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

Determine the momentum of an electron moving (a) at speed \(2.4 \times 10^{4} \mathrm{~m} / \mathrm{s}\) (about three times escape velocity) and (b) at speed \(2.4 \times 10^{8} \mathrm{~m} / \mathrm{s} .\) (c) In each case, by how much is the classical formula in error?

Short Answer

Expert verified
The classical and relativistic momenta of the electron moving (a) at speed 2.4 × 10^4 m/s are practically equal as the error is tiny (~0%). However, for speed 2.4 × 10^8 m/s, the classical momentum is significantly smaller than the relativistic momentum with a large error rate (~95.8% error).

Step by step solution

01

Classical momentum calculation

Calculate the momentum of the electron using the classical formula for each of the given speeds. This can be done by plugging in the mass of an electron \(9.11 * 10^{-31} kg\), and the given speed into the formula \(p = m*v \). Let's do this for both speeds: \(2.4 * 10^4 m/s\) and \(2.4 * 10^8 m/s\)
02

Relativistic momentum calculation

Now calculate the momentum using the relativistic formula. This is similar to the previous step, but we use the relativistic formula for momentum \(p = m*v/ \sqrt{1 - v^2/c^2}\). Again, plug the mass of the electron and each speed and calculate the total momentum 'p'. The speed of light to be used in the calculation is \(3 * 10^8 m/s\).
03

Calculate the error

Calculate the difference between the relativistic momentum and classical momentum for each speed. This is the error in the classical momentum calculation.

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.

Classical Momentum Formula
Understanding the classical momentum formula is essential when dealing with the motion of objects at speeds much less than the speed of light. Momentum in classical mechanics is defined as the product of an object's mass and velocity and is given by the equation \( p = m \times v \), where \( p \) represents the momentum, \( m \) is the mass of the object, and \( v \) is the velocity. This relationship is linear, suggesting that the momentum of an object increases directly with either mass or velocity.

For instance, when calculating the momentum of an electron moving at a speed significantly less than the speed of light, the classical formula would suffice. Using the classical formula imparts a straightforward approach, but it has its limitations. It does not account for the effects of relativity, which become appreciable at speeds approaching that of light. Consequently, the formula is likely to introduce significant errors under such conditions, errors which can be examined by comparing the classical momentum to its relativistic counterpart.
Relativistic Momentum Formula
When an object moves at a speed comparable to the speed of light, the classical concept of momentum no longer holds true. The relativistic momentum formula takes into account the relativistic effects that are ignored by the classical formula. In relativity, as an object's speed increases towards the speed of light, its mass effectively increases, requiring more force to continue accelerating. The formula for relativistic momentum is \( p = \frac{m \times v}{\sqrt{1 - v^2/c^2}} \), where \( c \) denotes the speed of light in a vacuum.

The denominator accounts for the relativistic effects by incorporating the speed of light. As the velocity \( v \) of the object approaches \( c \) in magnitude, the denominator approaches zero, causing the momentum to increase dramatically. This formula is critical for precise calculations in high-speed physics, such as in the behavior of subatomic particles in accelerators where they attain near-light speeds.
Mass of an Electron
The mass of an electron, a fundamental property of one of the elementary particles, is a cornerstone of many calculations in both classical and modern physics. The standard mass of an electron is approximately \( 9.11 \times 10^{-31} \) kilograms. Despite its tiny magnitude, the electron's mass is immensely significant when calculating its momentum, energy, and interactions with other particles and fields.

Since the mass of an electron remains constant at classical speeds, it simplifies many calculations. However, according to Einstein's theory of relativity, as an object approaches the speed of light, its relativistic mass increases infinitely. It's essential to note that in the realm of relativistic physics, the rest mass is used, and the increase in momentum with speed is accounted for by the relativistic momentum formula, rather than attributing it to a change in mass.
Speed of Light
One of the most fundamental constants in physics is the speed of light in a vacuum, denoted as \( c \). Its value is approximately \( 3 \times 10^{8} \) meters per second. This extraordinary speed has profound implications in physics, forming the basis of the theory of relativity and affecting how we understand space and time.

The speed of light serves as a cosmic speed limit; according to current scientific understanding, no matter or information can travel faster. It is also the critical factor in the relativistic momentum formula, shaping our understanding of how objects behave at high velocities. Due to its constancy and the role it plays in relativistic equations, the dynamics of elementary particles like electrons can only be accurately described by incorporating \( c \) in computations at velocities that are a substantial fraction of the speed of light.

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

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