1. No category

PHYS2332-Modern Physics II
Winter 2015, Assignment #7
Assigned on Sunday 29 March. Due on Tuesday 7 April.
Do Problem 6 of Chapter 14.
For the omega meson ω → Eω = mω c 2 = 782MeV = 7.82 × 10 8 eV . Using the uncertainty
!
!
principle ΔEΔt = → Δt =
, where we’ve used ΔE = Eω . Following example 14.1,
2
2Eω
we estimate the range
!c
6.5821 × 10 −16 eV is × 3 × 10 8 m / s
R = cΔt =
=
= 1.26 × 10 −16 m = 0.126 fm
2Eω
2 × 7.82 × 10 8 eV
Do Problem 8 Chapter 14.
a) µ + → e+ + ? . From table 14.3 µ + is an anti-particle with muon lepton number
Lµ = −1 on LHS. On the RHS the positron e+ has Le = −1 . In order to conserved
all lepton numbers, the neutinos on RHS of reaction must be ν e (electron
neutrino with Le = 1 to cancel Le = −1 of the positron) and ν µ (muon anti-neutrino
with Lµ = −1 to equate with Lµ = −1 of the muon on the RHS). This gives
µ + → e+ + ν e + ν µ .
b) ?+ p → n + e+ . Since n and p are both baryons so there’s no need to worry about
baryon number. The positron on RHS gives electron lepton number Le = −1 .
Lepton conservation requires the LHS to have an electron anti-neutrino ν e with
Le = −1 . The full reaction is ν e + p → n + e+ .
c) π − → µ − + ? On the LHS the pi meson is not a lepton, so all lepton number are
zero. On RHS µ − has Lµ = 1 , so we must have a muon anti-neutrino ν µ with
Lµ = −1 : π − → µ − + ν µ
d) K − → µ − + ? On the LHS the Kaon meson has zero lepton number so all lepton
number are zero. On RHS µ − has Lµ = 1 , so we must have a muon anti-neutrino
ν µ with Lµ = −1 : K − → µ − + ν µ
e) ?+ n → p + µ − Since n and p are both baryons so there’s no need to worry about
baryon number. The negative muon, µ − , on RHS gives muon lepton number
Lµ = 1 . Lepton conservation requires the LHS to have an muon neutrino ν µ with
Lµ = 1 . The full reaction is ν µ + n → p + µ − .
Do Problem 11 of chapter 14. Since the answer is given in the back, you
must show work to receive full marks.
From table 14.4, the lifetimes are as follow: τ ( J / ψ ) = 7.1× 10 −21 s ;
τ ( ϒ (1S )) = 1.2 × 10 −20 s , and τ (π ± ) = 7.1× 10 −8 s .
!
, where Γ is the fullwidth at half maximum
Γ
(FWHM). Γ ( J / ψ ) = ! / τ ( J / ψ ) = 6.5821× 10 −16 eV i s / 7.1× 10 −21 s = 9.27 × 10 4 eV .
Use Example 14.3, the lifetime is τ =
Γ ( ϒ (1S )) = ! / τ ( ϒ (1S )) = 6.5821× 10 −16 eV i s /1.2 × 10 −20 s = 5.49 × 10 4 eV
Γ (π ± ) = ! / τ (π ± ) = 6.5821× 10 −16 eV i s / 2.6 × 10 −8 s = 2.53 × 10 −8 eV
Γ ( J / ψ ) & Γ ( ϒ (1S )) ≫ Γ (π ± ) means that data on the half width such as in example
14.3 can be used to determine the particles produced by a reaction.
Do Problem 16 of Chapter 14
γ1 photon #1
Eγ1 energy, Pγ1 momentum
Eγ1 = Pγ1c
+x
π°, mπc2 = 135MeV
Kinetic Energy Kπ = 720 MeV
Eπ energy, Pπ momentum
Using Eqs 2.65 and 2.71
Eπ =
θ
θ
γ2 photon #2
Eγ2 energy, Pγ2 momentum
Eγ2 = Pγ2c
pπ2 c 2 + ( mπ c 2 ) = K π + mπ c 2
2
For π°, Eπ = K π + mπ c 2 = 720MeV + 135MeV = 855MeV , and
Eπ2 − ( mπ c 2 )
2
( 855MeV )2 − (135MeV )2
MeV
c
c
c
Due to the symmetry of the problem, the two photons have equal energy E γ 1 = E γ 1 = E γ
pπ =
=
and magnitude of momentum Pγ 1 = Pγ 1 = Pγ =
Eγ
c
= 844.3
, with the direction of the momentum
as shown above.
Conservation of energy
One π ! , Eπ = 2Eγ , two photons, gives Eγ = Eπ / 2 = 855MeV / 2 = 427.5MeV . This
MeV
.
c
Conservation of momentum Use above diagram.
Vertical (y-axis) π ! , Pπ ,Y = 0 = Pγ sin θ − Pγ sin θ = 0 , two photons.
gives the photon momentum P γ = 427.5
MeV ⎛
MeV ⎞
MeV
= ⎜ 427.5
cosθ , two
⎟⎠ cosθ + 427.5
⎝
c
c
c
MeV
MeV ⎞
844.3
⎛
= 2 ⎜ 427.5
cosθ → cosθ =
→ θ = 9.08! .
photons. This gives 844.3
⎟
⎝
c
c ⎠
855
Horizontal (x-axis) π ! , Pπ ,x = 844.3
Do problem 20 of Chapter 14.
