Problem#1
An electron with a total energy of 20.0 GeV collides with a stationary positron. (a) What is the available energy? (b) If the electron and positron are accelerated in a collider, what total energy corresponds to the same available energy as in part (a)?Answer:
(a) When the masses of the target and projectile particles are equal, this can be sim plified to
Ea2 = 2mc2(Em + mc2)
available energy, equal masses, Em << mc2
Ea2 = 2mc2Em
Ea2 = 2(0.511MeV)(20.0GeV)
Ea = 142 MeV
(b) For colliding beams of equal mass, each particle has half the available energy, so each has 71.5 MeV. The total energy is twice this, or 143 MeV.
Problem#2
Deuterons in a cyclotron travel in a circle with radius 32.0 cm just before emerging from the dees. The frequency of the applied alternating voltage is 9.00 MHz. Find (a) the magnetic field and (b) the kinetic energy and speed of the deuterons upon emergence.
Answer:
A deuteron is a deuterium nucleus 1H2. Its charge is q = +e. Its mass is the mass of the neutral 12H atom minus the mass of the one atomic electron:
m = 2.014102u – 0.0005486u = 2.013553u(1.66054 x 10-27 kg/u) = 3.34 x 10-27kg, then we use
B = 2πmf/│q│
B = 2π(3.34 x 10-27kg)(9.00 x 106 Hz)/(1.602 x 10-19 C) = 1.18 T
Problem#3
The magnetic field in a cyclotron that accelerates protons is 1.30 T. (a) How many times per second should the potential across the dees reverse? (This is twice the frequency of the circulating protons.) (b) The maximum radius of the cyclotron is 0.250 m. What is the maximum speed of the proton? (c) Through what potential difference would the proton have to be accelerated from rest to give it the same speed as calculated in part (b)?
Answer:
A proton has mass 1.67 × 10-27 kg.
(a) We use
f = ω/2π
This is twice the frequency of the circulating protons, then
2f = ω/π = eB/mπ
2f = (1.602 x 10-19 C)(1.30 T)/(π x 1.67 × 10-27 kg) = 3.97 x 107 Hz
(b) The maximum radius of the cyclotron is 0.250 m is
v = ωR = eBR/m
v = (1.602 x 10-19 C)(1.30 T)(0.250 m)/(1.67 × 10-27 kg) = 3.12 x 107 m/s
(c) For three-figure precision, the relativistic form of the kinetic energy must be used,
K = (γ – 1)mc2
eV = (γ – 1)mc2
where γ = [1 – v2/c2]-1/2 = [1 – (3.12 x 107/2.998 x 108)2]-1/2 = 1.0054596
(1.602 x 10-19 C)V = (0.0054596)(1.67 × 10-27 kg)(2.998 x 108 m/s)2
V = 5.11MV
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