#2  Calculate the kinetic energy of a muon produced by the decay of a pion at rest.
The muon and muon antineutrino produced must go in opposite directions with the
same momentum. Since the antineutrino is essentially massless, E=pc for it.
The kinetic energy of the muon is related to it by KE = Q -pc. The energy Q
released by the decay is equal to Q=139.57 MeV - 105.66 MeV = 33.9 MeV
For the muon, (pc)^2 = KE^2 + 2*KE*mc^2 from reducing the relativistic energy expression.
Substituting the expression for Q and pc into this gives
pc=(Q^2 + 2*Q*mc^2)/(2*(Q+mc^2))=(33.9^2+2*33.9*105.66)/(2*(33.9+105.66))=      29.78 MeV
The kinetic energy of the muon is then KE= Q-pc=33.9 - 29.78=    4.13 MeV

#12  The  omega minus baryon has the quark constitution sss, but the quarks are fermions
with spin 1/2, so no two of them can be identical states. This implies a fundamental property which has three possible values
rather than the two possible values of electric charge. It is the three distinct parameter values which suggested
the three primary colors as a visual metaphor, hence "color".

#13 A mystery baryon has strangeness -2 and charge zero.
a. The isospin projection mI = q/e - (S+B)/2 =1/2, corresponding to an isospin of 1/2
The particle is neutral and the s quark has a charge of -1/3, so the third quark has a charge +2/3
This would make it either an up or a charm, with u being the likely choice.
The uss is called a Xi baryon.

#16 Pick out the possible reactions:
a.   p + p -> pi zero + n    NO. violates conservation of baryon number, B=0 on left, B=1 on right.
b.   Pi minus + p ->Kzero + n   NO. Conserves charge and baryon number, no leptons involved, but the Kaon contains a
strange quark and strangeness is not conserved.
c. p + p -> pi plus + n+n    NO doesn't conserve charge
d. p + p -> pi zero + pi plus + pi minus   YES Conserves Baryon number, charge, doesn' t have leptons.
e. K- + p -> pi0 + lambda zero  YES with enough energy to create the lambda. Conserves Baryon number
charge and strangeness.
 
#17. Pick out the possible decays and indicate relevant force.
a. Sigma zero -> lambda zero + gamma   YES, electromagnetic. Conserves B, S, charge
b. n -> p + pi minus   NO Conserves all the relevant quantum numbers, but violates conservation of energy
c. Delta minus -> n + pi minus   YES Delta minus is excited state of ddd, so with enough energy it can give a ddu plus an ud. It goes by strong interaction.
 
#21 Estimate the mean free path of a 1 GeV neutrino in the Earth.
The mean free path can be estimated by calculating the cross section for interaction times the number of nucleons per unit volume and then take the reciprocal.
The density of the earth is about 5000 kg/m^3.
The number of nucleons per unit volume is approximately the density divided by the mass per nucleon
n= 5000 kg/m^3/ (1.7x10^-27 kg) = 3 x 10^30.
The cross section for interaction of a neutrino is 10^-42 m^2
Mean free path = 1/((1E-42 m^2)*(3E30 m^-3)) = 3x10^11 meters.