PHY 251 Fall 2009: homework problem set 11, due in the PHY 251 drop box in room A129 by
noon on Friday, Dec. 11.
1. Serway 14.3
Answer:
The reaction is
9
4
Be(
α
,n)
12
6
C or
9
4
Be
+
4
2
He
→
12
6
C
+
1
0
n
The mass change is
Δ
m
= 9
.
012 182 + 4
.
002 603
−
12
.
000 000 + 1
.
008 665 = 0
.
006 120
u
so
Q
= Δ
m c
2
= (0
.
006 120
u
)
·
(931
.
49
MeV
/c
2
) = 5
.
70
MeV.
2. Serway 14.7
Answer:
The value of
Q
for
4
2
He
+
14
7
N
→
17
8
O
+
1
1
H is
Q
= (4
.
002 603 + 14
.
003 074
−
16
.
999 131
−
1
.
007 825)
931
.
494
MeV
/c
2
amu
c
2
= 1
.
19
MeV
.
Therefore we need the incoming
α
particle to have a kinetic energy of at least 1.19 MeV to
drive the reaction. Next, we want to find the
Q
of
1
1
H
+
7
3
Li
→
4
2
He
+
4
2
He, which is
Q
= (1
.
007 825 + 7
.
016 004
−
4
.
002 603
−
4
.
002 603)
931
.
494
MeV
/c
2
amu
c
2
= 17
.
35
MeV
so the alpha particles that emerge are very energetic.
3. Serway 14.8
Answer:
The reaction
4
2
He
+
9
4
Be
→
12
6
C
+
1
0
n
has
Q
= (4
.
002 603 + 9
.
012 182
−
12
.
000 000
−
1
.
008 665
u
)
·
931
.
49
MeV
/c
2
u
= 5
.
70
MeV
as calculated in problem 14.3. The reaction
2
1
H
+
2
1
H
→
3
2
He
+
1
0
n
has
Q
= (2
.
014 102 + 2
.
014 102
−
3
.
016 029
−
1
.
008 665
u
)
·
931
.
49
MeV
/c
2
u
= 3
.
27
MeV
which is positive (meaning mass is lost) so it is exothermic (releases heat).
1
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4. Serway 14.22
Answer:
In example 14.4 on page 513 of Serway, we find that one
235
U fission event releases
208 MeV. Therefore the number
x
of fission events needed per day to generate 1000 MW is
x
= (
1
decay
208
×
10
6
eV
)
·
(
eV
1
.
602
×
10

19
Joules
)
·
(
10
9
Joules
sec
)
·
(
24
·
60
·
60
sec
1
day
)
or
x
= 2
.
59
×
10
24
decays per day. The mass
m
of the number of
235
U nuclei which must
undergo fission in a day is then
m
= (2
.
59
×
10
24
atoms
)
·
(235
.
043
amu
atom
)
·
(1
.
661
×
10

27
kg
amu
) = 1
.
01
kg
.
Because density
ρ
is mass per volume, we have
ρ
=
m
(4
/
3)
πr
3
r
=
parenleftBigg
3
m
4
πρ
parenrightBigg
1
/
3
=
parenleftBigg
3 (1010
g
)
4
π
(19
.
0
g/cm
3
)
parenrightBigg
1
/
3
= 2
.
33
cm
or a sphere with a diameter of less than two inches.
5. Serway 14.24
Answer:
The estimate for the total mass
m
of recoverable
235
U is
m
= (10
9
tons U
)
·
(
2000
pounds
ton
)
·
(
1
kg
2
.
204
pounds
)
·
(
0
.
007
235
U
U
) = 6
.
35
×
10
9
kg
The total amount of energy
E
that can be released from this is
E
= (6
.
35
×
10
9
kg
)
·
(
1
amu
1
.
661
×
10

27
kg
)
·
(
1
atom
235
.
043
amu
)
·
(
208
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 Fall '01
 Rijssenbeek
 Physics, Energy, Mass, Work, atoms, Joule, MeV, ev

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