/*! 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 39 (a) Through what potential diffe... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) Through what potential difference does an electron have to be accelerated, starting from rest, to achieve a speed of \(0.980 c ?\) (b) What is the kinetic energy of the electron at this speed? Express your answer in joules and in electron volts.

Short Answer

Expert verified
The potential difference necessary for the electron to reach a speed of 0.980c is \(-15.8 MV\), and its kinetic energy at this speed is \(8.69 \times 10^{-14} J\) or \(543 keV\).

Step by step solution

01

Calculating Accelerated Electron Speed

Commence by applying Einstein's special relativity equation due to the speed nearing the speed of light, which is \(\Delta V = \frac{m_0c^2(1 - \sqrt{1 - v^2/c^2})}{e}\), where \(\Delta V\) is the potential difference, \(m_0\) is the rest mass of the electron (\(9.11 \times 10^{-31}kg\)), \(c\) is the speed of light (\(2.998 \times 10^8 m/s\)), \(v\) is the required electron speed (\(0.980c\)), and \(e\) is the elementary charge (\(1.602 \times 10^{-19}C\)). Inputting these values yields potential difference.
02

Calculating Kinetic Energy in Joules

To establish kinetic energy in Joules, insert the values into the equation for computing kinetic energy under special relativity circumstances, \(K = m_0c^2(1 - \sqrt{1 - v^2/c^2})\), where \(K\) represents kinetic energy, \(m_0\) is the rest mass of the electron, \(c\) is the speed of light, and \(v\) is the speed of the electron. The yielded result will be kinetic energy in joules.
03

Converting Kinetic Energy to Electron Volts

The last step is converting the calculated kinetic energy from Joules to electron volts (eV). It is done by using the fact that \(1 eV = 1.602 \times 10^{-19} J\) by simply dividing the computed kinetic energy in joules by the value of electron volt. This will give kinetic energy in electron volts.

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.

Einstein's Special Relativity
To begin with, Einstein's special relativity is a branch of physics that was introduced by Albert Einstein in 1905. It consists of two postulates: firstly, the laws of physics are invariant (identical) in all inertial frames of reference (i.e., frames of reference with no acceleration), and secondly, the speed of light in a vacuum is the same for all observers, regardless of their relative motion or the motion of the light source.

Understanding special relativity is crucial when dealing with objects moving at significant fractions of the speed of light, referred to as 'relativistic speeds'. Traditional Newtonian mechanics fails to accurately describe motions at these high velocities since time, length, and even mass are no longer constant but vary with speed. In our exercise, we apply these principles to calculate the potential difference needed for an electron to achieve such high speeds.
Kinetic Energy Calculation
Kinetic energy is the energy that an object possesses due to its motion. Traditionally, its calculation for an object with mass 'm' and velocity 'v' is provided by the equation \( \frac{1}{2}mv^2 \). Nevertheless, this formula does not hold true at relativistic speeds, like those close to the speed of light (c).

Einstein's theory modifies the kinetic energy calculation for these situations. The correct formula, which takes into account relativistic effects, is given by \( K = m_0c^2(1 - \sqrt{1 - v^2/c^2}) \), where \( m_0 \) is the rest mass of the object and \( v \) is its velocity. The term \( m_0c^2 \) represents the rest energy of the object, and the complete equation accounts for the additional energy required to move the object from rest up to the velocity 'v'.
Electron Acceleration
In our context, electron acceleration refers to the process of increasing an electron's velocity. It is commonly achieved by applying a potential difference, which creates an electric field that impels the electron to accelerate. Since electrons have charge, they naturally respond to such electric fields.

When electrons are accelerated to high speeds approaching the speed of light, as in particle accelerators or certain electronic devices, relativistic effects become significant. The mass of an electron appears to increase as its speed increases, demanding more and more energy to continue accelerating. This concept is vital for the exercise at hand, as we calculate the potential difference needed to attain a given fraction of the speed of light.
Energy Conversion
Finally, the notion of energy conversion is pivotal in moving from one unit of energy to another. In the realm of atomic and subatomic particles, it is often convenient to express energies in electron volts (eV), rather than the standard Joules (J) used in classical mechanics. One electron volt is defined as the amount of kinetic energy gained by an electron when it's accelerated through an electric potential difference of one volt.

The conversion factor between Joules and electron volts is a key piece of information, with \( 1 eV = 1.602 \times 10^{-19} J \). Using this conversion factor, we can translate the kinetic energy determined through special relativity calculations in Joules to a more conveniently small and manageable figure in electron volts, which is done in the final step of our exercise.

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

Two events are observed in a frame of reference \(S\) to occur at the same space point, the second occurring \(1.80 \mathrm{~s}\) after the first. In a frame \(S^{\prime}\) moving relative to \(S\), the second event is observed to occur \(2.15 \mathrm{~s}\) after the first. What is the difference between the positions of the two events as measured in \(S^{\prime} ?\)

A space probe is sent to the vicinity of the star Capella, which is 42.2 light-years from the earth. (A light-year is the distance light travels in a year.) The probe travels with a speed of \(0.9930 c\). An astronaut recruit on board is 19 years old when the probe leaves the earth. What is her biological age when the probe reaches Capella?

Physicists and engineers from around the world came together to build the largest accelerator in the world, the Large Hadron Collider (LHC) at the CERN Laboratory in Geneva, Switzerland. The machine accelerates protons to high kinetic energies in an underground ring \(27 \mathrm{~km}\) in circumference. (a) What is the speed \(v\) of a proton in the \(\mathrm{LHC}\) if the proton's kinetic energy is \(7.0 \mathrm{TeV} ?\) (Because \(v\) is very close to \(c,\) write \(v=(1-\Delta) c\) and give your answer in terms of \(\Delta .\) ) (b) Find the relativistic mass, \(m_{\text {rel }}\), of the accelerated proton in terms of its rest mass.

A proton (rest mass \(1.67 \times 10^{-27} \mathrm{~kg}\) ) has total energy that is 4.00 times its rest energy. What are (a) the kinetic energy of the proton; (b) the magnitude of the momentum of the proton; (c) the proton's speed?

In the earth's rest frame, two protons are moving away from each other at equal speed. In the frame of each proton, the other proton has a speed of \(0.700 c\). What does an observer in the rest frame of the earth measure for the speed of each proton?

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.