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Ch. 43
Page 1
CHAPTER 43 – Nuclear Energy; Effects and Uses of Radiation
Note:
A factor that appears in the analysis of energies is
e
2
/4p
Å
0
= (1.60
×
10
–19
C)
2
/4p(8.85
×
10
–12
C
2
/N
·
m
2
)= 2.30
×
10
–28
J
·
m = 1.44 MeV
·
fm.
1.
We find the product nucleus by balancing the mass and charge numbers:
Z
(X) =
Z
(
27
Al) +
Z
(n) = 13 + 0 = 13;
A
(X) =
A
(
27
Al) +
A
(n) = 27 + 1 = 28, so the product nucleus is
Al
13
28
.
If
Al
13
28
were a
β
+
emitter, the resulting nucleus would be
Mg
12
28
, which has too many neutrons relative to
the number of protons to be stable.
Thus we have a
–
emitter
.
The decay is
Al
13
28
→
Si
14
28
+e
– 1
0
+
ν
, so the product is
Si
14
.
2.
For the reaction
H
1
2
(d,n) He
2
3
, we find the difference of the initial and the final masses:
?
M
=
M
(
2
H) +
M
(
2
H) –
m
(n) –
M
(
3
He)
= 2(2.014102 u) – (1.008665 u) – (3.016029 u) = + 0.003510 u.
Thus
no threshold energy
is required.
3.
For the reaction
U
92
238
(n,
γ
)U
92
239
with slow neutrons, whose kinetic energy is negligible, we find the
difference of the initial and the final masses:
?
M
=
M
(
238
U) +
m
(n) –
M
(
239
U)
= (238.050782 u) + (1.008665 u) – (239.054287 u) = + 0.005160 u.
Thus no threshold energy is required, so the reaction is
possible
.
4.
For the reaction
Li
3
7
(p,
α
)He
2
4
, we determine the
Q
value:
Q
= [
M
(
7
Li) +
M
(
1
H) –
M
(
4
He) –
M
(
4
He)]
c
2
= [(7.016004 u) + (1.007825 u) – 2(4.002603 u)]
c
2
(931.5 MeV/u
c
2
) = + 17.35 MeV.
Thus
17.35 MeV is released
.
5.
For the reaction
Be
4
9
(
,n) C
6
12
, we determine the
Q
value:
Q
= [
M
(
9
Be) +
M
(
4
He) –
m
(n) –
M
(
12
C)]
c
2
= [(9.012182 u) + (4.002603 u) – (1.008665 u) – (12.000000 u)]
c
2
(931.5 MeV/u
c
2
) = + 5.701 MeV.
Thus
5.701 MeV is released
.
6. (
a
) For the reaction
Mg
12
24
(n, d) Na
11
23
, we determine the
Q
value:
Q
= [
M
(
24
Mg) +
m
(n) –
M
(
2
H) –
M
(
23
Na)]
c
2
= [(23.985042 u) + (1.008665 u) – (2.014102 u) – (22.989770 u)]
c
2
(931.5 MeV/u
c
2
) = – 9.469 MeV.
Because (
K
+
Q
) > 0, the reaction
can occur
.
(
b
) The energy released is
K
+
Q
= 10.00 MeV – 9.469 MeV =
0.53 MeV
.
7. (
a
) For the reaction
Li
3
7
(p,
2
4
, we determine the
Q
value:
Q
= [
M
(
7
Li) +
M
(
1
H) –
M
(
4
He) –
M
(
4
He)]
c
2
= [(7.016004 u) + (1.007825 u) – 2(4.002603 u)]
c
2
(931.5 MeV/u
c
2
) = + 17.35 MeV.
Because
Q
> 0, the reaction
can occur
.
(
b
) The kinetic energy of the products is
K
=
K
i
+
Q
= 2.500 MeV + 17.35 MeV =
19.85 MeV
.
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View Full DocumentCh. 43
Page 2
8. (
a
) For the reaction
N
7
14
(
α
,p) O
8
17
, we determine the
Q
value:
Q
= [
M
(
14
N) +
M
(
4
He) –
M
(
1
H) –
M
(
17
O)]
c
2
= [(14.003074 u) + (4.002603 u) – (1.007825 u) – (16.999131 u)]
c
2
(931.5 MeV/u
c
2
) = – 1.191 MeV.
Because (
K
+
Q
) > 0, the reaction
can occur
.
(
b
) The kinetic energy of the products is
K
=
K
i
+
Q
= 7.68 MeV – 1.191 MeV =
6.49 MeV
.
9.
For the reaction
O
8
16
(
,
γ
)N
e
10
20
, we determine the
Q
value:
Q
= [
M
(
16
O) +
M
(
4
He) –
M
(
20
Ne)]
c
2
= [(15.994915 u) + (4.002603 u) – (19.992440 u)]
c
2
(931.5 MeV/u
c
2
) =
+ 4.730 MeV
.
10. For the reaction
C
6
13
(d, n) N
7
14
, we determine the
Q
value:
Q
= [
M
(
13
C) +
M
(
2
H) –
m
(n) –
M
(
14
N)]
c
2
= [(13.003355 u) + (2.014102 u) – (1.008665 u) – (14.003074 u)]
c
2
(931.5 MeV/u
c
2
) = + 5.326 MeV.
The kinetic energy of the products is
K
=
K
i
+
Q
= 36.3 MeV + 5.326 MeV =
41.6 MeV
.
11. (
a
) We find the product nucleus by balancing the mass and charge numbers:
Z
(X) =
Z
(
6
Li) +
Z
(
2
H) –
Z
(
1
H) = 3 + 1 – 1 = 3;
A
(X) =
A
(
6
Li) +
A
(
2
H) –
A
(
1
H) = 6 + 2 – 1 = 7, so the product nucleus is
Li
3
7
.
(
b
) It is a “stripping” reaction because a
neutron is stripped from the deuteron
.
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 Spring '09
 Fielding
 Physics, Energy, Radiation

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