/*! 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 12 Wavelength of a Bullet. Calculat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 39: Problem 12

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?

Short Answer

Expert verified
The de Broglie wavelength is approximately \(3.90 \times 10^{-34} \text{ m}\). The bullet will not exhibit wavelike properties.

Step by step solution

01

Understanding the de Broglie Wavelength Formula

The de Broglie wavelength \( \lambda \) can be calculated using the formula:\[ \lambda = \frac{h}{mv} \]where \( h \) is Planck's constant \(6.626 \times 10^{-34} \text{ m}^2\text{ kg/s} \), \( m \) is the mass of the object in kilograms, and \( v \) is the velocity of the object in meters per second.
02

Convert the Mass to Kilograms

The mass of the bullet is given as \(5.00 \text{ g}\). Convert this to kilograms by using the conversion factor \(1 \text{ g} = 0.001 \text{ kg}\):\[ m = 5.00 \text{ g} \times 0.001 \text{ kg/g} = 0.005 \text{ kg} \]
03

Substitute Values into the Formula

Substitute \( h = 6.626 \times 10^{-34} \text{ m}^2\text{ kg/s} \), \( m = 0.005 \text{ kg} \), and \( v = 340 \text{ m/s} \) into the de Broglie wavelength formula:\[ \lambda = \frac{6.626 \times 10^{-34} \text{ m}^2\text{ kg/s}}{0.005 \text{ kg} \times 340 \text{ m/s}} \]
04

Calculate the Wavelength

Perform the calculation:\[ \lambda = \frac{6.626 \times 10^{-34} \text{ m}^2\text{ kg/s}}{1.7 \text{ kg m/s}} \]\[ \lambda \approx 3.90 \times 10^{-34} \text{ m} \]
05

Evaluate the Wavelength for Wavelike Properties

The calculated de Broglie wavelength \(3.90 \times 10^{-34} \text{ m}\) is extremely small and far below the threshold of detectability in everyday macroscopic systems. As a result, the bullet will not exhibit wavelike properties in a meaningful way at this scale.

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 91影视!

Key Concepts

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

Wave-Particle Duality
Wave-particle duality is a fascinating concept in quantum mechanics. It suggests that every particle or quantum entity exhibits both wave and particle properties. This dual nature is evident in fundamental particles such as photons and electrons. However, it becomes less noticeable in larger objects like bullets or baseballs.
  • For small particles, just like photons, the wave aspect dominates and becomes significant as seen in phenomena like interference and diffraction.
  • In contrast, for larger objects, the associated wavelengths are so small that they don't appear to show any wave properties in practical terms.
This is why, when considering macroscopic objects like a 5 g bullet moving at 340 m/s, the calculated de Broglie wavelength is extremely small. Thus, wave-particle duality fades into the background and the bullet behaves as a classic particle.
Planck's Constant
Planck's constant, denoted as 'h', is a fundamental constant in quantum mechanics. It plays a crucial role in relating the energy of a photon to its frequency, and it's used in the calculation of the de Broglie wavelength.
  • Its value is approximately \(6.626 \times 10^{-34} \text{ m}^2\text{ kg/s}\).
  • This tiny constant is a bridge between the macroscopic and quantum worlds; it characterizes the scale at which quantum effects are significant.
In the context of the de Broglie wavelength, Planck's constant contributes to determining how the particle's momentum affects its wave characteristics. Because 'h' is so small, noticeable wavelike properties are typically only observed in very tiny particles, like electrons and other subatomic particles.
Wavelike Properties
Wavelike properties describe how particles can exhibit characteristics similar to waves. This includes behaviors such as interference and diffraction, typically seen in light waves. Particles' wavelike properties are quantified by their de Broglie wavelength, showing that momentum and wavelength are inversely related.
  • For a particle to exhibit wavelike properties, its wavelength must be comparable to its size.
  • In macroscopic objects, like a moving bullet, the calculated wavelength is far too small to exhibit any noticeable wavelike behavior.
Thus, while the de Broglie wavelength conceptually provides a wave aspect to particles, practical observations of wavelike properties demand conditions and scales at which these properties can manifest, such as in subatomic realms.

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

You want to study a biological spocimen by mcans of a wavelength of \(10.0 \mathrm{nm},\) and you have a choice of using electromagnetic waves or an electron microscope. (a) Calculate the ratio of the energy of a 10.0 -nm- wavelength photon to the kinetic energy of a 10.0 -nm-wavelength electron. (b) In view of your answer to part (a), which would be less damaging to the specimen you are studying; photons or electrons?

What is the de Broglie wavelength for an electron with speed (a) \(v=0.480 c\) and $(b) v=0.960 c ?(\text { Hint } \text { Use the correct rela- }tivistic expression for linear momentum if necessary.)

(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?

The \(\psi\) (psin) particle has a rest energy of 3097 MeV \(\left(1 \mathrm{MeV}=10^{6} \mathrm{eV}\right) .\) The \(\psi\) particle is unstable with a lifetime of \(7.6 \times 10^{-21} \mathrm{s}\) . Estimate the uncertainty in rest energy of the \(\psi\) particle. Express your answer in MeV and as a fraction of the rest energy of the particle.

For crystal diffraction experiments (discussed in Section \(39.2 ),\) wavelengths on the order of 0.20 \(\mathrm{nm}\) are often appropriate. Find the energy in electron volts for a particle with this wavclength if the particle is \((a)\) a photon; \((b)\) an clectron; \((c)\) an alpha particle \(\left(m=6.64 \times 10^{-27} \mathrm{kg}\right) .\)
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Electricity and Magnetism

Read Explanation

Atoms and Radioactivity

Read Explanation

Oscillations

Read Explanation

Linear Momentum

Read Explanation

Particle Model of Matter

Read Explanation
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.