ch 31 1 chapter 31 nuclear energy effects and uses of radiation © 2006, b.j. lieb some figures...
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Ch 31 1
Chapter 31
Nuclear Energy
Effects and Uses of Radiation
© 2006, B.J. LiebSome figures electronically reproduced by permission of Pearson
Education, Inc., Upper Saddle River, New Jersey Giancoli, PHYSICS,6/E © 2004.
Ch 31 2
Nuclear Reactions
Nuclear reactions occur when a nucleus is struck by a particle or other nucleus.
Examples:pCNn 14
6147
HONHe 11
178
147
42
•The second reaction was observed by Rutherford and is the first nuclear reaction ever observed.
•It should be noted that in the first reaction, the neutron can enter the nucleus with very little energy but in the second reaction the 4He is repelled by the nucleus and thus has to overcome the Coulomb barrier in order to come close enough to cause a nuclear reaction.
Ch 31 3
Strong Nuclear Force & Coulomb Barrier
•The above graph (from Ch 30) shows the potential energy felt by the 4He as a result the strong nuclear force and the Coulomb force
•It is repelled by the Coulomb force ( 1/r2)
•The strong nuclear force is strong but short range, so it is not a factor until the 4He is essentially in contact with the nucleus
•This Coulomb barrier is felt by any positively charged particle approaching the nucleus
Ch 31 4
Overcoming the Coulomb Barrier
• If an incoming positive particle has enough energy, it will overcome the Coulomb Barrier and cause a nuclear reaction
• If there is not enough energy, it can still get to the nucleus by a process called tunneling
• Tunneling occurs because Quantum Mechanics predicts that the Wave Function of the incoming 4He is not zero at the barrier and thus there is a small probability that it can come closer to the nucleus than classical physics permits.
• Tunneling also occurs in alpha decay because the alpha particle doesn’t have enough energy to get over the Coulomb barrier
• After many tries the alpha particle eventually tunnels through the barrier.
Ch 31 5
Q-value
For the nuclear reaction
a + X → Y + b
we define the Q of the reaction as
Q = ( Ma + MX – Mb – MY ) c2
if Q > 0 exothermic reaction (energy released)
if Q < 0 endothermic (total KE less than initial KE)
Ch 31 6
Nuclear Fission
The curve of binding energy shows that energy can be released if a very heavy nucleus (A ~ 236 ) splits in half. The binding energy per nucleon increases from 7.6 MeV to 8.5 MeV and this results in the release of ≈ 200 MeV.
Ch 31 7
Fission of 235UNucleus absorbs a neutron n + 235U → 236U (excited state)
It takes on barbell shape because this reduces effect of Coulomb repulsion
Nucleus splits with the release of an average of two neutrons and about 200 MeV energy
Nuclei produced are called fission fragments. The mass split is usually 60%-40% but different each time.
The number of neutrons also differs.
Ch 31 8
Chain Reaction
•Fission occurs spontaneously but very slowly
•The fact that several neutrons are released allows for chain reaction, where each fission can cause two more fissions.
•In a fission bomb the chain reaction occurs very quickly
•In nuclear reactor the process is controlled
Ch 31 9
Controlling a Chain Reaction
•moderator is used to slow down neutrons so they are more effective in causing fissions
•need enriched uranium which has greater % of 235U which is only 0.7 % in nature
•critical mass is enough uranium to have sustaining chain reaction
•multiplication factor: f average number of neutrons that cause additional fissions
Ch 31 10
Nuclear Reactor
Ch 31 11
•fuel rods: contain the fissionable materials- help to prevent release of radioactive materials
•control rods: absorb neutrons and so are used to control the reaction—try to maintain f ≈ 1.
•energy heats up the water that surrounds the rods – this heat is used to drive steam turbine
•radioactive waste: products of fission tend to be very radioactive because they have too many neutrons to be stable
•some reactor designs can produce bomb materials
Ch 31 12
Atomic Bomb•If the number of neutrons is not controlled, they will increase exponentially and an explosion will result.
•Since nuclear reactions generally produce a million times more energy than a chemical reaction, fission can make a very effective weapon
•A major task for a bomb-builder is to obtain enough fissionable material – this means separating 235U from 238U or producing 239Pu (plutonium).
•It is then necessary to bring together a critical mass of the fissionable material.
Ch 31 13
Plutonium
A nuclear reactor can be made to produce additional “fuel” as it operates. Most uranium is 238U which can produce plutonium by
UUn 23992
23892
-239
93
239
92 eNpU
•breeder reactor earth’s supply of 235U is limited to several hundred years but this method could increase fuel by factor of 100.
