   Chapter 11, Problem 33P ### University Physics Volume 3

17th Edition
William Moebs + 1 other
ISBN: 9781938168185

#### Solutions ### University Physics Volume 3

17th Edition
William Moebs + 1 other
ISBN: 9781938168185
Textbook Problem

# When both an electron and a positron are at rest, they can annihilate each other according to the reaction e − + e + → γ + γ . In this case, what are the energy, momentum, and frequency of each photon?

To determine

The energy, momentum and frequency of each photon.

Explanation

Given:

e+e+γ+γ

Formula used:

The momentum of a photon is calculated by

p=hλ

Frequency of photon is calculated by

v=Eh

Calculation:

As an electron and a positron is anti-matter of one another the energy of an electron is equal to the energy of a positron which is 0.511 MeV

The given relation is

e+e+γ+γ

The reactant consists of an electron and a positron. So the total energy will be twice the energy of 0.511MeV. According to the law of conservation of energy the energy of the reactant should be equal to the energy of the product. So the energy of the products will be 1.022 MeV and hence the energy of each photon will be 0.511MeV.

As 1.0MeV is equal to the energy value of 0.511 MeV in terms of Joules is

0.511MeV=0.511MeV(1.602× 10 13J1.0MeV)

=8.19×1014J

The relation between energy and frequency is

E=hv

Where E is the energy, h is the Plank's constant and v is the frequency. Using the values of energy obtained and the frequency is calculated as follows.

h=6.626×1034Js

v=Ehv=8.19× 10 14J6.626× 10 34Jsv=1.2×1020s1=1

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started 