PHY 251 Fall 2009: homework problem set 3, due PHY 251 drop box in room A129 by noon on
Friday, Sep. 25.
1. A proton with a kinetic energy of 10 GeV collides with an antiproton at rest. They annihilate
each other, and produce two identicalenergy gamma rays (very energetic photons) that come
out at angles
θ
above and below, respectively, the direction of the proton’s motion. Calculate
the energy
E
and momentum
p
of the gamma rays, and the angle
θ
.
Answer:
Conservation of energy gives
(
E
K
+
m
p
c
2
) +
m
p
c
2
= 2
E
γ
because we go from a proton with kinetic energy
E
K
= 10
GeV and the massenergy of
the proton plus the antiproton (same mass as proton), to two photons of identical energy
E
γ
.
Since
m
p
= 938
.
3
MeV/
c
2
we have
E
γ
= (10 + 0
.
9383 + 0
.
9383)
/
2 = 5
.
94
GeV. Note for
the proton we have
E
K
= (
γ
−
1)
m
p
c
2
γ
= 1 +
E
k
m
p
c
2
= 1 +
10
0
.
9383
= 11
.
66
from which we can find
β
=
r
1
−
1
/γ
2
= 0
.
996
. At the same time, conservation of
momentum gives
p
p
= 2
p
γ
cos
θ
γm
p
v
= 2
E
γ
c
cos
θ
γβm
p
c
2
= 2
E
γ
cos
θ
cos
θ
=
γβm
p
c
2
2
E
γ
θ
= cos
−
1
p
γβm
p
c
2
2
E
γ
P
= cos
−
1
p
11
.
66
·
0
.
996
·
0
.
9383
2
·
5
.
94
P
= cos
−
1
(0
.
703) = 23
.
5
◦
2. A atom of the radioactive isotope of carbon
14
C (mass
14
.
003 242
amu, where 1 amu=931.494
MeV/
c
2
) decays into an electron (mass
0
.
000 549
amu) and a singly ionized
14
N atom (mass
14
.
003 074
−
0
.
000 549
amu). If the
14
C atom were at rest, and no other particles were in
volved, calculate the total energy released by this decay, and the energy and momentum of
the electron and
14
N atoms after the decay.
Answer:
Let’s have
m
C
= 14
.
003 242
amu represent the mass of the
14
C atom before the
decay,
m
N
= 14
.
003 074
−
0
.
000 549
am represent the mass of the
14
N atom minus an elec
tron after the decay, and
m
e
= 0
.
000 549
amu represent the mass of the electron. The total
energy released in the decay is given by
Δ
E
c
2
= Δ
m
= [
m
C
−
m
N
−
m
e
] = [14
.
003 242
−
(14
.
003 074
−
0
.
000 549)
−
0
.
000 549]
amu
= [14
.
00 3242
−
14
.
003 074]
amu
= 0
.
000 168
amu
or
Δ
E
= (0
.
000 168
amu
)
·
(
931
.
494
MeV
/c
2
amu
)
·
c
2
= 0
.
156
MeV
1
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View Full Documentor 156 keV. Now even if all of that energy went to the electron it would have a value of
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 Fall '01
 Rijssenbeek
 Physics, Energy, Kinetic Energy, Work, Photon, Light, Compton

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