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Q & A: Proton+anti-proton: Minimum wavelength

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Most recent answer: 06/13/2011
Q:
quantum theory and nuclear physics ---------------------------------- Please comment on the following calculation: Two photons are produced when a proton and an anti proton annihilate each other. What is the minimum wavelength of each photon. calculation --------------- m(proton) = m(antiproton) = 938.27 MeV/c^2 .: delta_m = 2m = 1876.54 MeV/c^2 = 1.45E-10J since E = delta_m * c^2 => E = 1876.54 MeV since lambda = hc/E = (6.63E-34)(3E8)/1.4E-10 = 1.42E-15 m = 1.42 fm How does my calculation look? Thks Ken
- Ken Baratko (age 65)
Houston,Tx,77092
A:
Hi Ken,
If the particles are at rest relative to each other then your calculation is about right. ( I got 1.32 Fermis ).  However, if they have any relative motion then you have to take the kinetic energy into account.   For example at the proton-antiproton Tevatron Collider at Fermilab they have almost 1 TeV kinetic energy each.  Thus the minimum wavelength goes down by almost a factor of 1000. 

LeeH

(published on 06/13/2011)

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