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Questions & Solutions On Particle Physics
Q1. A photon with an energy Eγ = 2.09GeV creates a proton-antiproton pair in which
the proton has a kinetic energy of
antiproton ?
95.0 MeV . What is the kinetic energy of the
S1. An antiproton has the same mass as a proton , so it seems reasonable to expect that
both particles will have similar kinetic energies .
The total energy of each particle is the sum of its rest energy and its kinetic energy .
Conservation of energy requires that the total energy before this pair production event
equal the total energy after .
Eγ = (ERp + K p ) + (ERp + K p )
The energy of the proton is given as Eγ = 2.09GeV = 2.09 ⋅ 10 3 MeV .
From Table 46.2 ( S&B ) , we see that the rest energy of both the proton and the
antiproton is
E Rp = E Rp = m p c 2 = 938.3MeV .
If the kinetic energy of the proton is observed to be
the antiproton is
95.0 MeV , the kinetic energy of
K p = Eγ − E Rp − E Rp − K p = 2.09 ⋅ 10 3 MeV − 2(938.5MeV ) − 95.0 MeV
Or
K p = 118MeV
The kinetic energy of the antiproton is slightly ( ~ 20% ) greater than the proton . The
two particles most likely have different shares in momentum of the gamma ray , and
therefore will not have equal energies , either .
Q2. One of the mediators of the weak interaction is the
Z 0 boson , with mass
93.0GeV / c 2 . Use this information to find the order of magnitude of the range of the
weak interaction .
S2. The rest energy of the
Z 0 boson is E0 = 93GeV . The maximum time a virtual
Z 0 boson can exist is found from ∆E∆t ≥ h / 2 , or
h
1.055 ⋅ 10 − 34 J ⋅ s
− 27
=
3
.
55
⋅
10
s
∆t ≈
=
2∆E 2(93GeV ) 1.60 ⋅ 10 −10 J / GeV
(
)
The maximum distance it can travel in this time is
(
)(
)
d = c(∆t ) = 3.00 ⋅ 108 m / s 3.55 ⋅ 10 − 27 s ~ 10 −18 m
The distance
d is an approximate value for the range of the weak interaction .
Q3. A neutral pion at rest decays into two photons according to
π0 →γ +γ
.
Find the energy , momentum , and frequency of each photon .
S3. Since the pion is at rest and momentum is conserved , the two gamma – rays must
have equal momenta in opposite directions . So , they must share equally in the energy of
the pion . By Table 46.2 ( S&B ) ,
mπ = 135.0 MeV / c 2
0
Therefore , Eγ = 67.5MeV = 1.08 ⋅ 10 −11 J
p=
E
67.5MeV
− 20
=
=
⋅
kg ⋅ m / s
3
.
61
10
c 3.00 ⋅ 108 m / s
f =
E
= 1.63 ⋅ 10 22 Hz
h
Q4. Name one possible decay mode ( see Table 46.2 on S&B ) for
Ω + , K s0 , Λ0 and n .
S4. The particles in this problem are the antiparticles of those listed in Table 46.2 .
Therefore , the decay modes include the antiparticles of those shown in the decay modes
in Table 46.2 :
Ω + → Λ0 + K +
0
0
K s0 → π + + π − or π + π
Λ0 → p + π +
n → p + e+ + ν e
Q5. The following reactions or decays involve one or more neutrinos . In each case ,
supply the missing neutrino : ν e ,ν µ or ν τ .
(d)
?+ n → p + + e −
(b)
π − → µ− + ?
K+ → µ+ + ?
(e)
?+ n → p + + µ −
(c)
?+ p + → n + e +
(f)
µ − → e − + ?+ ?
(a)
π − → µ − + υµ
Lµ : 0 → 1 − 1
(b)
K + → µ + + υµ
Lµ : 0 → −1 + 1
(c)
Le : −1 + 0 → 0 − 1
(e)
υe + p + → n + e +
ν e + n → p + + e−
υµ + n → p + + µ −
(f)
µ − → e− + υe + υ µ
(a)
S5.
(d)
Le : 1 + 0 → 0 + 1
Lµ : 1 + 0 → 0 + 1
Lµ : 1 → 0 + 0 + 1
and Le : 0 → 1 − 1 + 0
Q6. Determine which of the following reactions can occur . For those that cannot occur ,
determine the conservation law ( or laws ) violated .
p →π + +π 0
(c) p + p → p + π +
(b)
p + p → p + p +π 0
(d)
π + → µ + + υµ
n → p + e − + υe
(f)
π + → µ+ + n
(a)
(e)
S6.
