Photon beam is incident on a proton target produces a particle

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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
 
TSny said:
That looks very good.
Woohoo! thanks!