Photon beam is incident on a proton target produces a particle

In summary, the photon beam has energy E_1 and the proton has energy E_2. The energy of the moving particle X is P*c + m_p*c^2 = γ*M*c^2.
  • #1
Elvis 123456789
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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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  • #2
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.
 
  • #3
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
 
  • #4
That looks very good.
 
  • #5
TSny said:
That looks very good.
Woohoo! thanks!
 

FAQ: Photon beam is incident on a proton target produces a particle

1. What is a photon beam?

A photon beam is a stream of particles of light known as photons. These photons have no mass and travel at the speed of light.

2. What is a proton target?

A proton target is a material or substance that is used to absorb and interact with protons. It is typically made of a heavy element, such as lead or gold.

3. How does a photon beam interact with a proton target?

When a photon beam is incident on a proton target, the photons will collide with the protons in the target. This collision can result in the production of other particles.

4. What particles are produced when a photon beam is incident on a proton target?

The particles produced when a photon beam is incident on a proton target can vary depending on the energy of the beam and the material of the target. Some possible particles include pions, kaons, and neutrons.

5. What is the significance of a photon beam incident on a proton target?

This phenomenon is important in the study of high energy physics and nuclear reactions. By analyzing the particles produced, scientists can gain a better understanding of the fundamental properties of matter and the interactions between particles.

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