(a)
p →π + +π 0
Baryon number is violated : 1 → 0 + 0
(b)
p + p → p + p +π 0
This reaction can occur .
(c)
p + p → p +π +
Baryon number is violated : 1 + 1 → 1 + 0
(d)
π + → µ + + υµ
This reaction can occur .
(e)
n → p + e − + υe
This reaction can occur .
(f)
π + → µ+ + n
Violates baryon number : 0 → 0 + 1
and violates muon–lepton number : 0 → −1 + 0
Q7. Determine whether strangeness is conserved in the following decays and reactions :
(a)
Λ0 → p + π −
(d)
(b)
π − + p → Λ− + K 0
(e)
(c)
p + p → Λ0 + Λ0
(f)
π − + p → π − + Σ+
Ξ − → Λ0 + π −
Ξ0 → p + π −
S7. We look up the strangeness quantum numbers in Table 46.2 ( S&B) .
(a)
Λ0 → p + π −
Strangeness : − 1 → 0 + 0
( -1 does not equal 0 : so strangeness is not conserved )
(b)
π − + p → Λ− + K 0
Strangeness : 0 + 0 → −1 + 1
( 0 = 0 : strangeness conserved )
(c)
p + p → Λ0 + Λ0
Strangeness : 0 + 0 → +1 − 1
( 0 = 0 : strangeness conserved )
(d)
π − + p → π − + Σ+
Strangeness : 0 + 0 → 0 − 1
( 0 does not equal -1 : so strangeness is not conserved )
(e)
Ξ − → Λ0 + π −
Strangeness : − 2 → −1 + 0
( -2 does not equal -1 : so strangeness is not conserved )
(f)
Ξ0 → p + π −
Strangeness : − 2 → 0 + 0
( -2 does not equal 0 : so strangeness is not conserved )
Q8. Analyze each reaction in terms of constituent quarks :
(a)
(b)
π − + p → K 0 + Λ0
π + + p → K + + Σ+
(c)
K − + p → K + + K 0 + Ω−
(d)
p + p → K0 + p +π + + ?
In the last reaction , identify the mystery particle .
S8. We look up the quark constituents of the particles in Table 46.4 and Table 46.5 in
S&B .
(a)
du + uud → ds + uds
(b)
d u + uud → us + uus
(c)
u s + uud → us + ds + sss
(d)
uud + uud → ds + uud + ud + uds
A
0
uds is either a Λ
or a
Σ0
Q9. Calculate the kinetic energies of the proton and pion resulting from the decay of a
Λ0 at rest :
Λ0 → p + π −
mΛ c 2 = 1115.6 MeV
S9. We look up the energy of each particle :
mP c 2 = 938.3MeV
mπ c 2 = 139.6 MeV
The difference between starting mass-energy and final mass-energy is the kinetic energy
of the products :
(K p + Kπ ) = (1115.6 − 938.3 − 139.6)MeV = 37.7 MeV
In addition , since momentum is conserved ,
p p = pπ = p
Applying conservation of relativistic energy :
( (938.3)
2
) ( (139.6)
+ p 2 c 2 − 938.3 +
Solving the algebra yields
2
)
+ p 2 c 2 − 139.6 = 37.7 MeV
pπ c = p p c = 100.4 MeV
Thus
Kp =
(m p c 2 )2 + (100.4)2 − m p c 2 = 5.35MeV
And
Kπ =
(139.6)2 + (100.4)2 − 139.6 2 = 32.3MeV
−10
K S0 meson at rest decays in 0.900 ⋅ 10 sec , how far will a K S0 meson
travel if it is moving at 0.960c through a bubble chamber ?
Q10. If a
S10. The motion of the
K S0 particle is relativistic . Just like the spaceman who leaves for
a distant star , and returns to find his family long gone , the kaon appears to us to have a
longer lifetime . That time-dilated lifetime is :
t = γt0 =
0.900 ⋅ 10 −10 sec
1 − γ 2 / c2
=
0.900 ⋅ 10 −10 sec
1 − (0.960 )2
During this time , we see the kaon travel at
[
(
)](
= 3.21 ⋅ 10 −10 sec
0.960c . It travels for a distance of
d = vt = (0.960 ) 3.00 ⋅ 108 m / s 3.21 ⋅ 10 −10 sec
d = 0.0926m = 9.26cm
)