What is the energy of each photon produced in the decay of the pion?

  • Thread starter Thread starter cashmoney805
  • Start date Start date
  • Tags Tags
    Photons Pion
Click For Summary

Homework Help Overview

The discussion revolves around the decay of a neutral pion into two photons, focusing on the energy and momentum involved in the process. The original poster presents a problem that includes calculating the velocity and momentum of the pion, as well as the energy of the resulting photons, given the pion's rest mass energy and kinetic energy.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculations for the pion's velocity and momentum, with some expressing confidence in their results. There is uncertainty regarding the total energy of the photons produced in the decay, with questions about whether to use momentum conservation or energy conservation principles. Some participants also explore different methods for calculating momentum.

Discussion Status

The discussion is ongoing, with participants sharing their calculations and questioning the validity of their approaches. There is recognition of differing interpretations of energy conservation and momentum conservation in the context of the decay process. Some participants are seeking clarification on the equations used and their applicability at relativistic speeds.

Contextual Notes

Participants note potential confusion regarding the conservation of momentum and energy in the decay process, as well as the implications of using different equations for calculating momentum. There is also mention of a diagram related to the problem, which may provide additional context.

cashmoney805
Messages
50
Reaction score
0

Homework Statement


The neutral pion is an unstable particle that decays very quickly after its creation into two photons (“particles” of light: v = c, mo = 0). The pion has a rest-mass energy of 135 MeV. Consider a pion that has a kinetic energy of 90 MeV
1) Determine the v of this pion
2) Determine the momentum of the pion
3) Determine the sum of the energies of the photons


Homework Equations


1) moc2\gamma = KE + Rest E
2) mov\gamma = p
3) Ephotons= pc


The Attempt at a Solution


I'm pretty sure I got 1 and 2. For 1 I added the KE + RE, converted to proper units, and finally got v = .80c
For 2 I got p = 9.6 x 10-20 kg m/s
Now I'm not so sure about #3. I think Ephotons = pc because momentum is conserved, and all of the pion's momentum gets turned into the photons' momentum. However, what about adding up the original rest mass energy + KE of the pion? If I do that I get a bigger energy then if I do pc. Thanks for all the help!
 
Physics news on Phys.org
cashmoney805 said:

Homework Statement


The neutral pion is an unstable particle that decays very quickly after its creation into two photons (“particles” of light: v = c, mo = 0). The pion has a rest-mass energy of 135 MeV. Consider a pion that has a kinetic energy of 90 MeV
1) Determine the v of this pion
2) Determine the momentum of the pion
3) Determine the sum of the energies of the photons


Homework Equations


1) moc2\gamma = KE + Rest E
2) mov\gamma = p
3) Ephotons= pc


The Attempt at a Solution


I'm pretty sure I got 1 and 2. For 1 I added the KE + RE, converted to proper units, and finally got v = .80c
For 2 I got p = 9.6 x 10-20 kg m/s
Now I'm not so sure about #3. I think Ephotons = pc because momentum is conserved, and all of the pion's momentum gets turned into the photons' momentum. However, what about adding up the original rest mass energy + KE of the pion? If I do that I get a bigger energy then if I do pc. Thanks for all the help!

Just using pure energy conservation, how much energy is there before and after the decay?
 
Nabeshin said:
Just using pure energy conservation, how much energy is there before and after the decay?
According to my calculations, E before = (90 + 135) MeV = 225 MeV
After E = pc = 181.25 MeV
 
I calculated p a different way this time, p = sqrt(2mKE) where m is the relativistic mass. When I do this then multiply p by c to get E, I get E = 201 MeV. Oh boy...

edit: actually I'm not sure if that equation works for speeds close to c...
 
Last edited:
cashmoney805 said:
According to my calculations, E before = (90 + 135) MeV = 225 MeV
After E = pc = 181.25 MeV

Hrm, you're letting the calculations get bogged down too much. The change in energy is zero, right? You know initial energy, so final energy must be.. the same.
 
I get what you're saying, but I don't understand why the equations don't work here. It seems to me that momentum isn't conserved.
 
Last edited:
and there is one more part to this problem which I thought I could get myself, but I can't. Here is a pic of the question/diagram

http://i44.tinypic.com/k12kja.gif
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 5 ·
Replies
5
Views
6K
  • · Replies 34 ·
2
Replies
34
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 1 ·
Replies
1
Views
4K