Photon beam is incident on a proton target produces a particle

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Homework Help Overview

The discussion revolves around a problem in particle physics involving a photon beam incident on a proton target, resulting in the production of a particle with a specified rest mass. Participants are tasked with determining the energy of the photon beam, the momentum and energy of the produced particle, and its speed, using principles of conservation of energy and momentum.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of conservation laws, specifically mass-energy and momentum conservation, to derive equations related to the problem. There is an exploration of using energy-momentum four-vectors, although one participant indicates a lack of familiarity with this concept. Attempts to rearrange equations to isolate variables such as the energy of the photon beam are noted.

Discussion Status

Some participants express agreement with the initial equations presented, indicating a positive direction in the discussion. There is acknowledgment of the validity of the equations used, and further calculations are attempted based on those equations. However, no consensus on final values or solutions has been reached yet.

Contextual Notes

One participant mentions not having learned about four-vectors, which may limit their approach to the problem. The discussion is framed within the context of homework constraints, focusing on understanding rather than providing complete solutions.

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Homework Statement


A photon beam is incident on a proton target (at rest). Particle X (and nothing else) with rest mass M=1.232GeV/c2 is then produced. Use m_p =0.938GeV/c2 as the proton mass.

a) What is the energy of the photon beam, in terms of GeV?

b) What is the momentum of the moving particle X, in terms of GeV/c? c) What is the energy of the moving particle X, in terms of GeV?

c) What is the speed of the moving particle X, in terms of c? (i.e, βX=?

Homework Equations


E = γ*m*c^2

P = γ*m*u

γ = 1/sqrt(1-β^2)

β = u/c

The Attempt at a Solution


Using the conservation of mass-energy law
E_1 = E_2

the energy before the light hits the proton is E_1 and the energy afterwards is E_2

E_1 = P*c(the energy due to the photon beam) + m_p*c^2(the rest energy of the proton) = γ*M*c^2(the energy of particle X) = E_2

P*c + m_p*c^2 = γ*M*c^2

can somebody let me know if i got this equation right so far?
 
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Yes, your energy equation looks good.

It might be easier to approach the problem using energy-momentum four-vectors, if you are familiar with them.
 
TSny said:
Yes, your energy equation looks good.

It might be easier to approach the problem using energy-momentum four-vectors, if you are familiar with them.
No I haven't gotten learn about 4-vectors yet, unfortunately. But since you said my equation looked good I tried the following...
Ok so then I can write

P*c + m_p*c^2 = sqrt( (P_x*c)^2 + (M*c^2)^2 )

using conservation of momentum

P + P_p = P_x where P, P_p, and P_x are the photon, proton, and X particle momentum's, repectively.

--> P + 0 = P_x
--> P = P_x

using the fact that the momentum of particle X is the same as the photon's

P*c + m_p*c^2 = sqrt( (P*c)^2 + (M*c^2)^2 )

--> E_γ + m_p*c^2 = sqrt( (E_γ)^2 + (M*c^2)^2 ) where E_γ is the energy of the photon

rearranging this equation for E_γ

E_γ = 0.5*( 1/m_p * (M*c)^2 -m_p*c^2 )

after plugging in the numbers I get

E_γ = 0.340 GeV
 
That looks very good.
 
TSny said:
That looks very good.
Woohoo! thanks!
 

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