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

High-speed electrons are used to probe the interior structure of the atomic nucleus. For such electrons the expression λ=h/p still holds, but we must use the relativistic expression for momentum, p=mv/1v2/c2. (a) Show that the speed of an electron that has de Broglie wavelength λ is v=c1+(mcλ/h)2 (b) The quantity h/mc equals 2.426 × 1012 m. (As we saw in Section 38.3, this same quantity appears in Eq. (38.7), the expression for Compton scattering of photons by electrons.) If λ is small compared to h/mc, the denominator in the expression found in part (a) is close to unity and the speed v is very close to c. In this case it is convenient to write v=(1Δ)c and express the speed of the electron in terms of rather than v. Find an expression for δ valid when λhmc. [Hint: Use the binomial expansion (1 + z)n=1+nz+[n(n1)z2/2]+ g, valid for the case 0 z 0 6 1.4 (c) How fast must an electron move for its de Broglie wavelength to be 1.00 × 1015 m, comparable to the size of a proton? Express your answer in the form v=(1Δ)c, and state the value of Δ

Short Answer

Expert verified
(a) For the speed, use v=c1+(mcλ/h)2. (b) Δ8.5×108. (c) The electron's speed is v(18.5×108)c.

Step by step solution

01

Relate de Broglie Wavelength to Relativistic Momentum

The de Broglie wavelength is given byλ=hp. Using the relativistic expression for momentum p=mv1v2/c2, plug this into the de Broglie equation to getλ=hmv1v2c2.
02

Solve for Velocity v in Terms of Wavelength

Rearranging the equation from the previous step gives:v=hmλ1v2c2.Isolating v, we obtain:v=c1+(mcλh)2,as required.
03

Evaluate the Limit When Wavelength is Much Smaller than h/mc

Since λh/mc, then (mcλh)21. Use the binomial expansion11+x1x2for small x.This implies:c1+(mcλh)2c(112(mcλh)2).Thus, v=(1Δ)c and Δ=12(mcλh)2.
04

Calculate Electron Speed for Given Wavelength

Given the wavelength λ=1.00×1015 m, compute Δ using previously derived expressionΔ=12(mcλh)2.Substitute hmc=2.426×1012 m, thenΔ=12(1.00×10152.426×1012)212(4.12×104)2=12×(1.7×107)8.5×108.

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.

de Broglie Wavelength
The de Broglie wavelength is an essential concept in quantum mechanics. It bridges the wave-like and particle-like properties of matter. This wavelength is defined specifically for any moving particle, such as an electron, through the relation: λ=hp, where λ is the de Broglie wavelength, h is Planck's constant, and p is the momentum of the particle. This formula tells us that every moving object has a wave associated with it. The smaller the object or the higher its speed, the more significant its wave nature becomes, creating implications for particles like electrons.
When electrons move at very high speeds, especially close to the speed of light, we must take relativistic effects into account while calculating their de Broglie wavelength. The usual momentum p=mv gets replaced by the relativistic momentum formula, emphasizing the need for understanding both the wave nature and relativistic aspects of fast-moving particles.
Relativistic Momentum
As particles move at speeds comparable to the speed of light, Einstein's theory of relativity alters how we understand momentum. Instead of using classical mechanics, where momentum is simply the product of mass and velocity, relativistic momentum must be considered:
  • The formula for relativistic momentum is p=mv1v2/c2.
  • Here, m is the mass of the particle, v is its velocity, and c is the speed of light.
As objects approach the speed of light, the denominator 1v2/c2 becomes much smaller, significantly increasing the value of p.
This adjustment in momentum is crucial for calculating properties like the de Broglie wavelength for high-speed electrons.
The influence of relativistic momentum becomes critical because it ensures that speeds above the speed of light are mathematically restricted, aligning with the principles of Einstein's relativity. Thus, understanding this momentum is pivotal when examining particles in high-energy physics.
Binomial Expansion
The binomial expansion is a mathematical method useful for approximating expressions involving terms raised to a power. It particularly helps when dealing with small quantities compared to unity, allowing the simplification of complex formulas. The binomial theorem states that:
  • The expression (1+z)n can be expanded to 1+nz+n(n1)2z2+.
In our context, when calculating relativistic speeds for electrons with a de Broglie wavelength much smaller than h/mc, the binomial expansion allows us to approximate terms like 1+x.
This approximation 11+x1x2 when x1, simplifies calculations by reducing the complexity of terms. For electron speeds near the speed of light, this simplification helps derive expressions for speed v as (1Δ)c.
In cases like these, simplicity and accuracy are key, making the binomial expansion a valuable tool in theoretical physics for reducing formidable equations to manageable calculations.

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 certain atom has an energy level 2.58 eV above the ground level. Once excited to this level, the atom remains in this level for 1.64 × 107 s (on average) before emitting a photon and returning to the ground level. (a) What is the energy of the photon (in electron volts)? What is its wavelength (in nanometers)? (b) What is the smallest possible uncertainty in energy of the photon? Give your answer in electron volts. (c) Show that ΔE/E∣=∣Δλ/λifΔλ/λ∣≪ 1. Use this to calculate the magnitude of the smallest possible uncertainty in the wavelength of the photon. Give your answer in nanometers.

An atom in a metastable state has a lifetime of 5.2 ms. What is the uncertainty in energy of the metastable state?

A CD-ROM is used instead of a crystal in an electrondiffraction experiment. The surface of the CD-ROM has tracks of tiny pits with a uniform spacing of 1.60 μm. (a) If the speed of the electrons is 1.26 × 104 m/s, at which values of θ will the m = 1 and m = 2 intensity maxima appear? (b) The scattered electrons in these maxima strike at normal incidence a piece of photographic film that is 50.0 cm from the CD-ROM. What is the spacing on the film between these maxima?

A 4.78-MeV alpha particle from a 226Ra decay makes a head-on collision with a uranium nucleus. A uranium nucleus has 92 protons. (a) What is the distance of closest approach of the alpha particle to the center of the nucleus? Assume that the uranium nucleus remains at rest and that the distance of closest approach is much greater than the radius of the uranium nucleus. (b) What is the force on the alpha particle at the instant when it is at the distance of closest approach?

Coherent light is passed through two narrow slits whose separation is 20.0 μm. The second-order bright fringe in the interference pattern is located at an angle of 0.0300 rad. If electrons are used instead of light, what must the kinetic energy (in electron volts) of the electrons be if they are to produce an interference pattern for which the second-order maximum is also at 0.0300 rad?

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.

Sign-up for free