Nuclear Binding Energy
Consider process of forming a C12 atom from protons, neutrons, electrons:
6p + 6n + 6e →
12
6
C
Mass of
12
6
C :
12.00000 u
(exactly 12)
or 12.0000
Masses of individual particles:
proton
1.00727 u or
neutron
1.00866 u or
electron
0.00055 u or
6p + 6n + 6e = 12.0989 u or
“Loss” of mass = 0.0989 u or
What is the explanation for this apparent violation of the Law of Conservation of Matter?
How
can mass be lost?
Energymass equivalence expressed by Einstein formula:
E = mc
2
.
E corresponds to energy released in forming the nucleus from its particles. This is the
binding
energy
of the nucleus
m is the “loss” of mass in the process.
What is the binding energy of
12
6
C according to this data?
E =
1000
0989
.
0
(2.998 x 10
8
)
2
= 8.89 x 10
12
=
8.89 x 10
9
Binding Energy could be expressed as MeV where
1 Mev = 1.602 x 10
13
Joule
Mev is on a per nucleus basis (not per mol)
8.89 x 10
12
x
nuclei
10
x
022
.
6
mol
1
23
x
J
10
x
602
.
1
Mev
1
13

= 92.1 MeV
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(atomic mass unit)
1.00 u x
u
kg
10
x
6607
.
1
27

x (2.998 x 10
8
)
2
= 1.4924 x 10
10
J
1.4924 x 10
10
J x
J
10
x
602
.
1
MeV
1
13

= 931.5 MeV
So each atomic mass unit is equivalent to 931.5 Mev of energy.
Knowing this conversion factor:
931.5
u
Mev
, we could directly solve for the binding energy of
12
6
C.
The “loss” of mass was 0.0989 u.
Therefore:
0.0989 u x 931.5
u
Mev
= 92.1 Mev
To compare the binding energy of different nuclei, we calculate
where “nucleon” means any nuclear
particle, proton or neutron.
For carbon12, there are 6 protons and 6 neutrons: 12 nucleons
For carbon12:
Nucleon
energy
Binding
=
nucleon
12
Mev
1
.
92
= 7.68
Figure 19.7 of the text shows a curve of binding energy, plotting binding energy per nucleon vs.
mass number.
The curve rises steeply at first, maximizing on mass number 56 (iron) which is the
nucleus with the highest binding energy.
The curve then gradually decreases, meaning that heavier
nuclei than iron are less strongly held together.
Consider an ordinary chemical reaction, such as the combustion of methane, CH
4
CH
4
(g) + 2O
2
(g) → CO
2
(g) + 2H
2
O(g)
∆
H = 802 kJ
Would there be any “loss” of mass due to E = mc
2
in this reaction?
The binding energy of
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 Spring '10
 Kimmel
 Nuclear Fission, Nuclear Fusion, Neutron, Binding energy

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