Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 35: Problem 74

An electron is accelerated from rest through a potential of \(1.0 \cdot 10^{6} \mathrm{~V}\). What is its final speed?

Short Answer

Expert verified
Answer: The final speed of the electron is approximately \(1.874 \cdot 10^{7} \mathrm{\frac{m}{s}}\).

Step by step solution

01

Identify given values and constants

In this problem, we have the following data: Potential, V = \(1.0 \cdot 10^{6} \mathrm{~V}\) (Voltage through which the electron is accelerated) Electron charge, e = \(-1.6 \cdot 10^{-19} \mathrm{~C}\) Electron mass, m = \(9.11 \cdot 10^{-31} \mathrm{~kg}\)
02

Calculate the work done on the electron

The work (W) done on the electron as it accelerates through the electric potential (V) is given by: \(W = e \times V\) Plug in the values for e and V: \(W = -1.6 \cdot 10^{-19} \mathrm{~C} \times 1.0 \cdot 10^{6} \mathrm{~V}\) \(W = -1.6 \cdot 10^{-13} \mathrm{~J}\)
03

Use the work-energy theorem to relate the work done to the final kinetic energy

According to the work-energy theorem, the work done on the electron equals the change in its kinetic energy: \(W = KE_{final} - KE_{initial}\) Since the electron starts from rest, its initial kinetic energy is 0. Thus, \(W = KE_{final}\)
04

Calculate the final kinetic energy of the electron

From step 2 and step 3, we have: \( KE_{final} = -1.6 \cdot 10^{-13} \mathrm{~J}\)
05

Calculate the final speed of the electron using the kinetic energy formula

The formula for kinetic energy is given by: \(KE = \frac{1}{2}mv^2\) Where m is the mass of the electron and v is its final speed. We can now solve for the final speed (v) by rearranging the formula: \(v = \sqrt{\frac{2 \times KE}{m}}\) Plug in the values for KE and m: \(v = \sqrt{\frac{2 \times -1.6 \cdot 10^{-13} \mathrm{~J}}{9.11 \cdot 10^{-31} \mathrm{~kg}}}\) \(v \approx 1.874 \cdot 10^{7} \mathrm{\frac{m}{s}}\) So, the final speed of the electron is approximately \(1.874 \cdot 10^{7} \mathrm{\frac{m}{s}}\).

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 rocket ship approaching Earth at \(0.90 c\) fires a missile toward Earth with a speed of \(0.50 c,\) relative to the rocket ship. As viewed from Earth, how fast is the missile approaching Earth?

A rod at rest on Earth makes an angle of \(10^{\circ}\) with the \(x\) -axis. If the rod is moved along the \(x\) -axis, what happens to this angle, as viewed by an observer on the ground?

You shouldn't invoke time dilation due to your relative motion with respect to the rest of the world as an excuse for being late to class. While it is true that relative to those at rest in the classroom, your time runs more slowly, the difference is likely to be negligible. Suppose over the weekend you drove from your college in the Midwest to New York City and back, a round trip of \(2200 .\) miles, driving for 20.0 hours each direction. By what amount, at most, would your watch differ from your professor's watch?

If a muon is moving at \(90.0 \%\) of the speed of light, how does its measured lifetime compare to when it is in the rest frame of a laboratory, where its lifetime is \(2.2 \cdot 10^{-6}\) s?

A square of area \(100 \mathrm{~m}^{2}\) that is at rest in the reference frame is moving with a speed \((\sqrt{3} / 2) c\). Which of the following statements is incorrect? a) \(\beta=\sqrt{3} / 2\) b) \(\gamma=2\) c) To an observer at rest, it looks like another square with an area less than \(100 \mathrm{~m}^{2}\) d) The length along the moving direction is contracted by a factor of \(\frac{1}{2}\)
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Kinematics Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Mechanics and Materials

Read Explanation

Classical Mechanics

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