Problem#1
A positronium atom consists of an electron and a positron. In the Bohr model the two particles orbit around their common center of mass. In the Bohr model, what is the ionization energy for a positronium atom when it is in its ground state?
Answer:
Use the Bohr model to calculate the ionization energy of positronium.
The reduced mass is
mr = mm/(m + m) = ½ m
For a hydrogen with an infinitely massive nucleus, the ground state energy is
E1 = –me4/8ϵ02n2h2 = –13.6 eV
For positronium,
E1 = – ½ [me4/8ϵ02n2h2] = –6.80 eV
The ionization energy is 6.80 eV.
Problem#2
In the LHC, each proton will be accelerated to a kinetic energy of 7.0 TeV. (a) In the colliding beams, what is the available energy in a collision? (b) In a fixed-target experiment in which a beam of protons is incident on a stationary proton target, what must the total energy (in TeV) of the particles in the beam be to produce the same available energy as in part (a)?
Answer:
For colliding beams the available energy is twice the beam energy. For a fixedtarget experiment only a portion of the beam energy is available energy.
(a) the available energy in a collision is
Ea = 2 x 7.0 TeV = 14.0 Tev
(b) we use
Ea2 = 2mc2(Em + mc2)
(14.0 x 106 MeV)2 = 2(938.3 MeV)[Em + 938.3 MeV]
Em = 1.0 x 1011 MeV = 1.0 x 105 TeV
Problem#3
A proton and an antiproton collide head-on with equal kinetic energies. Two rays with wavelengths of 0.780 fm are produced. Calculate the kinetic energy of the incident proton.
Answer:
The initial total energy of the colliding proton and antiproton equals the total energy of the two
photons.
For a particle with mass,
E = K + mpc2
hc/λ = K + mpc2
Where mpc2 = 938 MeV
(6.626 x 10-34 Js)(2.998 x 108 m/s)/(0.780 x 10-15 m x 1.6 x 10-19 C) = K + 938 MeV
K = 1592 MeV – 938 MeV = 654 MeV
Problem#4
Radiation Therapy with π– Mesons. Beams of π– mesons are used in radiation therapy for certain cancers. The energy comes from the complete decay of π– the to stable particles. (a) Write out the complete decay of a π– meson to stable particles. What are these particles? (b) How much energy is released from the complete decay of a single π– meson to stable particles? (You can ignore the very small masses of the neutrinos.) (c) How many π– mesons need to decay to give a dose of 50.0 Gy to 10.0 g of tissue? (d) What would be the equivalent dose in part (c) in Sv and in rem? Consult Table 43.3 and use the largest appropriate RBE for the particles involved in this decay.
Answer:
A charged pion decays into a muon plus a neutrino. The muon in turn decays into an electron or
positron plus two neutrinos.
(a) the complete decay of a π– meson to stable particles is
π– → µ–1 + neutrino + e–1 + three neutrinos.
(b) If we neglect the mass of the neutrinos, the mass decrease is
∆m = m(π–) – m(e–1) = 273me – me = 272me = 2.48 x 10-28 kg
Then,
E = ∆mc2 = 2.48 x 10-28 kg x (2.998 x 108 m/s)2 = 2.23 x 10-11 J = 139 MeV
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