Nuclear Energy, Stability,
Fisson & Fusion
Ordinary chemical reactions involve only the valence electrons. In a chemical reaction chemical bonds are formed broken and reformed. The difference in energy is usually expressed as heat. Apart from those changes that affect the valence electrons, the atoms themselves are unchanged. Since atoms are neither created nor destroyed there are the same number of each kind of atom present after the reaction as before.
In nuclear reactions changes occur in the nucleus of the atom. The nuclei of an atom many be transformed into nuclei of a different kind of atom. A new element or a new isotope of the original element may form as a result. Nuclear reactions occur naturally in radioactive elements whose nuclei are unstable. High speed particles as well as energy are emitted from the nucleus of a radioactive element. Nuclear transformations can also be made to occur artificially as the result of bombardment of atoms with protons, neutrons or other high energy particles
1. Nuclear Equations.
Nuclear equations are used to show nuclear transformations. Balanced nuclear equations require that both the atomic number and the mass number must be balanced.
Example 1:
When Beryllium 9 is bombarded with alpha particles (helium nuclei) a neutron is produced. The balanced nuclear reaction is given as:
9Be + 4He à 1n + 12C
4 2 0 6
The sum of the atomic numbers and the sum of the mass numbers must be equal on both sides of the equation.
Example 2:
When nitrogen 14 is bombarded with neutrons protons are produced. The balanced nuclear equation can be written as:
14N + 1n à 1p + 14C
7 0 1 6
2. Radioactive Particles
The French scientist Henri Becquerel accidentally discovered t radioactivity. Radioactivity is the ability of some isotopes to undergo reactions involving nuclear transformations. In naturally occurring radioactive substances, radioactive decay results in the
conversion of unstable nuclei to more stable nuclei. There are three basic types of radiation that occur naturally, alpha particles, beta particles, and gamma radiation. Their characteristics are described below
Name
Symbol Mass Charge Description
Alpha α 4 + Helium Nuclei
Beta β 0 - Electrons
3. Penetrating Power
Alpha Particlestravel at about 5-10% of the speed of light. Since they are relatively slow and massive their penetrating power is low. Several sheets of paper can block them.
Beta Particlestravel at velocities of up to 90% of the speed of light. They have moderate penetrating power. They can pass through skin but can be stopped by 1 mm thick sheet of aluminum.
Gamma Raystravel at the speed of light. Their penetrating power is very high, requiring at least an eight mm thick piece of lead to stop them.
Because of their relatively low penetrating power alpha and beta particles are relatively harmless, although high energy beta particles can cause skin burns. Gamma rays are very dangerous due to their great penetrating power and their high energy. Exposure to gamma rays can cause tissue damage resulting in various forms of cancer, cataracts, and any genetic damage to reproductive cells that lead to mutations.
4. Examples of Radioactive Decay:
Uranium 238, 238U, is an alpha emitter
238U à 4He à 234Th
2 2 90
Neptunium 239, 239Np, is an beta emitter
239Np à 0e à 239Pu
93 -1 94
During beta emissions a neutron splits into a proton and an electron. The electron is emitted as a beta particle, while the proton remains in the nucleus. The net effect is the conversion of a neutron to a proton. The atomic number increases by one while the mass number remains the same.
In natural radioactive decay gamma rays are always emitted with alpha or beta emission, never by itself. Radium 226 for example emits both alpha particles and gamma rays.
228Ra è 222Rn + 4He + 0
γ
88 86 2 0
There is no difference in the balanced nuclear equation if the gamma rays are not included.
5. The Half Life
The half life for a radioactive substance is the time required for one-half of the initial radioactive material to decay. Since it is a first order process, the half life is independent of the amount of the radioactive sample present.
Each radioactive isotope has its own unique half life. These range from a few seconds to billions of years. The radioactive isotope of carbon, C-14, has a half life of 5730 years. Plutonium, Pu-239 has a half life of 24,000 years. Plutonium is a highly toxic material that is used as a fuel in nuclear power plants. Because of its long half-life, waste plutonium will be around for a long time. It will take approximately 170,000 years for plutonium now present to decay to 1% of its current level and more like 250,000 years to reach a safe level.
The half lives of some radioactive isotopes
Isotope
Half life
Type of emission
Strontium 90 Uranium 238 Radon 222 Plutonium 239 Cesium 137 28.8 years Beta 4.5 x 109 years Alpha 3.825 days Alpha 24,000 years Alpha 30 years Beta
The first order rate laws can be used to predict the amount of a radioactive material remaining after or a given time, or the time it would take for a given amount of radioactive material to drop to a particular concentration.
6. First Order Rate equations
The same set of equations that are used to describe first order chemical reaction rates can also be applied to nuclear reactions. In a first order process the rate is independent of the initial concentration. Therefore the half life is only a function of the first rate
constant.
