Fission and Fusion Book pg 286 - 287 ©cgrahamphysics.com 2016
Fission and Fusion Book pg 286 - 287
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Review
• BE is the energy that holds a nucleus together. This is equal to the mass defect of the nucleus.
• Also called separation energy. The energy required to decompose an atom, or nucleus into its constituent particles, equal to the energy equivalent of the mass defect.
• Elements with a high binding energy per nucleon are very difficult to break up.
• Iron 56 is close to the peak of the curve and has one of the highest binding energies per nucleon of any isotope.
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Nuclear fission and nuclear fusion
Nuclear fission is the splitting of a large nucleus into
two smaller (daughter) nuclei.
An example of fission is
In the animation, 235U is hit
by a neutron, and capturing
it, becomes excited and
unstable:
It quickly splits into two
smaller daughter nuclei, and
two neutrons.
Nuclear fission
235U + 1n (236U*) 140Xe + 94Sr + 2(1n) 92
0 92 54 38 0
140Xe 94Sr ©cgrahamphysics.com 2016
Nuclear fission and nuclear fusion
Note that the splitting was triggered by a single neutron that had just the right energy to excite the nucleus.
Note also that during the split, two more neutrons were released.
If each of these neutrons splits subsequent nuclei, we have what is called a chain reaction.
Chain reaction
235U + 1n (236U*) 140Xe + 94Sr + 2(1n) 92
0 92 54 38 0
Primary Secondary Tertiary
1 2 4 8
Exponential Growth
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Critical Mass • The smallest possible
amount of fissionable material that will sustain a chain reaction
• Too few neutrons released - reaction will slow down
• Too many neutrons released - reaction becomes uncontrolled - causes explosion - nuclear meltdown
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Atomic bomb – nuclear fision
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Fission
• Nucleon loses energy when split mass decreases high BE p.n.
• High nucleon number elements undergo fission
• Low nucleon number nuclides may undergo fusion they all try to reach the nuclide that is most stable: iron - 56
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Stars start to shine - fission
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Fission
• Fission brought on by absorption of neutrons is called induced fission
• Fission that happens if a nucleus splits is called spontaneous fission
• Induced fission is called neutron activation
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Neutron Activation
• Radioactive decay is spontaneously
• If atoms are bombarded with different particles, these change into new elements
• Neutrons are particularly effective in inducing artificial transmutation to produce radioactive isotopes
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Neutrons are produced
• Beryllium is bombarded with 𝛼 - particles
• 𝛼 + 𝐵𝑒 → 𝐶 + 𝑛01
612
49
24
Identify the bombarding particle:
• 𝑁 + 𝑛 → 𝐻 + 𝐶614
11
01
714
• 𝐿𝑖 +36 𝑛 + 𝑝1
1 → 𝛼24
01 + 𝛼2
4
• 𝑇𝑙 +81205 𝑝1
1 + 𝑛 → 𝑃𝑏 + 𝑛01
82206
01
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Liquid drop model The balance of electrostatic forces b/w protons and the short range nuclear force hold the nucleus together Absorbing a slow neutron the unstable atom U-236 is formed and the balance b/w forces is disturbed The long range Coulomb force of repulsion dominates now over short ranged nuclear force of attraction
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A possible fission reaction
• 𝑛 + 𝑈 → 𝐵𝑎 + 𝐾𝑟 + 2 𝑛01
3690
56144
92235
01
• n = 1.009u
• U - 235 = 235.044u
• Ba – 144 = 143.923u
• Kr – 90 = 89.920u
• Find the mass defect and energy equivalent
• ∆𝑚 = 235.044 + 1.009 − 143.923 + 89.920 + 2 𝑥 1.009 = 236.053 − 235.861 = 0.192𝑢
• E = 0.192 x 931.5 = 178.85 ~ 179 MeV
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Conservation laws govern fission • Momentum is conserved (relativistically)
• Total charge is conserved (charge of products is charge of reactants)
• The number of nucleons remains constant
• Mass – energy is conserved (∆𝐸 = 𝑚𝑐2)
The KE after a reaction is greater than the KE of
the initial neutron – the mass defect ∆m has been
changed into energy ∆𝐸.
