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245
15
Particle Physics
151
The time for a particle traveling with the speed of light to travel a distance of

×
15
3 10
m is


×
∆
=
==
×
15
23
8
3 10
m
10
s
3 10 ms
d
t
v
.
152
The
ρ
ππ
+
→+
0
decay must occur via the strong interaction. The
0
K
decay
must occur via the weak interaction.
153
The minimum energy is released, and hence the minimum frequency photons are produced.
The proton and antiproton are at rest when they annihilate. That is,
=
0
EE
and
=
0
K
. To
conserve momentum, each photon must carry away onehalf the energy. Thus,
=
=
0
min
min
0
2
938.3 MeV
2
E
E
h
fE
. Thus,
(29
( 29


×
=
=×
×
19
23
min
34
938.3 MeV 1.6 10
J MeV
2.26 10
Hz
6.63 10
J s
f
and
λ

×
=
=
×
8
15
max
23
min
1.32 10
m
2.26 10
Hz
c
f
154
E
,
p
nucleus at rest
M
n
= 100 u
Before
100 u
proton
antiproton
K
E
n
,
p
n
After
(
29
(
photon
photon
n
E
pp
c
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CHAPTER 15
PARTICLE PHYSICS
(
29
(
==
photon
photon
n
E
pp
c
,
( 29
=
+
=+
2
photon
2
K
1 877 MeV K
p
nn
E
m
c
EE
. Assuming that
the kinetic energy of the recoiling nucleus, K
n
E
, is small compared to 1 877 MeV,
(
2245
photon
1 877 MeV
E
, and
2245
1 877 MeV
n
pc
. Since K
n
E
is small we can use the classical
expression
=
2
K
2
n
n
n
p
E
m
:
(
(29
(
(
λ


=
+=
=
=
=×
×
×
=
=
×
2
2
23
15
8
16
23
1 877 MeV
K
18.9 MeV
2 100 u 931.5 MeV u
photon
1 877 MeV 19 MeV
1 896 MeV
1 896 MeV
4.58 10
Hz
4.14 10
eV s
3.00 10
m s
6.55 10
m
4.58 10
Hz
n
c
E
c
E
E
f
h
c
f
155
The rest energy of the
0
Z
boson is
=
0
96 GeV
E
. The maximum time a virtual
0
Z
boson can
exist is found from
∆∆=
h
Et
.
( 29



×
∆
=
=
∆
×
h
34
27
10
1.055 10
J s
6.87 10
s
96 GeV 1.6 10
J GeV
t
E
.
The maximum distance it can travel in this time is
(
29
( 29( 29

=∆=×
×
=
×
8
27
18
3 10
m s 6.87 10
s
2.06 10
m
dct
.
The distance
d
is an approximate value for the range of the weak interaction.
156
μ
+
+→+
e vv
muonlepton number before reaction
( 29 (29
=+
10
electronlepton number before reaction
(29 (29
=+=
0 11 . Therefore, after the reaction, the
muonlepton number must be –1. Thus, one of the neutrinos must be the antineutrino
associated with muons or
μ
v
. Also, after the reaction, the electronlepton number must be 1.
Thus, one of the neutrinos must be the neutrino associated with electrons, or
e
v
. Thus,
μ
μ
+
+
→+
e
evv
.
157
Use Table 15.2 to find properties that can be conserved in the given reactions
Reaction 1
Reaction 2
(a)
Charge:
π

+
+ →
+Σ
pK
ππ

p
(29 (29 (29 (29
++→++
→
00
ǆ
→
ǆ
MODERN PHYSICS
247
(b)
Baryon number:
(
29(29 (
29(29
++ →
++
0101
(
+ →+
11
ǆ
ǆ
(c)
Strangeness:
(
29(
29 (29(29
+
→+ +
0
0
(29(29 (29( 29
+
→
0
001
→
00
ǆ
(29
→
01
X
Thus, the second reaction is not allowed since it does not conserve strangeness.
158
(a)
μ
πμ

→+
v
μ
L
:
0 11
(b)
μ
μ
Kv
μ
L
:
→+
0
(c)
+
+→+
pn
e
ve
e
L
:
+→
10 01
(d)

np
e
e
L
:
(e)
μ
μ

v
μ
L
:
(f)
μ
μ

→++
e
evv
μ
L
:
1 001, and
e
L
:
++
0110
159
(a)
ππ

0
p
(Baryon number is violated:
1 00 )
(b)
π
+→++
0
pp pp
(This reaction can occur)
(c)
π
+
ppp
(Baryon number is violated:
11 10 )
(d)
μ
v
(This reaction can occur)
(e)

e
ev
(This reaction can occur)
(f)
n
(Violates baryon number:
0 01, and violates
muonlepton number:
0
10 .
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This note was uploaded on 11/05/2009 for the course PHSX 313 taught by Professor Staff during the Fall '09 term at Kansas.
 Fall '09
 Staff

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