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438
Chapter 43
The Coulomb potential energy of the two reactant nuclei at this separation is
12
00
11
(
4
)
44
qq
e e
U
rr
ππ
==
PP
19
2
92
2
15
19
4(1.602 10
C)
(8.988 10 N m /C )
1.4 MeV
(4.0 10
m)(1.602 10
J/eV)
U
−
−−
×
=×
⋅
=
××
This is an estimate of the threshold energy for this reaction.
EVALUATE:
The reaction releases energy but the total initial kinetic energy of the reactants must be 1.4 MeV in
order for the reacting nuclei to get close enough to each other for the reaction to occur. The nuclear force is strong
but is very shortrange.
43.44.
IDENTIFY and SET UP:
0.7% of naturally occurring uranium is the isotope
235
U .
The mass of one
235
U nucleus
is about 235
m
p
.
EXECUTE: (a)
The number of fissions needed is
19
29
61
9
1.0 10 J
3.13 10
(200 10 eV)(1.60 10
J/eV)
−
×
. The mass of
235
U required is
29
5
p
(3.13 10 )(235
) 1.23 10 kg
m
×=
×
.
(b)
5
7
2
1.23 10 kg
1.76 10 kg
0.7 10
−
×
×
EVALUATE:
The calculation assumes 100% conversion of fission energy to electrical energy.
43.45.
IDENTIFY
and
SET UP:
The energy released is the energy equivalent of the mass decrease.
1 u is equivalent to
931.5 MeV.
The mass of one
235
U nucleus is 235
m
p
.
EXECUTE: (a)
235
1
144
89
1
92
0
56
36
0
Un
B
a K
r3
n
+→
+ +
We can use atomic masses since the same number of electrons are included on each side of the reaction equation
and the electron masses cancel.
The mass decrease is
( ) () ( ) ( ) ()
235
1
144
89
1
92
0
56
36
0
B
a
K
r
3
n
Mm
m
m
m
m
⎡⎤
Δ=
+
−
+
+
⎣⎦
235.043930 u 1.0086649 u 143.922953 u
88.917630 u
3(1.0086649 u)
M
+
−
−
−
0.1860 u
M
.
The energy released is (0.1860 u)(931.5 MeV/u) 173.3 MeV
=
.
(b)
The number of
235
U nuclei in 1.00 g is
3
21
p
1.00 10 kg
2.55 10
235
m
−
×
.
The energy released per gram is
21
23
(173.3 MeV/nucleus)(2.55 10 nuclei/g)
4.42 10 MeV/g
×
.
43.46.
(a)
28
24
14
12
Si
Mg
X.
24
28 so
4.
12 14 so
2.
A
Z
AA
ZZ
X
γ
+
⇒
++
=
=
+
=
=
is an
α
particle.
(b)
2
(23.985042 u
4.002603 u
27.976927 u) (931.5 MeV u)
9.984 MeV
γ
Em
c
=−Δ
=
+
−
=
43.47.
The energy liberated will be
347
224
( He)
( He)
( Be)
(3.016029 u
4.002603 u
7.016929 u)(931.5 MeV u) 1.586 MeV.
MMM
+−=
+
−
=
43.48.
(a)
3205
a
n
d
4711
0
.
ZA
=+−=
(b)
The nuclide is a boron nucleus, and
3
He
Li
n
B
3.00 10 u,
mmm
m
−
+−−=
− ×
and so 2.79 MeV of energy is absorbed.
43.49.
Nuclei:
4(2
) 4
2
22
XYH
e
AZ
A
Z
+−−
+
+
−
→+
.
Add the mass of
Z
electrons to each side and we find:
(X
)
( Y
)
(H
e
)
,
mM
M
M
−
−
−
−
where now we have the mass of the neutral atoms. So as long as the mass of
the original neutral atom is greater than the sum of the neutral products masses, the decay can happen.
43.50.
Denote the reaction as
1
XY
e
.
−
+
The mass defect is related to the change in the neutral atomic masses by
XeY
e
eX
Y
[]
[
(
1
)
]
(
)
,
mZ
m m Z mm mm
−
+−
=
−
where
X
m
and
Y
m
are the masses as tabulated in, for instance, Table (43.2).
43.51.
(1
)
1
A Z
β
+
+
−
.
Adding (
Z
–1) electrons to both sides yields
1
−
.
So in terms of masses:
()(
)
(
)
()
(
)
1e
e
1
e
X
Y
X
Y
2
.
A
A
A
A
Z
Z
Z
Z
m
M
M
m
M
M
m
+
−
Δ
=
=
−
=
So the decay will
occur as long as the original neutral mass is greater than the sum of the neutral product mass and two electron masses.
43.52.
IDENTIFY and SET UP:
.
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 Spring '06
 Buchler
 Physics, Energy, Potential Energy

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