• KE is due to energy of fired neutron
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Fusion • Two lighter nuclei fuse together
• Very high pressure and temperature are needed
• Nucleus must overcome Coulomb repulsion
• Nucleus must come under the influence of the strong nuclear force
If two nuclei with low atomic numbers could join together they would have - higher BE p.n - overall mass decrease - lose energy
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Example • For this reaction find the mass defect and the energy released
Mass of 𝐻2 = 2.014102u
Mass of 𝐻3 = 3.016049u
Mass of 𝐻𝑒4 = 4.002604u
Mass of n = 1.008665u ∆𝑚 = 2.014102 + 3.016049 − 4.002604 + 1.008665
= 5.030151 – 5.011269
= 0.01882u
BE = 0.01882 x 931.5 = 17.6MeV
• The energy released = KE of helium nucleus and neutron
• Advantage: no radioactive elements are produced
• Disadvantage: obtaining and maintaining high temperature and pressure to initiate fusion
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Revisit BE curve
- Nuclides with nucleon number ~ 60 are most stable
- High nucleon number nuclides undergo fission - Low nucleon number nuclides undergo fusion
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To reach most stable nuclide
• Fission: - loosing energy - increases BE of fission nuclei - becomes more stable
• Fusion - PE of new nucleus is less than that of individual atoms - increases BE of fused nucleus - becomes more stable
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Example 𝑈 + 𝑛 → 𝑆𝑟 + 𝑋𝑒 + 3 𝑛01
54146
3890
01
92238
Use the graph to compare the BE
238-U: 7.6 x 238 =1800MeV 90 – Sr: 8.7 x 90 = 780MeV 146 – Xe: 8.2 x 146 = 1200MeV 1200 + 780 = 1980MeV
The BE of fission nuclei is greater than BE of U – 238 losing energy increases BE nucleus more stable
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Fusion in the Sun • 𝑝:+ 𝑝:
𝐻12
• 𝐻12 + 𝑝:
𝐻23 𝑒
• 𝐻23 𝑒 + 𝐻2
3 𝑒 𝐻24 𝑒 +2 𝑝:
• For a complete cycle the first two
reactions must occur twice • Result:
- 1 helium nucleus - 2 positrones - 2 protons, which are available for future fusions - 2 neutrinos
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Example • 𝐻2
4 𝑒= 4.00260u
• 𝐻12 = 2.01410u
• 𝑛01 = 1.00867u
• 𝑝11 = 1.00728u
• 𝑒10 = 0.00055u
a) What is the mass of the 𝐻24 𝑒 nucleus?
b) What is the BE of the 𝐻24 𝑒 nucleus?
c) What is the BE p.n. for a 𝐻24 𝑒 nucleus?
d) What is the BE p.n. for a 𝐻12 nucleus?
e) When two deuterium nuclei fuse to form a helium nucleus how much energy is released?
a) 𝑚𝐻𝑒 = 4.00260 × 931.5 = 3728.4𝑀𝑒𝑉𝑐;2
b) 2p + 2n 𝐻𝑒24
∆𝑚 = 2 × 1.00728 + 2 × 1.00867 × 931.5 − 3728.4 = 3755.7 − 3728.4 = 27.3 𝑀𝑒𝑉𝑐;2 BE = 27.3 MeV c) BE =
𝐸
𝐴=
27.3
4= 6.8𝑀𝑒𝑉
d) 1p + 1n 𝐻12
∆𝑚 = 1.00728 + 1.00867 − 2.01410 = 2.01595 − 2.01410 = 0.00185𝑢
BE= 0.00185 x 931.5 = 1.723MeV and BE p.n. = 𝐸
𝐴=
1.273
2= 0.86𝑀𝑒𝑉
e) 𝐻12 + 𝐻1
2 𝐻24 𝑒
∆𝑚 = 2 𝑥 2.01410 − 4.00260 = 0.0256𝑢
BE = 0.0256 x 931.5 = 23.8 ~ 24MeV
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Example • Determine the number x of neutrons produced and calculate the energy
released in the following fission reaction:
𝑈 + 𝑛 → 𝐵𝑎 + 𝐾𝑟 + 𝑋 𝑛01
3690
56144
01
92235
• 235 – U = 235.043929u 144 – Ba = 143.922952u 90 – Kr = 89.919516u 1 – n = 1.008665u
• Solution
• Mass left side: 236
• Mass right side: 144 + 90 = 234
• Hence x = 2
• ∆𝑚 =235.043929 + 1.008665 −143.922952 + 89.919516 + 2 × 1.008665
= 236.052594 − 235.859798 = 0.192796u
• E = 0.192796 x 931.5 = 179.6MeV ~ 180 MeV
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