Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

Answers without the blur.

Just sign up for free and you're in.

Short Answer

A positron (charge +e, mass equal to the electron mass) is moving at 1.0x10⁷ m/s in the positive direction of an x axis when, at x=0, it encounters an electric field directed along the x axis. The electric potential V associated with the field is given in Fig. 24-57. The scale of the vertical axis is set by V_s=500.0 V. (a) Does the positron emerge from the field at x=0 (which means its motion is reversed) or at x=0.50 m (which means its motion is not reversed)? (b) What is its speed when it emerges?

The positron emerges from the field at x=0, which means its motion is reversed.
The speed of the positron when it emerges is $1.0 \times 10^{7} m / s$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: The given data

The speed of the positron at x=0, $v = 1.0 \times 10^{7} m / s$
The mass of the positron, $m = 9.1 \times 10^{- 31} kg$
The scale of the vertical axis, data-custom-editor="chemistry" $V_{s} = 500 V$

Step 2: Understanding the concept of energy

Using the concept of energy, we can get the kinetic energy of the positron. This will help in determining the potential height or energy of the system and hence, we can check the emerging point of the positron. Now, comparing the concept with the given situation, we can get the speed of the positron.

Formulae:

The potential energy of the system in terms of potential difference and charge is given by, U=q(deltaV) (i)

The kinetic energy of the system, $K . E = \frac{1}{2} {mv}^{2}$ (ii)

Step 3: a) Calculation of the position at which the positron emerges

The initial kinetic energy of the positron at x = 0 is given using equation (ii) as given:

$\begin{array}{rcl} KE & = & \frac{1}{2} (9.11 \times 10^{- 31} kg) {(1.0 \times 10^{7} m / s)}^{2} \\ = & 4.55 \times 10^{- 17} J \\ = & (4.55 \times 10^{- 17} J) (\frac{1 eV}{1.6 \times 10^{- 19} J}) \\ = & 284.37 eV \\ \approx & 284 e V \end{array}$

The height of the potential energy barrier is given using equation (i) as follows:

data-custom-editor="chemistry" $\begin{array}{rcl} U & = & e (500 V) \\ = & 500 eV \end{array}$

The height of the potential energy barrier is greater as compared with the initial kinetic energy of the positron.

So, the positron emerges from the barrier at x=0, which means its motion is reversed.

Step 4: b) Calculation of the speed of the positron

When the positron emerges from the field, the positron is moving in the opposite direction.

So, the final velocity of the positron is directed along the negative x-direction but its magnitude does not change.

Therefore, the speed of the positron is $1.0 \times 10^{7} m / s$

Find Study Materials for

Create Study Materials

Select your language

Fundamentals Of Physics

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of energy

Step 3: a) Calculation of the position at which the positron emerges

Step 4: b) Calculation of the speed of the positron

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Select your language

Fundamentals Of Physics

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: The given data

Step 2: Understanding the concept of energy

Step 3: a) Calculation of the position at which the positron emerges

Step 4: b) Calculation of the speed of the positron

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Work Energy and Power

Radiation

Astrophysics

Physics of Motion

Scientific Method Physics

Famous Physicists

Fluids

Torque and Rotational Motion

Nuclear Physics

Fields in Physics

Measurements

Space Physics

Electricity

Physical Quantities and Units

Medical Physics

Electrostatics

Further Mechanics and Thermal Physics

Mechanics and Materials

Engineering Physics

Energy Physics

Force

Particle Model of Matter

Waves Physics

Magnetism

Modern Physics

Atoms and Radioactivity

Dynamics

Electricity and Magnetism

Conservation of Energy and Momentum

Fundamentals of Physics

Kinematics Physics

Circular Motion and Gravitation

Linear Momentum

Rotational Dynamics

Oscillations

Translational Dynamics

Geometrical and Physical Optics

Thermodynamics

Magnetism and Electromagnetic Induction

Electromagnetism

Electric Charge Field and Potential

Turning Points in Physics