Question 26.9: Stationary Targets versus Colliding Beams Compare the energi...
Stationary Targets versus Colliding Beams
Compare the energies available from a 1.0-TeV proton accelerator when (a) two 1.0-TeV protons collide head-on and (b) a 1.0-TeV proton strikes another proton in a stationary target.
ORGANIZE AND PLAN For colliding particles each with kinetic energy K, the total available energy is 2K. For the stationary-target experiment, the energy is given by Equation 26.3:
E=\sqrt{\left(m_{1} c^{2}+m_{2} c^{2}\right)^{2}+2 m_{2} c^{2} K}.
\text { Known: Proton rest energy } m c^{2}=938 MeV =0.938 GeV \text {; } K=1.0 TeV =1000 GeV.
Here it’s convenient to express all energies in GeV. The total energy of the colliding beams is 2K = 2000 GeV. For the stationary-target experiment,
E=\sqrt{\left(m_{1} c^{2}+m_{2} c^{2}\right)^{2}+2 m_{2} c^{2} K}.
E=\sqrt{(0.938 GeV +0.938 GeV )^{2}+2(0.938 GeV )(1000 GeV )}.
= 43.4 GeV.
REFLECT The colliding beams provide almost 50 times the energy of the stationary-target experiment!