For a first order nuclear decay t 1/2 = 0.693
k
Where t1/2= the half life and k = first order rate constant and
ln ( Xo/X ) = ktWhere k = first order rate constant, t = the time elapsed
Xo= initial concentration of the isotope and X = concentration of the isotope at time, t Example
After 2 hours a solution containing 1.30 x 10-6 mol dm-3 of 240AmCl3 contained only 1.27 x 10-6 mol dm-3 of the radioactive substance. What is the half-life for 240Am?
ln(1.30 x 10 -6 ) = k (2.0 hours) (1.27 x 10-6 )
k = 1.17 x 10-2 hours-1 then t1/2= 0.693 = 59 hours
1.17 x 10-2 hours-1
7. Nuclear Stability
Spontaneous radioactivity involves unstable nuclei which decay with the eventual formation of a stable nuclei. Thus some nuclei are stable and others are not, for example 12C and 14C , 3H and 40K. The isotopes in these examples have the same number of protons but differ in the number of neutrons. Two important factors determine nuclear stability (1) The mass number (which is the total number of nucleons in the nucleus) and (2) the neutron to proton ratio.
In the nucleus positively charged protons repel each other. As the number of protons in the nucleus increases the forces of repulsion between the protons increases substantially. Therefore a greater
number of neutrons is required for the nucleus to be stable. This is evident from the graph of the number of neutrons against the number of protons present in stable nuclei.
A neutron to proton ration of one to one holds true for the stable nuclei of the first twenty elements in the periodic table. This ratio increases to 1.2 to 1 around number 60 and 1.5 to 1 around atomic number 80. Elements with atomic numbers above 83 and mass numbers greater than 209 do not exist as stable isotopes. Polonium for example has 84 protons. The repulsive forces that result from these 84 protons are great that the nucleus is unstable that regardless of the number of neutrons. All isotopes of polonium are radioactive. Whenever the neutron to proton ratio is too large or too small, the nucleus is unstable. It is then called radionuclide and it undergoes radioactive decay.
If a radionuclide has a higher neutron to proton ratio than for a stable nucleus, It falls to the left of the stable nucleus To reduce the number of neutrons back to the stable ratio it undergoes beta decay. A neutron disintegrates with the emission of a high speed electron known as a beta particle:
1 n à 1 p + 0 e
0 1 -1
As a result a proton is produced from the neutron. Since this raises the number of protons by one and reduces the number of neutrons by one. The neutron to proton ratio is lowered until it reaches a stable value. Then no further radioactive occurs
If a radionuclide has a lower neutron to proton ratio, it has fewer neutrons and therefore to the right of the stable value. A proton can be transformed into a neutron either by position emission or by electron
capture: Positron emission 1 P à 0 e+ à 1 n 1 1 0 Electron Capture 1 P + 0 e à 1 n 1 -1 0
If the total number of nucleons exceeds 209, the limit for stable nuclei, the nucleus always lies beyond the stable limit, and the nucleus is always radioactive. Several kinds of radioactive decay are involved in order to reach the stability. For example, the radioisotope 238U undergoes a sequence of fourteen radioactive decay steps before forming the final product 206Pb. This sequence is called the 238U decay series
The tremendous amount of energy that results from radioactive fission comes from converting small amounts of mass are converted to energy Einstein's equation predicts that the energy released is equal to the mass times the speed of light squared.
E= mc2
where E is the energy, m the mass and c the velocity of light equal to 2.998x108 m s-1. A very small amount of matter should produce a tremendous amount of energy. If the mass is measured in kg and the velocity of light in meters per second, the units of energy will be expressed in Joules.A unit of energy can therefore be considered equivalent to a unit of mass.
Since one atomic mass unit is equal to 1.661x10-27 kg, Then
E = mc2 = ( 1.661x10-27 kg) x (2.998x108 m s-1)2 = 1.493x10-10 J
Since 1 eV = 1.602 x10-19J
E = 1.493x10-10 J / 1.602x10-19 J eV-1 = 932x106 eV = 932 MeV
Thus 1 atomic mass unit has an energy equivalent of 932 MeV.
Example 1:
Thorium 228 is an alpha particle emitter and that produces radium-224:
228 Th à 224Ra + 4 He 90 88 2 In this reaction
the mass of the products = 228.022800 amu (atomic mass units ) the mass of the reactants = 228.028726. amu
The difference equals = 0.005926 amu
In kilograms this is = 0.005926 amu x 1.661 x 10-27 kg amu-1
The energy released by one atom of thorium decaying is
E= (0.005926 amu x1.661x10-27kg amu-1) x (2.998x10-8m/s)2 = 8.847x10-13 J per atom.
Since 1 mole of 228Th has a molar mass = 228 g mol-1and contains 6.022x1023 atoms, it would produce ( 8.847x10-13 J per atom ) x (6.022x1023 atoms) = 5.329x1011 J
or about 533 billion Joules of energy!
