Nuclear Reactions
When two nucleon combine to form two or
more nuclear particles
+→ +
Laws
1. Conservation of nucleons
2. Conservation of charge
3. Conservation of energy
4. Conservation of momentum
Q of the reaction
• Recall: 1u = 1.66 x 10-24g = 932 MeV
• So
( = + − ! + " 932,• If Q>0 : Net increase in KE reaction is
exothermic
• If Q<0 : Net decrease in KE reaction is
endothermic
• Ex… Find the Q of the reaction
• Solution..
M(N14) = 14.0067
M(n) = 1.0086
15.0153 u
M(C14)=14.0032
M(H1) = 1.0079
15.0111 u
• Q = (+0.0042)932MeV = +3.9 MeV exothermic
Conservation of Energy
+ + + =
=
! +! + " + " • Rearranging
(! +" ) − + = [ + − ! + " ] • Define the Q of the reaction as the change in
the rest mass energy in MeV
( = [ + − ! + " ] • Ex… Complete the following reaction
/0. + /1 → ? + /3
1. Z (atomic #) for Nitrogen = 7
2. Z for n = 0
3. Z for H = 1
4. Z for ? = 7+0=?+1 => ? = 6=C
5. A for N = 14
6. A for n = 1
7.A for H = 1
8. A for ? = 14+1=?+1 => ? =14
/0. + /1 → /0 + /3
Binding Energy and Mass Defect
• The mass of all nuclei are somewhat smaller
than the sum of the masses of the neutrons
and protons contained in them. This mass
defect (∆) is
• ∆ = 56 + .7 − 8
mass of
protons
mass of
neutrons
mass of
nucleus
• Mass is converted to KE
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BE and mass defect
Binding Energy per Nucleon (BE/A)
• When ∆ is expressed in energy units, it is equal to the
energy that is necessary to break the nucleus into its
component nucleons. This amount of energy is called
the Binding Energy (BE) since it represents the energy
necessary to hold the nucleus together.
• Binding Energy (BE) = Mass Defect (∆) expressed in
energy units.
• Note ∆ = 5 6 + 9: + .7 − (8 + 59: )
mass of
mass of
neutral /3
neutral atom
Thus we can use the tabulated masses of neutral atoms
to compute ∆, Q and BE.
• The total BE of a nuclei is an increasing function of
the atomic number A. However, it does not increase
at a constant rate.
• ;<⁄8 = ∆⁄8 is the average BE per nucleon. It
decreases for large A.
BE/A
Q in terms of BE
• BE/A is a measure of stability of the nucleus. A
large BE/A means strong bonds and stability.
• Because of the shape of the BE/A curve it is
possible for a heavy nuclide to split (fission)
into two lighter fragments that each have a
larger BE/A and release energy in the process.
• Recall + → + • where ∆ = 5 ( /3) + . 7 − so
= 5 ( /3) + . 7 − ∆()
and = 5 ( /3) + . 7 − ∆()
and ! = 5! ( /3) + .! 7 − ∆()
and " = 5" ( /3) + ." 7 − ∆()
Since ( = [ + − ! + " ] Thus
( = > + > BE and ∆
( = > + > − [> + > ]
• If the total BE of the products (c + d) is greater
than BE of the reactants (a +b), then Q is
positive and the reaction is exothermic.
• Therefore, whenever a more stable
configuration is achieved by combining less
stable nucleons energy is released!
− [> + > ]
Fission
• In the region of large A, a more stable configuration
is formed when the heavy nucleus splits.
• Consider BE/A(U-238) = 7.5 MeV
and
BE/A(?-119) = 8.4 MeV
@A? → //B? + //B?
so if
Then there will be a gain of ~0.9 MeV per nucleon for a
total of 238 x 0.9MeV = 214 MeV per fission.
This is the source of energy in nuclear reactors.
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Chapter 2 Summary
Fusion
@
/
• Consider 2 3 → 3 + 3
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deuterium
tritium
Two deuterons have BE(H-2)= 2 x 2.23 MeV
Form @3 with BE(H-3) = 8.48 MeV
and BE(H-1) = 0 (since only one nucleon)
Q = 8.48 – 2 x 2.23 = +4.02 MeV which
appears as the KE of @3 and H.
Two light less stable nuclei produce a heavier
but more stable nucleus.
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Fundamental Particles
Nuclides and isotopes
Atom percent and weight percent
Atom density
Mass and energy
Radioactive decay
Half life
Nuclear reactions
Mass defect, Binding energy , & Q of reaction
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