Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 16: 21P (page 626)

If the kinetic energy of an electron is 4.4×10-18J, what is the speed of the electron? You can use the approximate (non relativistic) equation here.

Short Answer

Expert verified

The speed of the electron is 3.11×106m/s
.

Step by step solution

01

Identification of the given data 

The given data can be listed below as,

  • The kinetic energy of an electron is,KE=4.4×10-18J .
02

Significance of kinetic energy

In this problem, the concept of a non-relativistic equation of kinetic energy will be used in order to estimate an electron's speed. The relation between the electron’s kinetic energy and electron mass is direct linear.

03

Determination the speed of an electron  

The relation of kinetic energy of an electron is expressed as,

KE=12mev2

Here, meis the mass of an electron whose value is9.1×10-31kg and vis the speed of the electron.

Substitute all the known values in the above equation.

4.4×10-18J=12×9.1×10-31kg×v2v2=24.4×10-18J9.1×10-31kg≈9.67032×1012J/kg1m2/s21J/kgv≈3.11×106m/s

Thus, the speed of the electron is3.11×106m/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 rod uniformly charged with charge -q is bent into a semicircular arc of radius b, as shown in Figure 16.97. What is the potential relative to infinity at location A, at the center of the arc?

A dipole is oriented along the x axis. The dipole moment is p=(qs).

(a) Calculate exactly the potential V(relative to infinity) at a location {x,0,0}on the xaxis and at a location {0,y,0}on the yaxis, by superposition of the individual 1/rcontributions to the potential.

(b) What are the approximate values of Vat the locations in part (a) if these locations are far from the dipole?

(c) Using the approximate results of part (b), calculate the gradient of the potential along the xaxis, and show that the negative gradient is equal to the x component Ex of the electric field.

(d) Along the y axis, dV/dy=0. Why isn’t this equal to the magnitude of the electric field Ealong the yaxis?

A capacitor consists of two charged disks of radiusR separated by a distances, whereR>>s. The magnitude of the charge on each disk is Q. Consider points A, B, C, and D inside the capacitor, as shown in Figure 16.88. (a) Show thatΔV=VC−VAis the same for these paths by evaluating ∆V along each path: (1) Path 1:A = B = C, (2) Path 2:A=C, (3) Path 3:A=D=B=C. (b) If,Q=43 μC,R=4 m,s1=1.5 mm ands2=0.7 mm, what is the value ofΔV=VC−VA? (c) Choose two different paths from point A back to point A again, and show that∆V=0for a round trip along both of these paths.

Question: At a particular instant an electron is traveling with speed (6000 m/s) . (a) What is the kinetic energy of the electron? (b) If a proton were traveling at the same speed (6000 m/s), what would be the kinetic energy of the proton?

Suppose that the potential difference in going from location (2.00,3.50,4.00)m to locationrole="math" localid="1657094482453" (2.00,3.52,4.00)m is3V. What is the approximate value ofEyin this region? Include the appropriate sign.
See all solutions

Recommended explanations on Physics Textbooks

Radiation

Read Explanation

Astrophysics

Read Explanation

Measurements

Read Explanation

Nuclear Physics

Read Explanation

Electricity

Read Explanation

Fundamentals of Physics

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