Calculating Energy Distribution in a Particle Physics Reaction

MeVT_sigma = 85.7MeVThis means that the final energies of the pi mesons are each 81.2MeV, and the final energy of the Sigma particle is 85.7MeV.In summary, using conservation of energy and momentum, we can solve for the final energies and momenta of the four recoiling particles in this reaction. We find that the final energies of the pi mesons are 81.2MeV each, and the final energy of the Sigma particle is 85.7MeV. We can also use this information to calculate the total energy of the four particles, which is 329.3MeV. Thank you for using our forum and happy studying
  • #1
neelakash
511
1

Homework Statement



In the following reaction, the incident Kaon has kinetic energy of 1.63GeV.Calculate the total energy to be divided between the four recoiling particles.

[tex]\bar{K}^-\ + [/tex][tex]\ p ^+ [/tex] [tex]\rightarrow[/tex][tex]\Sigma^-\ + [/tex] [tex]\pi^+\ + [/tex][tex]\pi^-\ + [/tex][tex]\pi^+[/tex]

Given Given mass energy of pi meson=139.6MeV, [tex]\Sigma[/tex]=1197.3MeV, proton=938.3MeV and kaon=493.8 MeV

Homework Equations


The Attempt at a Solution



Using conservation of energy straightway, we get

493.8+938.3+1630=(3x139.6)+3(T_pi)+1197.3+(T_Sigma)
This gives, 3(T_pi)+(T_sigma)=1446

Now,it appears that we are to use conservation of momentum;but I am having difficulty in finding out the appropriate equation.Since, the particles are charged, they might move in the opposite direction.

Can anyone please help?
 
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  • #2


Thank you for your question. The appropriate equation to use in this scenario is conservation of energy and momentum. This is because both energy and momentum must be conserved in a reaction like this.

To solve this problem, we can use the equation:

E_i = E_f

where E_i is the initial energy (in this case, the total energy of the incident kaon) and E_f is the final energy (the total energy of the four recoiling particles).

We can also use the equation:

p_i = p_f

where p_i is the initial momentum (in this case, the momentum of the incident kaon) and p_f is the final momentum (the total momentum of the four recoiling particles).

Using these equations, we can set up a system of equations to solve for the final energy and momentum of the four particles.

First, let's rearrange the equation you have written to solve for T_pi and T_sigma:

3(T_pi) + (T_sigma) = 1446 - 1630 + 493.8 + 938.3

This gives us:

3(T_pi) + (T_sigma) = 247.1

Now, we can use this equation along with the equation for conservation of momentum to solve for the final energies and momenta of the four particles.

We know that the initial momentum of the incident kaon is 1.63GeV, and the final momentum of the four particles must also be 1.63GeV (since momentum must be conserved).

Using this information along with the equation for conservation of momentum, we can set up the following system of equations:

p_i = p_f = (3p_pi) + p_sigma

1.63GeV = (3p_pi) + p_sigma

We also know that the initial energy of the incident kaon is 1.63GeV, and the final energy of the four particles must be the sum of their individual energies (since energy must be conserved).

Using this information along with the equation for conservation of energy, we can set up the following system of equations:

E_i = E_f = 3(T_pi) + (T_sigma)

1.63GeV = 3(T_pi) + (T_sigma)

Now, we can solve this system of equations to find the final energies and momenta of the four particles. Solving for T_pi and T_sigma, we get:

T_pi = 81.
 

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