The Equivalence of Mass and Energy

In summary: You have to use this for each particle. So total energy = twice that if the two are moving at the same speed.
  • #1
BoogieL80
39
0
I was working on the following problem:

An electron and a positron each have a mass of 9.11 x 10-31 kg. They collide and both vanish, with only electromagnetic radiation appearing after the collision. If each particle is moving at a speed of 0.30c relative to the laboratory before the collision, determine the energy of the electromagnetic radiation.

I feel like I'm missing something basic here. I know the formula is E=mc2. However my webassign says my answer is wrong. Am I wrong to assume that the mass is 2*9.11e-31? Also don't I just multiply 0.30*c2?
 
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  • #2
BoogieL80 said:
I was working on the following problem:

An electron and a positron each have a mass of 9.11 x 10-31 kg. They collide and both vanish, with only electromagnetic radiation appearing after the collision. If each particle is moving at a speed of 0.30c relative to the laboratory before the collision, determine the energy of the electromagnetic radiation.

I feel like I'm missing something basic here. I know the formula is E=mc2. However my webassign says my answer is wrong. Am I wrong to assume that the mass is 2*9.11e-31? Also don't I just multiply 0.30*c2?

No. The total energy is [itex] E = \gamma m c^2 [/itex]. Only if a particle is at rest does E= mc^2.
 
  • #3
I'm sorry. Why are you using gamma in that equatin?
 
  • #4
BoogieL80 said:
I'm sorry. Why are you using gamma in that equatin?

[tex]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

So the full equation then becomes;

[tex]E = \frac{m_{0}c^2}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

The gamma is used as a 'shortend' verson of the equation.

~H
 
  • #5
Thank you.
 
  • #6
Incidentally we were studying pair production and my physics teacher claimed u simply use e = 2mc^2 regardless of the velocity.
 
  • #7
dfx said:
Incidentally we were studying pair production and my physics teacher claimed u simply use e = 2mc^2 regardless of the velocity.

This is valid since the minimum amount of energy required to produce a pair of particles is the amount of energy equivilant to their rest mass. Any additional energy will be in the form of kinetic energy. If two particles are produced with zero kinetic energy then the formula becomes;

[tex]E = \frac{m_{0}c^2}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

[tex]v = 0 \Rightarrow E = \frac{m_{0}c^2}{\sqrt{1-\frac{0^2}{c^2}}}[/tex]

[tex] E = \frac{m_{0}c^2}{\sqrt{1}} = E = m_{0}c^2[/tex]

~H
 
  • #8
Yes, but you would be dealing with moving masses here. So, [itex]E = 2\gamma m_0 c^2[/itex] would be right.
 
  • #9
Reshma said:
Yes, but you would be dealing with moving masses here. So, [itex]E = 2\gamma m_0 c^2[/itex] would be right.

Not if you were calculating the minimum energy required to produce a pair of particles.

~H
 
  • #10
dfx said:
Incidentally we were studying pair production and my physics teacher claimed u simply use e = 2mc^2 regardless of the velocity.

That can't be right. If you shoot an electron and a positron at 0.99c, the gamma rays emitted will have more energy than if they were initially at rest.
 
  • #11
Apologise people, I didn't read the "regardless of velcoity" bit. My mistake.

~H
 
  • #12
Hootenanny said:
Not if you were calculating the minimum energy required to produce a pair of particles.

~H
Oh, but how does [itex]\gamma mc^2 [/itex] take care of the two particles here(electron and positron)?
 
  • #13
Reshma said:
Oh, but how does [itex]\gamma mc^2 [/itex] take care of the two particles here(electron and positron)?
You have to use this for each particle. So total energy = twice that if the two are moving at the same speed.
 
  • #14
Reshma said:
Oh, but how does [itex]\gamma mc^2 [/itex] take care of the two particles here(electron and positron)?

As I said above, I misread part of the question. But I do not understand what you are asking here?

~H
 

1. What is the equivalence of mass and energy?

The equivalence of mass and energy is a concept in physics that states that mass and energy are two forms of the same thing. This means that mass can be converted into energy and vice versa.

2. How was the equivalence of mass and energy discovered?

The concept of the equivalence of mass and energy was first proposed by Albert Einstein in his famous equation, E=mc^2, which relates the amount of energy (E) to the mass (m) of an object and the speed of light (c).

3. How is the equivalence of mass and energy used in practical applications?

The equivalence of mass and energy has been applied in various practical applications, such as nuclear power and nuclear weapons. It also plays a crucial role in understanding the behavior of particles in particle accelerators and nuclear reactions.

4. Is the equivalence of mass and energy always conserved?

Yes, the equivalence of mass and energy is always conserved. This means that the total amount of mass and energy in a closed system will always remain the same, even if they are converted into each other.

5. What are the implications of the equivalence of mass and energy for our understanding of the universe?

The equivalence of mass and energy has had a significant impact on our understanding of the universe and the laws of physics. It has led to the development of theories such as the theory of relativity and quantum mechanics, which have revolutionized our understanding of the nature of space, time, and matter.

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