Problem#1
(a) A high-energy beam of alpha particles collides with a stationary helium gas target. What must the total energy of a beam particle be if the available energy in the collision is 16.0 GeV? (b) If the alpha particles instead interact in a colliding-beam experiment, what must the energy of each beam be to produce the same available energy?
Answer:
(a) The masses of the target and projectile particles are equal, we use
Ea2 = 2mc2(Em + mc2)
The mass for the alpha particle can be calculated by subtracting two electron masses from the 2He4 atomic mass:
m = mα = 4.002603u – 2(0.0005486u) = 4.001506u
then, mc2 = (4.001506u)(931.5 MeV/u) = 3.727 GeV
so,
(16.0 GeV)2 = 2(3.727 GeV)[Em + 3.727 GeV]
34.344 GeV = Em + 3.727 GeV
Em = 30.6 GeV
(b) Each beam must have
½ Ea = 8.0 GeV
Problem#2
(a) What is the speed of a proton that has total energy 1000 GeV? (b) What is the angular frequency of a proton with the speed calculated in part (a) in a magnetic field of 4.00 T? Use both the nonrelativistic and the correct relativistic expression, and compare the results.
Answer:
Given: a proton has rest energy mpc2 = 938.3 MeV
(a) the speed of a proton that has total energy E = 1000 GeV, we use
E = γmc2 where γ = [1 – v2/c2]-1/2
1000 GeV = γ(0.9383 GeV)
γ = 1065.8
[1 – v2/c2]-1/2 = 1065.8
1 – v2/c2 = 9.383 x 10-4
v = 0.99953c
(b) Nonrelativistic, we use
ω = eB/m = (1.602 x 10-19 C)(4.00T)/(1.67 × 10-27 kg) = 3.83 x 108 rad/s
Relativistic, we get
ω = eB/mγ = (3.83 x 108 m/s)/1065.8
ω = 3.59 x 105 rad/s
Problem#3
In Example 44.3 it was shown that a proton beam with an 800-GeV beam energy gives an available energy of 38.7 GeV for collisions with a stationary proton target. (a) You are asked to
design an upgrade of the accelerator that will double the available energy in stationary-target collisions. What beam energy is required? (b) In a colliding-beam experiment, what total energy of each beam is needed to give an available energy of 2(38.7 GeV) = 77.4 GeV?
design an upgrade of the accelerator that will double the available energy in stationary-target collisions. What beam energy is required? (b) In a colliding-beam experiment, what total energy of each beam is needed to give an available energy of 2(38.7 GeV) = 77.4 GeV?
Answer:
Given: Ea = 77.4 GeV.
For a proton beam on a stationary proton target and since Ea is much larger than the proton rest energy we can use
Ea2 = 2mc2Em
(77.4 GeV)2 = 2(0.938 GeV)Em
Em = 3200 GeV
(b) For colliding beams the total momentum is zero and the available energy Ea is the total energy for the two colliding particles.
For proton-proton collisions the colliding beams each have the same energy, so the total energy of each beam is
½ Ea = 38.7 GeV
Problem#4
Calculate the minimum beam energy in a proton–proton collider to initiate the p + p → p + p + η0. reaction. The rest energy of the η0 is 547.3 MeV.
Answer:
Only part of the initial kinetic energy of the moving electron is available. Momentum conservation tells us that there must be nonzero momentum after the collision, which means that there must
also be left over kinetic energy.
also be left over kinetic energy.
To create the η0, the minimum available energy must be equal to the rest mass energy of the
products, which in this case is the η0 plus two protons. In a collider, all of the initial energy is available,
products, which in this case is the η0 plus two protons. In a collider, all of the initial energy is available,
so the beam energy is the available energy.
The minimum amount of available energy must be rest mass energy
Ea = 2mpc2 + mηc2
Given: EP = 2mpc2 = 938.3 MeV and Eη = mηc2 = 537.3 MeV
Ea = 2(938.3 MeV) + (537.3 MeV) = 2420 MeV
Each incident proton has half of the rest mass energy, or 1210 MeV = 1.21 GeV
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