Problem#1
A K+ meson at rest decays into two π mesons. (a) What are the allowed combinations of π0, π+, and π- as decay products? (b) Find the total kinetic energy of the π mesons.Answer:
Given: m(K+) = 493.7 MeV/c2, m(π0) = 135.0 MeV/c2 and m(π+) = 139.6 MeV/c2.
(a) Charge must be conserved, so K+ → π0 + π+ is the only possible decay.
(b) the total kinetic energy of the π mesons is
E = [m(K+) – m(π0) – m(π+)]c2
E = [493.7 MeV/c2 – 135.0 MeV/c2 – 139.6 MeV/c2]c2 = 219.1 MeV
Problem#2
How much energy is released when a µ- muon at rest decays into an electron and two neutrinos? Neglect the small masses of the neutrinos.
Answer:
Given: m(µ–) = 105.7 MeV/c2, m(e–) = 0.511 MeV/c2 .
The mass decrease is
∆m = m(µ–) – m(e–) = 105.7 MeV/c2 – 0.511 MeV/c2 = 105.2 MeV/c2
So, the energy equivalent is
E = ∆mc2 = 105.2 MeV
Problem#3
What is the mass (in kg) of the Z0? What is the ratio of the mass of the Z0 to the mass of the proton?
Answer:
Given: m(Z0) = 91.2 GeV/c2.
E = mc2
91.2 x 109 x 1.602 x 10-19 J = m(2.988 x 108 m/s)2
m(Z0) = 1.63 x 10-25 kg
then, the ratio of the mass of the Z0 to the mass of the proton is
m(Z0)/m(e-) = 1.63 x 10-25 kg/1.67 x 10-27 kg = 97.2
Problem#4
Table 44.3 shows that a ∑0 decays into a Λ0 and a photon. (a) Calculate the energy of the photon emitted in this decay, if the Λ0 is at rest. (b) What is the magnitude of the momentum of the photon? Is it reasonable to ignore the final momentum and kinetic energy of the Λ0? Explain.
Answer:
Gives the rest energies to be 1193 MeV for the ∑0 and 1116 MeV for the Λ0.
(a) We shall assume that the kinetic energy of the Λ0 is negligible. In that case we can set the value of the photon’s energy equal to Q:
Q = E(∑0) – E(Λ0) = 1193 MeV – 1116 MeV = 77 MeV = Epoton
(b) The momentum of this photon, we use
p = Ephoton/c = (77 x 106 V x 1.602 x 10-19 J)/(2.998 x 108 m/s) = 4.1 x 10-20 kg.m/s
To justify our original assumption, we can calculate the kinetic energy of a Λ0 that has this value of momentum
KΛ = p2/2m = E2/2mc2 (77 MeV)2/[2 x 1116MeV] = 2.7 MeV << Q = 77 MeV
Thus, we can ignore the momentum of the Λ0 without introducing a large error.
Problem#5
If a ∑+ at rest decays into a proton and a π0 what is the total kinetic energy of the decay products?
Answer:
The mass decrease is
∆m = m(∑+) – m(π0) – m(p) = 1189 MeV/c2 – 938.3 MeV/c2 – 135.0 MeV/c2 = 116 MeV/c2
So, the energy released is
E = ∆mc2 = 116 MeV
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