Consider first the D 0 meson, which Table 14.6 the charmed number C = 1, and the
strangeness number is zero. We know that the meson must be made of a quark-antiquark
pair. Using table 14.5 it is clear that to have C = 1, the quark must be c , with charge
+2e/3, which has charmed number 1. The antiquark must
! have charge -2e/3 to make the
final particle neutral, so the possible pairs are cu or ct . These pairs would give the
required zero strangeness and charmed quantum number C =1. However the top quark is
really heavy (178GeV/c2) compare to the D 0 (1.8 GeV/c2). Since the c quark has mass
(1.15 to 1.3 Gev/c2) and the u quark has 0.0015 to 0.004 GeV/c2, the cu is the most likely
combination for D 0 .
For D + (really DS+ ) charmed meson, using table 14.6 the charmed number C = 1, and the
strangeness number S = 1. Using table 14.5 it is clear that to have C = 1, the quark must
be c , with charge +2e/3, which has charmed number 1. The antiquark uhuhimust have
charge e/3 to make the final total charge +e, so the possible pairs are cs or cd . However
on the pair cs would give final strangeness of 1. The only possible combination for D + is
cd .
Do problem 50 of Chapter 14.
A) π + → µ + + n . On the LHS the anti pi meson is not a lepton, so all lepton number
are zero. On RHS µ + has Lµ = −1 (see table 14.3), and n is not a lepton, so all lepton
numbers are zeros. Hence this reaction (process) is not allowed since lepton numbers
is not conserved.
B) µ − → e− + γ . On the LHS is µ − has Lµ = +1 (see table 14.3), and all other lepton
number zero, including Le = +1 . On RHS e− has Le = +1 (see table 14.3) and Lµ = 0 ,
and γ is not a lepton, so all lepton numbers are zeros. Hence this reaction (process) is
not allowed since not all lepton numbers are conserved.
C) Λ → p + π − . Using table 14.4 we can see that charge and baryon number are
conserved. Spin is conserved, with both sides ½. Lepton numbers are conserved since
none of the particles are leptons. None of the particles are charm particles. For
strangeness number LHS is −1, and on RHS it is zero since the particles have no
strangeness. ΔS = SRHS − SLHS = 0 − ( −1) = +1 , which is allowed by weak decay.
Indeed appendix14.4 indicate that this decay does occur.
D) p → π + π 0 . The LHS is a baryon and the RHS are two mesons. Hence baryon
numbers are not conserved, and this process is not allowed. As discussed in class and
in the textbook, free protons are stable and do not decay to other particles.
QUESTION 7 Cyclotron
A positively charge elementary particle (rest energy 1777 MeV) is moving
with 2206 MeV of kinetic energy in a circular path perpendicular to a
uniform 1.270 T magnetic field. (a) Calculate the momentum of the tau in
kilogram-meters per second. Relativistic effects must be considered. (b)
What is the radius of the path?
(a) In SI units, the kinetic energy of the positive tau particle is
K = 2260MeV × 1.6 × 10 −13 J i MeV −1 = 3.616 × 10 −10 J
Similarly, mc 2 = 2.84 × 10 −10 J for a positive tau. The total energy is E = K + mc 2 ,
K = E − mc 2 = p 2 c 2 + m 2 c 4 − mc 2 , we have K + mc 2 = p 2 c 2 + m 2 c 4 , which gives
2
1
1
p=
K 2 + 2mc 2 K =
3.616 × 10 −10 J ) + 2 ( 2.84 × 10 −10 J ) ( 3.616 × 10 −10 J )
(
8
c
2.998 × 10 m / s
−18
= 1.93 × 10 kg i m i s −1
(b) The radius should be calculated with the relativistic momentum:
γ mv p 1.93 × 10 −18 kg i m i s −1
r=
=
=
= 9.5m
qB eB 1.6 × 10 −19 C × 1.270T
QUESTION 8 Elementary Particle Path
A neutral pion ( π 0 ) has an initial kinetic energy of 106 MeV and decays
after one mean lifetime (see table 14.4 for data). What is the longest possible
track this particle could leave in a detection chamber? From our class
discussion, would a neutral pion π 0 leave any track in the Wilson’s cloud
chamber? Explain your answer in one sentence. NOTE: 1) Take relativity
into account; 2) The lifetime in table 14.4 are for particles at rest.
The kinetic energy is K = γ mc 2 − mc 2 , with γ = 1 / 1− ( v / c ) . Solving we get
2
γ = 1+
K
106MeV
2
= 1+
= 1.785 . Now γ = 1 / 1− ( v / c ) → v = c 1− (1 / γ 2 ) , which
2
mc
135MeV
gives v = 2.998 × 10 8 m i s −1 1− (1.785 ) = 2.48 × 10 8 m i s −1 .
−2
The resting lifetime is Δt 0 = 8.4 × 10 −17 s . In the laboratory frame the lifetime of the
moving π 0 is the dilated time Δt = γΔt 0 = 1.785 × 8.4 × 10 −17 s = 1.5 × 10 −16 s .
No, only charge particles can leave tracks in a Wilson’s cloud chamber.
Finally use x = vΔt = 3.72 × 10 −8 m .