What is the de Broglie wavelength (in \(\mathrm{nm}\) ) associated with a 2.5 -g Ping-Pong ball traveling 35 mph?

Short Answer

Expert verified
The de Broglie wavelength of the Ping-Pong ball is approximately \(2.68418 * 10^{-23} nm\).\n

Step by step solution

01

Convert Units

Convert the mass from grams to kilograms (as 1 gram = 0.001 kilograms) and the speed from miles per hour to meters per second (as 1 mph = 0.44704 m/s). So the mass and speed are \(m = 2.5g = 2.5 * 0.001kg = 0.0025 kg\) and \(v = 35 mph = 35 * 0.44704 m/s = 15.6464 m/s\).\n
02

Use de Broglie Wavelength Formula

Now, use the de Broglie wavelength formula \(\lambda = h / (mv)\). Substitute \(h = 6.62607015 * 10^{-34} J•s\), \(m = 0.0025 kg\), and \(v = 15.6464 m/s\) into the formula. It should look like this: \(\lambda = 6.62607015 * 10^{-34} / (0.0025 * 15.6464)\).\n
03

Calculation

Calculate the above expression to attain the value of wavelength in meters. The result is \(2.68418 * 10^{-32}\) meters.\n
04

Convert Units

The answer needs to be in nanometers, so convert meters to nanometers (as \(1 m = 10^{9} nm\)). The wavelength in nm is \(λ = 2.68418 * 10^{-32} * 10^9 = 2.68418 * 10^{-23} nm\).\n

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Unit Conversion
Understanding unit conversion is essential for solving physics problems, especially in quantum mechanics where the scales of measurement differ greatly from everyday experiences. To start, we need to comprehend that different physical quantities, like mass and speed, are often measured in various units, and converting them to a standard unit of measurement is crucial for accurate calculations.

When dealing with mass, the standard unit in the International System of Units (SI) is the kilogram (kg), but in everyday scenarios, we might encounter grams (g). To convert grams to kilograms, we multiply by 0.001, since there are 1000 grams in a kilogram. Similarly, speed is typically measured in meters per second (m/s) in scientific contexts, though it may be quoted in miles per hour (mph) in daily life. To convert mph to m/s, we multiply by the conversion factor 0.44704.

In the given exercise, we converted a Ping-Pong ball's mass from grams to kilograms and its speed from mph to m/s before applying the de Broglie wavelength formula. Accurate unit conversion is crucial to ensure that the computed wavelengths are correct and in the expected order of magnitude.
Quantum Mechanics
Quantum mechanics is a fundamental branch of physics that provides the theoretical framework for understanding the behavior of particles at the atomic and subatomic levels. It operates under principles that are strikingly different from classical physics, describing phenomena that classical physics can't, such as the wave-particle duality of matter.

This duality is where the concept of de Broglie wavelength arises. Proposed by Louis de Broglie, it states that particles also have wave-like properties, and the wavelength of this 'matter wave' is inversely proportional to the particle's momentum. The smaller and faster a particle is, the shorter its de Broglie wavelength will be. Although in everyday objects like a Ping-Pong ball, its de Broglie wavelength is so small due to its large mass and relatively slow speed that it has no observable wave-like behavior. However, understanding these concepts in quantum mechanics is vital for interpreting the behavior of electrons, photons, and other subatomic particles that exhibit both wave-like and particle-like properties depending on the experiment.
Wavelength Formula
The wavelength formula presented in the context of quantum mechanics is associated with the de Broglie hypothesis, which connects wave-like properties to particles of matter. The de Broglie wavelength of any particle is given by the formula \(\lambda = \frac{h}{mv}\), where \(\lambda\) represents the wavelength, \(h\) is the Planck constant (a value of roughly \(6.626 × 10^{-34} J•s\)), \(m\) stands for mass, and \(v\) is the velocity of the particle.

Understanding the de Broglie wavelength formula and how to apply it is crucial in quantum mechanics. It enables the calculation of the wavelength of any moving particle, providing insights into the wave-like behavior of matter. By plugging in the converted mass and velocity of the Ping-Pong ball into the formula, we get the de Broglie wavelength in SI units of meters, which can then be converted to nanometers or other desired units. Remember that this calculation is primarily applicable to microscopic particles, where their wave properties are significant enough to influence their physical behavior.

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

How is the concept of electron density used to describe the position of an electron in the quantum mechanical treatment of an atom?

How many photons at \(660 \mathrm{nm}\) must be absorbed to melt \(5.0 \times 10^{2} \mathrm{~g}\) of ice \(?\) On average, how many \(\mathrm{H}_{2} \mathrm{O}\) molecules does one photon convert from ice to water? (Hint: It takes \(334 \mathrm{~J}\) to melt \(1 \mathrm{~g}\) of ice at \(0^{\circ} \mathrm{C}\).)

(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 1 percent, what is the uncertainty in knowing 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 3.2-g Ping-Pong ball moving at 50 mph has a momentum of \(0.073 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s} .\) If the uncertainty in measuring the momentum is \(1.0 \times 10^{-7}\) of the momentum, calculate the uncertainty in the Ping-Pong ball's position.

When two atoms collide, some of their kinetic energy may be converted into electronic energy in one or both atoms. If the average kinetic energy is about equal to the energy for some allowed electronic transition, an appreciable number of atoms can absorb enough energy through an inelastic collision to be raised to an excited electronic state. (a) Calculate the average kinetic energy per atom in a gas sample at \(298 \mathrm{~K}\). (b) Calculate the energy difference between the \(n=1\) and \(n=2\) levels in hydrogen. (c) At what temperature is it possible to excite a hydrogen atom from the \(n=1\) level to \(n=2\) level by collision? [The average kinetic energy of 1 mole of an ideal gas is \(\left.\left(\frac{3}{2}\right) R T .\right]\)

All molecules undergo vibrational motions. Quantum mechanical treatment shows that the vibrational energy, \(E_{\mathrm{vib}},\) of a diatomic molecule like \(\mathrm{HCl}\) is given by $$ E_{\mathrm{vib}}=\left(n+\frac{1}{2}\right) h \nu $$ where \(n\) is a quantum number given by \(n=0,1,2,\) \(3, \ldots\) and \(\nu\) is the fundamental frequency of vibration. (a) Sketch the first three vibrational energy levels for \(\mathrm{HCl}\). (b) Calculate the energy required to excite a HCl molecule from the ground level to the first excited level. The fundamental frequency of vibration for \(\mathrm{HCl}\) is \(8.66 \times 10^{13} \mathrm{~s}^{-1} .\) (c) The fact that the lowest vibrational energy in the ground level is not zero but equal to \(\frac{1}{2} h v\) means that molecules will vibrate at all temperatures, including the absolute zero. Use the Heisenberg uncertainty principle to justify this prediction. (Hint: Consider a nonvibrating molecule and predict the uncertainty in the momentum and hence the uncertainty in the position.)

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