In a nuclear explosive the critical mass is separated into sub-critical masses until detonation. It can produce 5 to 10 million 0C on explosion! An atomic explosion does not take place in a uranium mine because the fissionable isotope of uranium, 235U is not found pure in nature.Less than 1% uranium found in nature is fissionable 235U. The rest is 238U which is not fissionable by slow neutrons. Thus the fissionable 235U has to be separated from 238U before is can be used for producing energy.
8. Fission and Fusion
Nuclear fission is the splitting (by impact of neutrons) of a heavy nucleus into two or more lighter nuclei (called fission fragments) with the simultaneous release of neutrons and large amount of energy. The additional neutrons released can induce further fission. Two common fissionable substances include uranium 235 and plutonium 239.
235 U + 1n à 92 Kr + 141 Ba + 3 1 n 92 0 36 56 0 239 Pu + 1 n à 90 Sr + 147 Ba + 3 1 n 94 0 38 56 0
During the fission process tremendous amounts of energy are released. The energy, E, is produced is the result of the mass defect. The total mass of the fission products in the above reactions is slightly less than the fission reactants. The difference in mass is converted into energy The amount of energy produced is given by Einstein’s equation:
E = mc2 where c is the velocit
y of light.
A certain concentration of fissionable material must be present if the fission reaction is to occur. The critical mass is defined as the mass of the fissionable material in a certain volume needed to sustain a chain reaction. A chain reaction occurs when the uranium sample is large enough for most of the neutrons emitted, to be captured by another nucleus before passing out of the sample. Also, for a chain reaction, there must be a balance between net production of neutrons
and the loss of neutrons caused by:
(i) the capture of neutrons by uranium atoms without fission happening (ii) the capture of neutrons by other materials in the sample and
(iii) the escape of neutrons without being captured. Thus critical mass is very important.
9. Nuclear Fusion
Nuclear fusion occurs when the nuclei of small or lighter elements combine to form heavier elements with the release of energy. Because the heavier nucleus is more stable than the 2 lighter nuclei that are fused together, there is a net loss of mass and energy is released: An example of a fusion reaction:
2 H + 2 H à 3 He + 1 n + energy 1 1 2 0
The energy produced by the sun is the result of by fusion of protons; Hydrogen and Helium comprise approximately 99% of the sun’s mass. Nuclei must overcome electric repulsion between the positive nuclei in order to fuse. As a result, extremely high temperatures (approximately 108 K) are required. Fusion is also called a thermonuclear reaction.
The hydrogen bomb is an example of uncontrolled nuclear fusion. Controlled fusion is not yet possible but scientists are predicting that the appropriate technology will be available soon after the year 2000.Technical problems with fusion are related to the fact that the positive nuclei repel each other and pushing these together and holding them together requires enormous amounts of force. Thus, the problems with fusion are:
(i) the production of plasma. At very high temperatures the atoms are stripped of their electrons and the intensely hot mixture of positive nuclei and free electrons is called plasma. This process needs upwards of 40 million oC! (between 40 and 100 x106 oC)
(ii) that the plasma must be held together long enough (about 1 second) to fuse the nuclei, (iii) that enough energy must be produced to make the process profitable.
Obviously, a “magic bottle” is needed to hold the plasma. Two possibilities are being considered:
a. Magnetic Containment MethodThe magnetic containment method uses a magnetic field to confine the plasma and allows it to be heated electrically to achieve fusion. A super-cooled magnet is used. Lithium is then used to absorb and transfer the fusion heat to a generator. Thus an engineering problem includes having a temperature in the region of 40-100 million oC at the center of the plasma and yet just two meters away the temperature of the magnet at close to –273oC!.
b. Laser Ignition Method
In the second method, focused laser light is used to cause the nuclei to fuse and again molten lithium is used to absorb the energy from fusion and transfer it as heat to a steam generator to produce electricity.
The choice of material for making a fusion reactor has to be such that (I) it must not react with extremely hot lithium (at about 1000 oC) and it must last continuous neutron bombardment because high energy neutrons are produced as part of the product!
Nonetheless, the advantages of fusion as a power generating source are many:
1. The fuel deuterium (2 H or 2 D) is abundant (there are about 1022 deuterium atoms in 1.0 dm3 of sea water!). In fact, the fuel is limitless and cheap.
2. Fusion is considered much less dangerous than fission with regard to radioactive materials.
3. Massive shipments of radioactive fuel would not be required as they are in fission. Also, far less waste has to be stored. 4. Theft of fuel material is eliminated unlike in fission (of plutonium).
5. Fusion energy could be used to electrolyze water into H2(g. Hydtrogen is a much cleaner replacement for natural gas and petroleum.
Besides the technical problems at the moment, tritium, 3H (an isotope of hydrogen) is produced and is very radioactive. If 3H can be isolated it can then be used as a fuel as well:
2 H + 3 H à 4 He + 1 n + energy 1 1 2 0
