# Conservation principles and particle-particle reactions

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Please can someone help explain these to me? I have completed a-d but I'm not how e works. I thought gamma was an exchange particle so should it then decay further???

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hilbert2
Gold Member
The $\gamma$ is a photon that forms in electron-positron annihilation in part (e). You're supposed to check that charge conservation, lepton number conservation, energy-momentum conservation and others are satisfied in these reactions. For example, the charge conservation in the reaction (e) is okay because both the reactants and products have total electric charge of zero.

• Jimmy Ridley
The $\gamma$ is a photon that forms in electron-positron annihilation in part (e). You're supposed to check that charge conservation, lepton number conservation, energy-momentum conservation and others are satisfied in these reactions. For example, the charge conservation in the reaction (e) is okay because both the reactants and products have total electric charge of zero.
okay so for (e)

charge is conserved
baryon number is conserved
and lepton number is conserved?

hilbert2
Gold Member
Yes, charge is conserved, and there are no baryons in the reaction so the baryon number is zero on both sides. The electron and positron have electron lepton numbers of 1 and -1, so they also sum to 0. Now you should also show somehow that the four-momentum can be conserved in that relativistic collision where an electron and positron turn into two photons.

• Jimmy Ridley
Yes, charge is conserved, and there are no baryons in the reaction so the baryon number is zero on both sides. The electron and positron have electron lepton numbers of -1 and 1, so they also sum to 0. Now you should also show somehow that the four-momentum can be conserved in that relativistic collision where an electron and positron turn into two photons.
okay thank you so much, am I also right is saying that (f) is not possible, as

charge is conserved
but baryon number is not conserved as it has baryon number 0 on the left, and 1 on the right? (the neutron has baryon number 1 and the photon has baryon number 0)?

hilbert2
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