/*! 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 53 (II) What is the speed of an ele... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(II) What is the speed of an electron just before it hits a television screen after being accelerated from rest by the \(28,000 \mathrm{~V}\) of the picture tube?

Short Answer

Expert verified
The electron's speed is approximately \(3.13 \times 10^7 \text{ m/s}\).

Step by step solution

01

Understanding the Energy Change

When the electron is accelerated by a potential difference of 28,000 V, it gains kinetic energy. According to the principle of energy conversion, the electric potential energy gained by the electron is completely converted into kinetic energy.
02

Calculating Potential Energy

The potential energy gained by the electron is calculated using the formula: \[ PE = q imes V \]where \( q \) is the charge of the electron \( (q = 1.6 imes 10^{-19} ext{ C}) \) and \( V \) is the potential difference \( (V = 28000 ext{ V}) \).Thus,\[ PE = 1.6 imes 10^{-19} imes 28000 \text{ C} \approx 4.48 imes 10^{-15} \text{ J} \]
03

Relating Potential Energy to Kinetic Energy

Since the potential energy (PE) is converted into kinetic energy (KE), we set \( KE = PE \). Therefore, the kinetic energy of the electron is also \( 4.48 \times 10^{-15} \text{ J} \).
04

Applying the Kinetic Energy Formula

Kinetic energy is given by the formula:\[ KE = \frac{1}{2} m v^2 \]where \( m \) is the mass of an electron \( (9.11 \times 10^{-31} \text{ kg}) \).So:\[ 4.48 \times 10^{-15} = \frac{1}{2} \times 9.11 \times 10^{-31} \times v^2 \]
05

Solving for the Velocity

Rearrange the kinetic energy equation to solve for velocity \( v \):\[ v^2 = \frac{2 imes 4.48 \times 10^{-15}}{9.11 \times 10^{-31}} \]Calculating the above, we find \[ v^2 = 9.82 \times 10^{15} \]Thus, \[ v = \sqrt{9.82 \times 10^{15}} \approx 3.13 \times 10^{7} \text{ m/s} \]

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.

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.

Kinetic Energy
Kinetic energy is a form of energy that an object possesses due to its motion. For objects moving at relatively low speeds compared to the speed of light, the kinetic energy (KE) can be calculated with the equation:\[ KE = \frac{1}{2} m v^2 \]In this equation, \( m \) stands for the mass of the object, and \( v \) is its velocity. The notable part here is that kinetic energy is directly proportional to both mass and the square of velocity. This means if you double the speed of an object, its kinetic energy increases four times. For an electron, which is very small and light, even a tiny velocity results in significant kinetic energy. This concept is crucial in studying fast-moving charged particles like electrons.Kinetic energy was fully converted from the original potential energy of the electron. As the electron is accelerated through the picture tube, it is crucial to understand the transition it undergoes from potential energy to kinetic energy.
Potential Energy
Potential energy is the energy stored in an object due to its position relative to other objects. In the case of an electron in an electric field, this position is described by the potential difference (or voltage) it experiences.The potential energy (PE) for an electron accelerated through a voltage \( V \) is calculated as:\[ PE = q \times V \]where \( q \) is the charge of the electron (approximately \( 1.6 \times 10^{-19} \text{ C} \)) and \( V \) is the voltage the electron is subjected to. This formula encompasses the principle that when an electron moves through an electric field, it transforms the potential energy gained from the field into kinetic energy. For the electron in the television picture tube, it was exposed to a potential difference of 28,000 volts and thus gained potential energy that was later transformed into kinetic energy when calculating its speed.
Velocity Calculation
Velocity calculation is an essential part of understanding how quickly an object is moving towards unknown or known outcomes, especially when kinetic energy is involved. In the context of the electron in a television screen, once we determine its kinetic energy from the potential energy gained, we can find its velocity using the kinetic energy equation. After rearranging the formula:\[ v^2 = \frac{2 \times KE}{m} \]By solving for \( v \) (velocity), we take the square root:\[ v = \sqrt{\frac{2 \times KE}{m}} \]When we plug in the values for the kinetic energy and the mass of the electron, we arrive at the final velocity. This process emphasizes how energy transformations directly affect the motion and speed of particles. This approach of calculating velocity can be applied to any particle where forces and energies are quantifiable, revealing insights into fundamental principles of physics and motion.

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

(II) Escape velocity from the Earth is \(11.2 \mathrm{~km} / \mathrm{s}\). What would be the percent decrease in length of a 65.2 -m-long spacecraft traveling at that speed as seen from Earth?

(II) When it is stationary, the half-life of a certain subatomic particle is \(T_{0}\). That is, if \(N\) of these particles are present at a certain time, then a time \(T_{0}\) later only \(N / 2\) particles will be present, assuming the particles are at rest. A beam carrying \(N\) such particles per second is created at position \(x=0\) in a high-energy physics laboratory. This beam travels along the \(x\) axis at speed \(v\) in the laboratory reference frame and it is found that only \(N / 2\) particles per second travel in the beam at \(x=2 c T_{0},\) where \(c\) is the speed of light. Find the speed \(v\) of the particles within the beam.

(II) A certain star is 18.6 light-years away. How long would it take a spacecraft traveling \(0.950 c\) to reach that star from Earth, as measured by observers: \((a)\) on Earth, \((b)\) on the spacecraft? (c) What is the distance traveled according to observers on the spacecraft? (d) What will the spacecraft occupants compute their speed to be from the results of \((b)\) and \((c) ?\)

(II) The americium nucleus, \({ }_{95}^{241} \mathrm{Am},\) decays to a neptunium nucleus, \({ }^{237} \mathrm{~Np},\) by emitting an alpha particle of mass \(4.00260 \mathrm{u}\) and kinetic energy \(5.5 \mathrm{MeV}\). Estimate the mass of the neptunium nucleus, ignoring its recoil, given that the americium mass is \(241.05682 \mathrm{u} .\)

(II) A spaceship leaves Earth traveling at \(0.61 c .\) A second spaceship leaves the first at a speed of \(0.87 c\) with respect to the first. Calculate the speed of the second ship with respect to Earth if it is fired \((a)\) in the same direction the first spaceship is already moving, \((b)\) directly backward toward Earth.

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.