# Proton+anti-proton: Minimum Wavelength

*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

- 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

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)*