44.1:
a)
2
2
2
2
1547
.
0
1
1
1
mc
c
v
mc
K
=


=
J
10
27
.
1
so
kg,
10
109
.
9
14
31


×
=
×
=
K
m
b) The total energy of each electron or positron is
=
=
+
=
2
2
1547
.
1
mc
mc
K
E
J.
10
46
.
9
14

×
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
J.
10
46
.
9
14

×
c)
pm
10
.
2
)
J
10
46
.
9
(
λ
so
λ
14
=
×
=
=
=

hc
E
hc
hc
E
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
MeV.
5.51
MeV
0.511
MeV
00
.
5
2
=
+
=
+
=
mc
K
E
We can calculate the speed of the positron from Eq. 37.38
.
996
.
0
MeV
5.51
MeV
511
.
0
1
1
1
2
2
2
2
2
2
=

=

=
⇒

=
E
mc
c
v
mc
E
c
v
44.3:
Each photon gets half of the energy of the pion
ray.
gamma
m
10
8
.
1
Hz
10
7
.
1
s
m
10
00
.
3
λ
Hz
10
7
.
1
s)
J
10
63
.
6
(
)
eV
J
10
6
.
1
(
)
eV
10
9
.
6
(
MeV
69
MeV)
511
.
0
(
)
270
(
2
1
)
270
(
2
1
2
1
14
22
8
22
34
19
7
2
e
2
γ



×
=
×
×
=
=
⇒
×
=
⋅
×
×
×
=
=
⇒
=
=
=
=
f
c
h
E
f
c
m
c
m
E
π
44.4:
a)
)
s
m
10
(3.00
kg)
10
(9.11
(207)
s)
J
10
626
.
6
(
λ
8
31
34
2
×
×
⋅
×
=
=
=
=


c
m
h
c
m
hc
E
hc
μ
μ
pm.
0.0117
m
10
17
.
1
14
=
×
=

In this case, the muons are created at rest (no kinetic energy).
b) Shorter wavelengths
would mean higher photon energy, and the muons would be created with nonzero kinetic
energy.
44.5:
a)
e
e
e
63
207
270
m
m
m
m
m
m
=

=

=
∆
+
+
μ
π
MeV.
32
MeV)
511
.
0
(
63
=
=
⇒
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.
44.6:
a) The energy will be the proton rest energy, 938.3 MeV, corresponding to a
frequency of
Hz
10
27
.
2
23
×
and a wavelength of
m.
10
32
.
1
15

×
b) The energy of
each photon will be
MeV,
1768
MeV
830
MeV
3
.
938
=
+
with frequency
Hz
10
8
.
42
22
×
and wavelength
m.
10
02
.
7
16

×
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44.7:
J.
10
20
.
7
)
s
m
10
00
.
3
(
)
kg
400
kg
400
(
)
(
19
2
8
2
×
=
×
+
=
∆
=
c
m
E
44.8:
n
C
Be
He
1
0
12
6
9
4
4
2
+
→
+
We take the masses for these reactants from Table 43.2, and use Eq. 43.23
reaction.
exoergic
an
is
This
MeV.
701
.
5
)
u
MeV
(931.5
u)
1.008665
u
12.000000
u
9.012182
u
002603
.
4
(
=


+
=
Q
44.9:
He
Li
B
n
4
2
7
3
10
5
1
0
+
→
+
MeV
79
.
2
)
u
MeV
(931.5
u)
(0.002995
u;
0.002995
u
11.018607
u
4.002603
u
7.016004
He)
Li
(
u
11.021602
u
10.012937
u
1.008665
B)
n
(
4
2
7
3
10
5
1
0
=
=
∆
=
+
=
+
=
+
=
+
m
m
m
The mass decreases so energy is released and the reaction is exoergic.
44.10:
a) The energy is so high that the total energy of each particle is half of the
available energy, 50 GeV.
b) Equation (44.11) is applicable, and
MeV.
226
a
=
E
44.11:
a)
q
mf
q
m
B
m
B
q
π
ϖ
ϖ
2
=
=
⇒
=
T
18
.
1
C
10
60
.
1
Hz)
10
(9.00
)
u
kg
10
(1.66
u)
01
.
2
(
2
19
6
27
=
⇒
×
×
×
=
⇒


B
B
π
b)
J
10
47
.
5
)
u
kg
10
66
.
1
(
)
u
01
.
2
(
2
m)
(0.32
T)
(1.18
C)
10
60
.
1
(
2
13
27
2
2
2
19
2
2
2



×
=
×
×
=
=
m
R
B
q
K
.
s
m
10
81
.
1
)
u
kg
10
(1.66
u)
(2.01
J)
10
47
.
5
(
2
2
and
MeV
3.42
eV
10
3.42
7
27
13
6
×
=
×
×
=
=
=
×
=


m
K
v
44.12:
a)
)
c
s.
m
10
12
.
3
)
b
s.
10
97
.
3
2
7
7
×
=
=
×
=
=
=
m
eBR
R
m
eB
f
ϖ
π
π
ϖ
For three
figure precision, the relativistic form of the kinetic energy must be used,
,
)mc
(γ
eV
2
1

=
V.
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 Fall '07
 Nearing
 Physics, Energy

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