Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 29: Q19CQ (page 1061)

Which formula may be used for the momentum of all particles, with or without mass?

Short Answer

Expert verified

De-Broglie’s formula could be used for mass and mass-less particles:

\[{\rm{p = }}\frac{{\rm{h}}}{{\rm{\lambda }}}\]

Step by step solution

01

Formula to describe the momentum of particles

As per De-Broglie's hypothesis which is all microscopic particles, whether mass-less, like photons or having mass, like electrons, have wave properties. Where the relationship between momentum and wavelength is fundamental for all particles. So, we can use this formula to describe the momentum of all particles with or without mass:

\[p = \frac{h}{\lambda }\]

02

De- Broglie’s formula

Where, \[{\rm{p}}\]is the momentum, \[{\rm{h}}\] is Planck's constant and \[{\rm{\lambda }}\] is the wavelength.

Therefore, De-Broglie’s formula could be used for mass and mass-less particles:

\[{\rm{p = }}\frac{{\rm{h}}}{{\rm{\lambda }}}\]

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.

Start your free trial

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

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

(a) What is the momentum of a\({\bf{0}}{\bf{.0100 - nm}}\)-wavelength photon that could detect details of an atom? (b) What is its energy in MeV?

Question: (a) What is the binding energy of electrons to a material from which \({\rm{4}}{\rm{.00 - eV}}\) electrons are ejected by \({\rm{400 - nm}}\]) EM radiation? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

Question: Prove that, to three-digit accuracy, \({\mathbf{h = 4}}{\mathbf{.14 \times 1}}{{\mathbf{0}}^{{\mathbf{ - 15}}}}{\mathbf{ eV}} \cdot {\mathbf{s}}\), as stated in the text.

The velocity of a proton emerging from a Van de Graaff accelerator is 25% of the speed of light. (a) What is the proton’s wavelength?

(b) What is its kinetic energy, assuming it is nonrelativistic?

(c) What was the equivalent voltage through which it was accelerated?

An electron microscope produces electrons with a \(2.00 - {\rm{pm}}\) wavelength. If these are passed through a \({\rm{1}}{\rm{.00 - nm}}\) single slit, at what angle will the first diffraction minimum be found?
See all solutions

Recommended explanations on Physics Textbooks

Mechanics and Materials

Read Explanation

Atoms and Radioactivity

Read Explanation

Physical Quantities And Units

Read Explanation

Rotational Dynamics

Read Explanation

Work Energy and Power

Read Explanation

Thermodynamics

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.

Sign-up for free