•Problems: in addition to its radioactivity, plutonium is a serious chemical health hazard
•plutonium can be used to make a bomb
-23994
23993 ePuNp
Ch 31 14
Fusion•The curve of binding energy shows that energy is usually released if additional nucleons are added to nuclei that are lighter than Fe (iron).
•The strong nuclear force has a very short range and only acts between the closest neighbors of a particular nucleus
•The Coulomb repulsion felt by all protons is very long range
•For nuclei heavier than Fe, the Coulomb force dominates and the binding energy decreases so there is a loss of energy
•A key factor in fusion is that it only occurs if the the nucleons come within the range of the strong nuclear force which is ≈ 1.0 x 10-15m and this requires high energy (or temperature) if the particles are positively charged.
Ch 31 15
Fusion Reactions in the SunThe following fusion reactions are occurring at the center of the sun:
)42.0(2
1
1
1
1
1 MeVQeHHH
•This reaction occurs because of the weak nuclear force and thus is very slow (even if two protons are within the range of the strong nuclear force they don’t always react because of the weakness of the weak nuclear force)
•It is too hot at the center of the sun for nuclei to have electrons attached.
)49.5(32
21
11 MeVQHeHH
).( MeVQHHHeHeHe 861211
11
42
32
32
Ch 31 16
Making Carbon
BeHeHe 84
42
42
In certain conditions carbon can be made via:
CHeBe 126
42
84
•Since 84B is unstable, this must occur quickly
•Once a star has 12C, then elements up to iron can be made by the carbon cycle shown in the book
•Beyond iron, reactions are endothermic so heavier elements are not produced
Ch 31 17
Making Heavier Elements
•Stars more massive than our sun will explode (supernova) and produce elements up to uranium
•Since the earth has such heavy elements, our solar system was formed from the remnants of a supernova
•Since elements heavier than 4He did not exist in the early universe, they must have been produced in nuclear reactions in stars.
Ch 31 18
Controlled Nuclear Fusion•To achieve net energy production, must achieve conditions similar to center of sun for short periods of time
•a gas at this temperature would be a completely ionized collection of nuclei and electrons called a plasma.
•In order to achieve controlled fusion, it is necessary to achieve a high ion density for a long enough time. It has been shown that is necessary for a self sustaining fusion process
• Most efficient reaction would use deuterium and tritium which releases 17.59 MeV in the reaction:
nHeHH 42
31
21
Ch 31 19
Example 31-1. Calculate the energy released in the fusion reaction:
v eHHH 2
1
1
1
1
1
22
1
1
1)(22 cemmmQ HH
25.931)]00054868.0(2014102.2007825.1)(2[( cMeVu
MeV42.0
Note: We are working with atomic masses, so each 11H includes an electron mass, but the
product 21H only includes one electron. Thus it is necessary to subtract two electron masses:
one for the positron created and the other for the extra electron on the left hand side of the equation.
Ch 31 20
Example 31-2 Calculate the kinetic energy the protons in Example 31-1 must have to overcome the coulomb barrier and approach close enough for fusion to occur.
mmrH
153115 102.11)102.1(11
3115 1)102.1)(2( m m15104.2
In order to undergo fusion, the distance must be the sum of their radii.
r
QQkPEKE 21
m
c
c
Nm15
219
2
29
104.2
104100.9
EK
J
VJ
19
14
106.1
1106.9
e
HHH 11
11
11
2 rrrr
MeV60.0
First calculate the nuclear radius:
Ch 31 21
MeVKEKEavg 30.02
1
kTEK avg
2
3
The above KE is shared by each particle, so that
k
KET avg
3
)(2
KT 9103.2
)1038.1)(3(
106.1)30.0)(2(
23
19
KJ
eVJ
MeV
Example 31-2 continued: Assume that fusion occurs in a “gas” of protons with enough energy to overcome the coulomb barrier. What must be the temperature of the gas?
Ch 31 22
Fusion Methods
Magnetic Confinement: uses a magnetic field to contain and heat the plasma. Most research centers on the tokamak design first developed in Russia.
Inertial Confinement: uses a collection o laser beams from all directions to compress and heat a pellet of deuterium and tritium. A future system might implode 100 pellets per second
Ch 31 23
Future of Controlled Fusion•Progress has been much slower than expected but there are many advantages to fusion over fission
•far less leftover radioactive materials
•can’t result in an out of control situation where radiation is released, which could happen in a nuclear reactor
•doesn’t produce materials that could be used to make a bomb
•Is important to stress that fission and fusion do not contribute greenhouse gases, so they do not increase global warming