44
1
P
ARTICLE
P
HYSICS AND
C
OSMOLOGY
44.1.
(a)
I
DENTIFY
and
S
ET
U
P
:
Use Eq.(37.36) to calculate the kinetic energy
K
.
E
XECUTE
:
2
2
2
2
1
1
0.1547
1
/
K
mc
mc
v
c
⎛
⎞
=
−
=
⎜
⎟
−
⎝
⎠
31
14
9.109
10
kg, so
1.27
10
J
m
K
−
−
=
×
=
×
(b)
I
DENTIFY
and
S
ET
U
P
:
The total energy of the particles equals the sum of the energies of the two photons.
Linear momentum must also be conserved.
E
XECUTE
:
The total energy of each electron or positron is
2
2
13
1.1547
9.46
10
J.
E
K
mc
mc
−
=
+
=
=
×
The total
energy of the electron and positron is converted into the total energy of the two photons. The initial momentum of
the system in the lab frame is zero (since the equalmass particles have equal speeds in opposite directions), so the
final momentum must also be zero. The photons must have equal wavelengths and must be traveling in opposite
directions. Equal
λ
means equal energy, so each photon has energy
14
9.46
10
J.
−
×
(c)
I
DENTIFY
and
S
ET
U
P
:
Use Eq. (38.2) to relate the photon energy to the photon wavelength.
E
XECUTE
:
/
E
hc
λ
=
so
14
/
/(9.46
10
J)
2.10 pm
hc E
hc
λ
−
=
=
×
=
E
VALUATE
:
The wavelength calculated in Example 44.1 is 2.43 pm. When the particles also have kinetic energy,
the energy of each photon is greater, so its wavelength is less.
44.2.
The total energy of the positron is
2
5.00 MeV
0.511MeV
5.51MeV.
E
K
mc
=
+
=
+
=
We can calculate the speed of the positron from Eq.(37.38):
2
2
2
2
2
2
0.511 MeV
1
1
0.996.
5.51 MeV
1
mc
v
mc
E
c
E
v
c
⎛
⎞
⎛
⎞
=
⇒
=
−
=
−
=
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
−
44.3.
I
DENTIFY
and
S
ET
U
P
:
By momentum conservation the two photons must have equal and opposite momenta.
Then
E
pc
=
says the photons must have equal energies. Their total energy must equal the rest mass energy
2
E
mc
=
of the pion. Once we have found the photon energy we can use
E
hf
=
to calculate the photon frequency
and use
/
c
f
λ
=
to calculate the wavelength.
E
XECUTE
:
The mass of the pion is
e
270
,
m
so the rest energy of the pion is
270(0.511 MeV)
138 MeV.
=
Each
photon has half this energy, or 69 MeV.
6
19
22
34
(69
10
eV)(1.602
10
J/eV)
so
1.7
10
Hz
6.626
10
J s
E
E
hf
f
h
−
−
×
×
=
=
=
=
×
×
⋅
8
14
22
2.998
10
m/s
1.8
10
m
18 fm.
1.7
10
Hz
c
f
λ
−
×
=
=
=
×
=
×
E
VALUATE
:
These photons are in the gamma ray part of the electromagnetic spectrum.
44.4.
(a)
The energy will be the proton rest energy, 938.3 MeV, corresponding to a frequency of
23
2.27
10
Hz
×
and a
wavelength of
15
1.32
10
m.
−
×
(b)
The energy of
each photon will be
938.3 MeV
830 MeV
1768 MeV,
+
=
with frequency
22
42.8
10
Hz
×
and
wavelength
16
7.02
10
m.
−
×
44.5.
(a)
e
e
e
270
207
63
m
m
m
m
m
m
π
μ
+
+
Δ
=
−
=
−
=
63(0.511MeV)
32 MeV.
E
⇒
=
=
(b)
A positive muon has less mass than a positive pion, so if the decay from muon to pion was to happen, you
could always find a frame where energy was not conserved. This cannot occur.
44
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442
Chapter 44
44.6.
(a)
34
14
2
31
8
(6.626
10
J s)
1.17
10
m
0.0117 pm
(207)(9.11 10
kg)(3.00
10 m s)
μ
μ
hc
hc
h
E
m c
m c
λ
−
−
−
×
⋅
=
=
=
=
=
×
=
×
×
In this case, the muons are created at rest (no kinetic energy).
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 Spring '08
 Jabbour
 Materials Science And Engineering

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