Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(a) An electron in the ground state of the hydrogen atom moves at an average speed of \(5 \times 10^{6} \mathrm{~m} / \mathrm{s}\). If the speed is known to an uncertainty of 20 percent, what is the minimum uncertainty in its position? Given that the radius of the hydrogen atom in the ground state is \(5.29 \times 10^{-11} \mathrm{~m},\) comment on your result. The mass of an electron is \(9.1094 \times 10^{-31} \mathrm{~kg} .\) (b) A \(0.15-\mathrm{kg}\) baseball thrown at 100 mph has a momentum of \(6.7 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\). If the uncertainty in measuring the momentum is \(1.0 \times 10^{-7}\) of the momentum, calculate the uncertainty in the baseball's position.

Short Answer

Expert verified
(a) Minimum position uncertainty for electron is \(5.785 \times 10^{-10} \text{ m}\). (b) Minimum position uncertainty for baseball is \(7.87 \times 10^{-29} \text{ m}\).

Step by step solution

Achieve better grades quicker with Premium

  • Unlimited AI interaction
  • Study offline
  • Say goodbye to ads
  • Export flashcards

Over 22 million students worldwide already upgrade their learning with Vaia!

01

Calculate the Uncertainty in Velocity for the Electron

The uncertainty in velocity ( \( \Delta v \)) is given as 20% of the average velocity. The average velocity \( v \) is \( 5 \times 10^6 \text{ m/s} \). Thus, the uncertainty can be calculated as follows:\[ \Delta v = 0.2 \times 5 \times 10^6 \text{ m/s} = 1 \times 10^6 \text{ m/s}. \]
02

Use Heisenberg's Uncertainty Principle for the Electron

Heisenberg's Uncertainty Principle states that the product of the uncertainties in position and momentum is at least \( \frac{\hbar}{2} \), where \( \hbar = \frac{h}{2\pi} \approx 1.054 \times 10^{-34} \text{ m}^2 \text{ kg/s} \). The momentum \( p \) is \( mv \), where \( m \) is the electron's mass and \( v \) is its velocity.Therefore, for the electron:\[ \Delta p = m \times \Delta v = 9.1094 \times 10^{-31} \text{ kg} \times 1 \times 10^6 \text{ m/s} = 9.1094 \times 10^{-25} \text{ kg m/s}. \]Using the principle:\[ \Delta x \cdot \Delta p \geq \frac{\hbar}{2} \]\[ \Delta x \geq \frac{1.054 \times 10^{-34} \text{ m}^2 \text{ kg/s}}{2 \times 9.1094 \times 10^{-25} \text{ kg m/s}}. \]Solving gives:\[ \Delta x \geq 5.785 \times 10^{-10} \text{ m}. \]
03

Compare the Uncertainty in Position with the Hydrogen Atom Radius

The calculated uncertainty in position \( \Delta x \approx 5.785 \times 10^{-10} \text{ m} \) is substantially larger than the hydrogen atom's ground state radius of \( 5.29 \times 10^{-11} \text{ m} \). This indicates that the position is highly uncertain compared to the size of the atom.
04

Calculate the Uncertainty in Momentum for the Baseball

The uncertainty in momentum ( \( \Delta p \)) for the baseball is given as \(1.0 \times 10^{-7}\) of its momentum. The given momentum \( p \) is \(6.7 \text{ kg m/s}\). Thus, the uncertainty is:\[ \Delta p = 1.0 \times 10^{-7} \times 6.7 \text{ kg m/s} = 6.7 \times 10^{-7} \text{ kg m/s}. \]
05

Use Heisenberg's Uncertainty Principle for the Baseball

Use Heisenberg's Uncertainty Principle again:\[ \Delta x \cdot \Delta p \geq \frac{\hbar}{2} \]\[ \Delta x \geq \frac{1.054 \times 10^{-34} \text{ m}^2 \text{ kg/s}}{2 \times 6.7 \times 10^{-7} \text{ kg m/s}} \]Solving gives:\[ \Delta x \geq 7.87 \times 10^{-29} \text{ m}. \]

