#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.