.5}{1.187 \times 10^7} \approx 2.11 \times 10^{-7}\, \text{s} \]
04

Calculate Uncertainty in y-Momentum

The uncertainty in the electron's \( y \)-momentum is determined using the width of the central diffraction pattern. For a single slit, the angular width of the central maximum \( \theta \) is approximately \( \lambda / w \), where \( w = 5.0 \times 10^{-6}\, \text{m} \). The momentum in the \( y \)-direction can be related to the angular width by \( \Delta p_y = mv\sin\theta \). Approximate \( \sin\theta \approx \theta \) for small angles:\[ \theta = \frac{\lambda}{w} = \frac{6.02 \times 10^{-11}}{5.0 \times 10^{-6}} \approx 1.20 \times 10^{-5} \]Thus, \( \Delta p_y \approx mv\theta \): \[ \Delta p_y = 9.11 \times 10^{-31} \times 1.187 \times 10^7 \times 1.20 \times 10^{-5} \approx 1.301 \times 10^{-28}\, \text{kg m/s} \]
05

Calculate Uncertainty in y-position

Using the Heisenberg Uncertainty Principle \( \Delta y \Delta p_y \geq \frac{h}{4\pi} \), solve for \( \Delta y \):\[ \Delta y \geq \frac{6.626 \times 10^{-34}}{4\pi \times 1.301 \times 10^{-28}} \approx 4.05 \times 10^{-7}\, \text{m} \]Compare this with the width of the slit, \( 5.0 \times 10^{-6}\, \text{m} \). The uncertainty \( \Delta y \) is smaller than the slit width.

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Key Concepts

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

De Broglie Wavelength
In quantum mechanics, the concept of wave-particle duality is beautifully illustrated by the de Broglie wavelength. Louis de Broglie postulated that particles like electrons have wave-like properties. The de Broglie wavelength is given by the formula \( \lambda = \frac{h}{mv} \), where \( \lambda \) represents the wavelength, \( h \) is Planck's constant \(6.626 \times 10^{-34}\, \text{J s}\), \( m \) is the mass of the particle, and \( v \) is its velocity.
When calculating the de Broglie wavelength of an electron with known velocity, use these values: the mass of an electron \( 9.11 \times 10^{-31} \text{ kg} \) and the velocity derived from its kinetic energy. This wavelength is incredibly small, which means that electrons exhibit noticeable wave-like behavior only under specific conditions, such as small scales or high velocities. This calculation allows us to explore phenomena like diffraction, highlighting the dual nature of particles in quantum mechanics.
Heisenberg Uncertainty Principle
The Heisenberg Uncertainty Principle is a fundamental concept in quantum mechanics, formulated by Werner Heisenberg. It asserts that there are limits to how precisely we can simultaneously measure pairs of conjugate properties, such as position and momentum. To illustrate, consider an electron passing through a slit.
When the electron's path is confined by the slit, its position becomes more certain. This increased certainty in position means an increased uncertainty in its momentum, particularly in the perpendicular direction (\( y \)-axis in this case). The uncertainty principle is mathematically expressed as \( \Delta y \Delta p_y \geq \frac{h}{4\pi} \), where \( \Delta y \) is the uncertainty in position and \( \Delta p_y \) is the uncertainty in momentum. This principle is not about measurement errors, but rather describes a fundamental property of quantum systems, indicating how our knowledge of nature has inherent limits. The constraints arise due to the wave-like nature of particles, encapsulating the core of quantum reality.
Electron Diffraction
Electron diffraction occurs when a beam of electrons, exhibiting wave-like properties, passes through a slit or grating and displays a pattern of constructive and destructive interference. This phenomenon is a direct outcome of the wave-particle duality proposed by de Broglie.
In a typical diffraction experiment (such as the one described in the original exercise), electrons travel through a narrow slit several micrometers wide. As they pass through, their wavefronts spread out, leading to interference patterns that can be observed on a detection screen some distance away. The central maximum and the surrounding pattern are due to differences in path length that electrons traveling through different parts of the slit experience. The angular width of the central diffraction maximum can be approximated by \( \theta = \frac{\lambda}{w} \), where \( w \) is the width of the slit. This gives clues about the electron's momentum distribution as it relates to the beam's spread and is foundational in demonstrating the quantum mechanical nature of particles.
Kinetic Energy Calculation
Kinetic energy is the energy possessed by a moving object. For electrons, it's calculated using the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass and \( v \) is the velocity. In quantum mechanics, knowing an electron's kinetic energy allows us to derive its speed, further enabling calculations of properties like the de Broglie wavelength.
In problems where kinetic energy is given in electron volts (eV), conversion to joules is necessary. Remember, \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \). Once the kinetic energy is in joules, you can rearrange the kinetic energy formula to solve for the electron's velocity: \( v = \sqrt{\frac{2 \times KE}{m}} \). This velocity is crucial for subsequent calculations involving an electron's wave properties, impacting phenomena such as diffraction and uncertainty predictions. Understanding these relationships highlights how fundamental principles of energy and motion intertwine with advanced quantum concepts.

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Most popular questions from this chapter

The wave nature of particles results in the quantum- mechanical situation that a particle confined in a box can assume only wavelengths that result in standing waves in the box, with nodes at the box walls, (a) Show that an electron confined in a one- dimensional box of length \(L\) will have energy levels given by $$E_{n}=\frac{n^{2} h^{2}}{8 m L^{2}}$$ (Hint: Recall that the relationship between the de Broglic wave- length and the speed of a nonrelativistic particle is \(m v=h / \lambda\) . The energy of the particle is \(\frac{1}{2} m v^{2}, )(b)\) If a hydrogen atom is modeled as a one-dimensional box with length equal to the Bohr radius, what is the energy (in electron volts) of the lowest energy level of the electron?

(a) If a photon and an electron each have the same energy of 20.0 eV, find the wavelength of each. (b) If a photon and an electron each have the same wavelength of 250 \(\mathrm{nm}\) , find the energy of each. (c) You want to study an organic molecule that is about 250 \(\mathrm{nm}\) long using either a photon or an electron microscope. Approximately what wavelength should you use, and which probe, the electron or the photon, is likely to damage the molecule the least?

Wavelength of a Bullet. Calculate the de Broglie wavelength of a \(5.00-\) g bullet that is moving at 340 \(\mathrm{m} / \mathrm{s}\) . Will the bullet exhibit wavelike properties?

(a) A nonrelativistic free particle with mass \(m\) has kinetic energy \(K\) . Derive an expression for the de Broglie wavelength of the particle in terms of \(m\) and \(K .\) (b) What is the de Broglic wave- length of an 800 -eV electron?

An atom in a metastable state has a lifetime of 5.2 \(\mathrm{ms}\) . What is the uncertainty in energy of the metastable state?

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