Key Concepts

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

Quantum Mechanics
Quantum mechanics is a fascinating area of physics that deals with the behavior of particles on a microscopic scale. Unlike classical physics, which provides precise predictions, quantum mechanics introduces the concept of probability and uncertainty.
This is where Heisenberg's Uncertainty Principle comes into play, stating that it's impossible to know both the position and momentum of a particle, like an electron, with absolute certainty at the same time.
- This principle is fundamental to understanding atomic and subatomic systems. - Quantum mechanics has led to revolutionary technologies such as semiconductors and lasers. The principle is represented mathematically and is vital in explaining phenomena within quantum mechanics, providing insights into atomic structures that are not possible with classical theories.
Electron Motion
The motion of electrons within an atom is both intriguing and complex. Due to their small mass and high speed, electrons do not follow traditional paths around the nucleus like planets around the sun.
Instead, they exist in areas of probability known as orbitals, where they are likely to be found at any given time.
- Electrons move rapidly, reaching speeds like those noted in the exercise, around \(5 \times 10^6 \text{ m/s}\).- Their motion affects how they bond with other atoms, influencing the chemical properties of the elements.Understanding electron motion is crucial, as it helps explain why elements have specific colors, odors, and reactivity levels. It further explains how light and energy interact with matter on a fundamental level.
Momentum Uncertainty
Momentum uncertainty is a crucial part of quantum mechanics. According to Heisenberg's Uncertainty Principle, the more precisely we know a particle's momentum, the less precisely we know its position, and vice versa. This principle fundamentally limits the precision of simultaneous measurements of key properties.
- For tiny particles like electrons, this uncertainty can significantly affect their behavior and measurement outcomes. - It has minimal impact on larger objects, such as a baseball, which operate under classic physics rules. This concept reveals the limitations of measurement at quantum levels and illustrates the underlying randomness of quantum systems. It also explains why observing or measuring a quantum system inevitably alters it.
Atomic Structure
Atomic structure refers to the composition and arrangement of particles, including protons, neutrons, and electrons, within an atom. The nucleus is made up of protons and neutrons, while electrons orbit this central core.
- The electron's role is critical because their arrangement dictates an atom's chemical properties and behavior. - Understanding atomic structure is crucial for comprehending how atoms bond to form molecules. In atomic theory, electrons are not fixed in one place but spread out over an area around the nucleus. This arrangement forms what we know as electron clouds or orbitals. Different elements have varying numbers of these particles, which accounts for their diverse attributes and interactions.
Hydrogen Atom
The hydrogen atom is the simplest and most fundamental atom in the universe, consisting of just one proton and one electron. This simplicity makes it an excellent model for understanding basic quantum mechanics and atomic physics.
- The electron in a hydrogen atom occupies orbitals that are spherical in the simplest model.- Its ground state radius, given as \(5.29 \times 10^{-11} \text{ m}\), is crucial to calculations in quantum mechanics.Learning about the hydrogen atom is pivotal because it serves as a foundation for understanding more complex atoms and molecules. Its study has provided substantial insights into atomic interaction, electromagnetic radiation, and the quantization of energy levels.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An electron in a hydrogen atom is excited from the ground state to the \(n=4\) state. Comment on the correctness of the following statements (true or false). (a) \(n=4\) is the first excited state. (b) It takes more energy to ionize (remove) the electron from \(n=4\) than from the ground state. (c) The electron is farther from the nucleus (on average) in \(n=4\) than in the ground state. (d) The wavelength of light emitted when the electron drops from \(n=4\) to \(n=1\) is longer than that from \(n=4\) to \(n=2\) (e) The wavelength the atom absorbs in going from \(n=1\) to \(n=4\) is the same as that emitted as it goes from \(n=4\) to \(n=1\)

What is the de Broglie wavelength (in \(\mathrm{cm}\) ) of a 12.4-g hummingbird flying at \(1.20 \times 10^{2} \mathrm{mph}\) \((1\) mile \(=1.61 \mathrm{~km}) ?\)

The retina of a human eye can detect light when radiant energy incident on it is at least \(4.0 \times 10^{-17} \mathrm{~J}\). For light of 575 -nm wavelength, how many photons does this correspond to?

Photodissociation of water $$ \mathrm{H}_{2} \mathrm{O}(l)+h \nu \longrightarrow \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) $$ has been suggested as a source of hydrogen. The \(\Delta H_{\mathrm{rxn}}^{\circ}\) for the reaction, calculated from thermochemical data, is \(285.8 \mathrm{~kJ}\) per mole of water decomposed. Calculate the maximum wavelength (in \(\mathrm{nm}\) ) that would provide the necessary energy. In principle, is it feasible to use sunlight as a source of energy for this process?

In 1996 , physicists created an anti-atom of hydrogen. In such an atom, which is the antimatter equivalent of an ordinary atom, the electric charges of all the component particles are reversed. Thus the nucleus of an anti-atom is made of an antiproton, which has the same mass as a proton but bears a negative charge, while the electron is replaced by an anti-electron (also called a positron) with the same mass as an electron, but bearing a positive charge. Would you expect the energy levels, emission spectra, and atomic orbitals of an antihydrogen atom to be different from those of a hydrogen atom? What would happen if an anti-atom of hydrogen collided with a hydrogen atom?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free