Determine the energy of the electromagnetic radiation

Click For Summary
SUMMARY

The discussion focuses on calculating the energy of electromagnetic radiation resulting from the annihilation of an electron and positron, both with a rest mass of 9.11 x 10^-31 kg, moving at 0.20c. The total energy (Et) is derived using the formula Et=mc^2/sqrt(1-v^2/c^2), yielding a value of 8.37 x 10^-14 J. Additionally, the discussion addresses the energy released during a supernova explosion and the mass-energy equivalence principle, specifically using E=mc^2 to determine the mass converted to energy. The calculations presented yield results that differ from textbook answers, indicating potential errors in the application of relativistic formulas.

PREREQUISITES
  • Understanding of relativistic energy equations, specifically Et=mc^2 and kinetic energy calculations.
  • Familiarity with the concept of mass-energy equivalence and its application in particle physics.
  • Knowledge of basic astrophysics, particularly regarding star formation and energy release in supernovae.
  • Proficiency in unit conversions, especially between joules and MeV (mega-electronvolts).
NEXT STEPS
  • Study the derivation and application of the relativistic energy equation Et=mc^2.
  • Learn about the annihilation process of particles and the minimum photon energy involved (0.511 MeV).
  • Research the mass-energy conversion during supernova explosions and its implications for stellar evolution.
  • Explore advanced topics in particle physics, including the behavior of particles at relativistic speeds.
USEFUL FOR

Students and professionals in physics, particularly those focusing on particle physics, astrophysics, and energy calculations related to high-energy interactions.

F.B
Messages
81
Reaction score
0
Please i really need help I am sorry to have had to make a second thread but i really need to finish this.


2. The electron and positron each have a rest mass of 9.11 x 10^-31 kg. In a certain experiment, an electron and positron collide and vanish, leaving only electromagnetic radiation after the interaction. Each particle is moving at a speed of 0.20c relative to the laboratory before the collision. Determine the energy of the electromagnetic radiation.

First of all do i have to solve this question like a collision problem, if i do there is no after situation.

Anyways i think i have to solve for Ek.

So Et=Ek + Erest

Et=mc^2/sqrt(1-v^2/c^2)
Et=9.11 x 10^-31 x (3.00 x 10^8)^2/sqrt(0.96)
Et=8.37 x 10^-14
Erest = 8.2 x 10^-14
Ek= 8.37 x 10^-14 - 8.2 x 10^-14
Ek= 1.7 x 10^-15

But this answer isn't right. The books has 0.615 Mev

2. The Big Bang, which is a theory predicting the origin of the universe, is estimated to have released 1.00 x 10^68 J of energy. How many stars could half this energy create, assuming the average star's mass is 4.00 x 10^30 kg.

I did this but again my answer is different from the back of the book.

E=mc^2
m=1.00 x 10^68/(3.00 x 10^8)^2
=1.11 x 10^51

=1.11 x 10^51/4.00 x 10^30
=2.78 x 10^20 kg

The book's answer is 1.39 x 10^20 stars. What did i do wrong.

3. A supernova explosion of a star with a rest mass of 1.97 x 10^31 kg, produces 1.02 x 10^44 J of kinetic energy and radiation.
a) How many kilograms of mass are converted to energy in the explosion?

This one i have no idea how to do.

Please i really need help with all these questions i really have to get these done soon.
 
Physics news on Phys.org
1. The rest mass of the electron/positron is 0.511 MeV, and that is the minimum photon energy when annihilation occurs. Just at the relativistic kinetic energy from each electron to each photon. Think total energy.

2. How many stars could half this energy create, . . . ?

3. Again assume total energy. Some of supernova's rest mass is converted to radiation and kinetic energy of the material.

What is the mass equivalent of 1.02 x 10^44 J of kinetic energy and radiation?
 

Similar threads

Replies
2
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
6
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
4K