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nuclear binding energyemre özyetis 10-U
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The binding energy is the energy required to
decompose the nucleus into protons and neutrons.
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In chemical reactions, the nucleus remains the
same no new atoms are formed. The number and
kinds of atoms are conserved. There is no extra
energy given off from nucleus besides the energy
given off or taken in by the breakage or forming
of bonds.
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However, in nuclear reactions, divisions and
combinations of nuclei occur. New atoms are
formed rays and particles with energy are
emitted.
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When new atoms are formed, binding energy is
given off to keep the neutrons and protons
together and to become stable. This energy can
be calculated by Einsteins famous formula.
Emc2 energymass(speed of light) 2
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According to The Relativity Theory of Einstein,
some of the mass is converted into energy and
emitted so that more stable atoms are formed.
This is why in nuclear reactions energy and mass
are conserved together.
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?E?mc2
Where ? E is binding energy ?m is mass defect
(m n0 m p)- m element itself c is speed of
light in vacuum which is 2.99108 ms-1
When units are kg for mass defect and m.s-1 for
speed of light, then we get kgm2s-2 which is
Joule.
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To make comparison between different nuclides
easily, we express binding energies on a
per-nucleon. Example.1b will give information
about BE/nucleon
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Example.1 Calculate a) the change in energy if
1 mol O nuclei were formed from neutrons and
protons. b) Binding Energy per nucleon.
16 8
1 0
Mass of n 1.6749310-24 g Mass of p
1.6726210-24 g Mass of O nucleus is
2.6553510-23g
1 1
16 8
?For ?m we ought to know the equation of
formation 8 n 8 p
O
16 8
1 0
1 1
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In nucleus the difference is (Mass of O) (Mass
of 8p and 8n) -2.26910-25g The difference in
1 mole is (6.0221023nuclei/mol)
-2.26910-25g/nucleus ?m -0.1366 g/mol ?E
?mc2 ?E -0.1366 g/mol (3.00108m/s)2 1
kg/1000g ?E -1.231013 J/mol
a)
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b)
In order to find ?E per nucleon, we need to know
?E per nucleus first
-1.231013 J/mol
?E per O nucleus
-2.0410-11
J/nucleus
6.0221023nuclei/mol
In terms of a more convenient energy unit, a
million electron volts where 1 MeV 1.60
10-13 J
-2.0410-11 J/nucleus -1.28 102 MeV/nucleus
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At last we can calculate ?E per nucleon by
dividing by the sum of neutrons and protons
which is 16 for O ?E per nucleon for O
is
16 8
-1.28 102 MeV/nucleus

16 nucleons/nucleus
16 8
-7.98 MeV/nucleon
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Example.2 In the radioactive decomposition of
radium-226, the equation for the nuclear process
is Ra Rn He How much mass is
converted into energy during this radioactive
decay process? Ra 226.025360 amu Rn 222.017530
amu He 4.00260361 amu
226 88
222 86
4 2
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?m m produced-m reacted ?m( 222.017530amu
4.00260361amu ) - 226.025360amu ?m -5.22610-3
amu is lost in this process. This means that
amount of mass is converted into energy
?E?mc2 ?E (-5.22610-3 g/mol) 1 kg/1000g
(3.00108 m/s)2 ?E -4.701011 J/mol Negative
sign indicates that the process is exothermic
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The binding energy curve is obtained by dividing
the total nuclear binding energy by the number of
nucleons. The fact that there is a peak in the
binding energy curve in the region of stability
near iron means that either the breakup of
heavier nuclei (fission) or the combining of
lighter nuclei (fusion) will yield nuclei which
are more tightly bound (less mass per nucleon).