The Equivalence of Mass and Energy

[BoogieL80]

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?

 

[nrqed]

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 E = \gamma m c^2. Only if a particle is at rest does E= mc^2.

 

[BoogieL80]

I'm sorry. Why are you using gamma in that equatin?

 

[Hootenanny]

I'm sorry. Why are you using gamma in that equatin?

\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}

So the full equation then becomes;

E = \frac{m_{0}c^2}{\sqrt{1-\frac{v^2}{c^2}}}

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

~H

 

[BoogieL80]

Thank you.

 

[dfx]

Incidentally we were studying pair production and my physics teacher claimed u simply use e = 2mc^2 regardless of the velocity.

 

[Hootenanny]

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;

E = \frac{m_{0}c^2}{\sqrt{1-\frac{v^2}{c^2}}}

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

E = \frac{m_{0}c^2}{\sqrt{1}} = E = m_{0}c^2

~H

 

[Reshma]

Yes, but you would be dealing with moving masses here. So, E = 2\gamma m_0 c^2 would be right.

 

[Hootenanny]

Yes, but you would be dealing with moving masses here. So, E = 2\gamma m_0 c^2 would be right.

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

~H

 

[nrqed]

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.

 

[Hootenanny]

Apologise people, I didn't read the "regardless of velcoity" bit. My mistake.

~H

 

[Reshma]

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

~H

Oh, but how does \gamma mc^2 take care of the two particles here(electron and positron)?

 

[nrqed]

Oh, but how does \gamma mc^2 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.

 

[Hootenanny]

Oh, but how does \gamma mc^2 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