/*! 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 8 What is the de Broglie wavelengt... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

What is the de Broglie wavelength for an electron with speed (a) \(v=0.480 c\) and \((b) v=0.960 c ?\) (Hint: Use the correct relativistic expression for linear momentum if necessary.)

Short Answer

Expert verified
The de Broglie wavelengths for an electron traveling at \(v = 0.480c\) and \(v = 0.960c\) can be found by simply understanding the de Broglie equation and the relativistic definition of momentum, then substituting the corresponding values into the equation and performing the division.

Step by step solution

01

Understanding de Broglie's Equation

The de Broglie equation is a basic equation in quantum physics and it relates the wavelength of a wave (\(\lambda\)) with its associated particle's momentum (\(p\)). The equation is \(\lambda = h / p\), where \(h\) is Planck's constant (\(6.62607015 \times 10^{-34} \, m^2 kg / s\))
02

Finding Momentum for \(v = 0.480c\)

If \(v < 0.1c\), Newtonian physics can be used. However for \(v = 0.480c\), it's necessary to use relativistic definitions. The relativistic momentum equation is \(p = \gamma m v\), where \(m\) is electron's mass (\(9.11 \times 10^{-31}\,kg\)), and \(\gamma\) is the Lorentz factor. The Lorentz factor can be calculated as \(\gamma = 1 / \sqrt{1 - (v/c)^2}\).
03

Calculate the Wavelength for \(v = 0.480c\)

Having the momentum, we can simply substitute \(p\) into de Broglie's equation to find out the associated wavelength.
04

Finding Momentum for \(v = 0.960c\)

We follow the same procedure to calculate the relativistic momentum for the electron when \(v = 0.960c\).
05

Calculate the Wavelength for \(v = 0.960c\)

Again, substitute \(p\) in de Broglie's equation to find the associated wavelength.

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 91Ó°ÊÓ!

Key Concepts

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

Relativistic Momentum
When particles travel at high speeds, close to the speed of light (\(c\)), we can't use the regular formula for momentum due to the effects of relativity. Traditional momentum is calculated as \(p = mv\), where \(m\) is mass and \(v\) is velocity. However, at speeds close to the speed of light, we must account for relativistic effects using the formula: \(p = \gamma m v\).
Relativistic momentum is crucial when dealing with particles moving at a significant fraction of light speed, such as electrons in this problem. The reason is their mass effectively increases due to relativistic effects.
The "\(\gamma\)" in the equation stands for the Lorentz factor, which plays a pivotal role in ensuring the momentum accounts for these high-speed tweaks. By using this adjusted momentum, physicists can accurately predict behaviors and interactions at tiny scales and high velocities.
Planck's Constant
Planck's constant is fundamental in linking the world of the tiny particles to wave phenomena. Represented by \(h\), its value is approximately \(6.626 \times 10^{-34} \, \text{m}^2 \text{kg/s}\). This constant appears in the de Broglie equation \(\lambda = h/p\), connecting wave properties with particle momentum.
  • It's a bridge between classical physics and quantum mechanics.
  • Quantifies the action required to shift the energy level for particles such as photons and electrons.
In this exercise, Planck’s constant helps us find the wavelength of electrons moving at relativistic speeds. By dividing Planck’s constant by the momentum calculated using the relativistic formula, we can determine the de Broglie wavelength, revealing crucial insights into wave-particle dual nature.
Lorentz Factor
The Lorentz factor, symbolized by \(\gamma\), modifies calculations of time, length, and other physical quantities for particles moving close to the speed of light. It's calculated as \(\gamma = 1 / \sqrt{1 - (v/c)^2}\), where \(v\) is velocity and \(c\) is speed of light.
In the realm of special relativity:
  • The Lorentz factor quantifies how much relativistic effects alter physical parameters.
  • For example, it shows how time dilation and length contraction happen as \(v\) approaches \(c\).
In this problem, \(\gamma\) ensures the momentum values are accurate at high velocities. For velocities such as \(0.480c\) and \(0.960c\), using the Lorentz factor helps account for the difference between everyday Newtonian expectations and what really happens at near-light speeds.

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

CALC Zero-Point Energy. Consider a particle with mass \(m\) moving in a potential \(U=\frac{1}{2} k x^{2},\) as in a mass-spring system. The total energy of the particle is \(E=\left(p^{2} / 2 m\right)+\frac{1}{2} k x^{2}\). Assume that \(p\) and \(x\) are approximately related by the Heisenberg uncertainty principle, so \(p x=h\) (a) Calculate the minimum possible value of the energy \(E,\) and the value of \(x\) that gives this minimum \(E .\) This lowest possible energy, which is not zero, is called the zero-point energy. (b) For the \(x\) calculated in part (a), what is the ratio of the kinetic to the potential energy of the particle?

(a) What accelerating potential is needed to produce electrons of wavelength \(5.00 \mathrm{nm} ?\) (b) What would be the energy of photons having the same wavelength as these electrons? (c) What would be the wavelength of photons having the same energy as the electrons in part (a)?

DATA In the crystallography lab where you work, you are given a single crystal of an unknown substance to identify. To obtain one piece of information about the substance, you repeat the DavissonGermer experiment to determine the spacing of the atoms in the surface planes of the crystal. You start with electrons that are essentially stationary and accelerate them through a potential difference of magnitude \(V_{a c}\) The electrons then scatter off the atoms on the surface of the crystal (as in Fig. \(39.3 \mathrm{~b}\) ). Next you measure the angle \(\theta\) that locates the first-order diffraction peak. Finally, you repeat the measurement for different values of \(V_{\mathrm{ac}}\). Your results are given in the table. $$\begin{array}{l|rrrrrr} \boldsymbol{V}_{\mathbf{a c}}(\mathbf{V}) & 106.3 & 69.1 & 49.9 & 25.2 & 16.9 & 13.6 \\ \hline \boldsymbol{\theta}\left({ }^{\circ} \mathbf{)}\right. & 20.4 & 24.8 & 30.2 & 45.5 & 59.1 & 73.1 \end{array}$$ (a) Graph your data in the form \(\sin \theta\) versus \(1 / \sqrt{V_{\mathrm{ac}}} .\) What is the slope of the straight line that best fits the data points when plotted in this way? (b) Use your results from part (a) to calculate the value of \(d\) for this crystal.

In the Bohr model of the hydrogen atom, what is the de Broglie wavelength of the clectron when it is in (a) the \(n=1\) level and (b) the \(n=4\) level? In both cases, compare the de Broglie wavelength to the circumference \(2 \pi r_{n}\) of the orbit.

Consider the Bohr-model description of a hydrogen atom. (a) Calculate \(E_{2}-E_{1}\) and \(E_{10}-E_{9} .\) As \(n\) increases, does the energy separation between adjacent energy levels increase, decrease, or stay the same? (b) Show that \(E_{n+1}-E_{n}\) approaches \((27.2 \mathrm{eV}) / n^{3}\) as \(n\) becomes large. (c) How does \(r_{n+1}-r_{n}\) depend on \(n ?\) Does the radial distance between adjacent orbits increase, decrease, or stay the same as \(n\) increases?

See all solutions

Recommended explanations on Physics 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.