Introductory Chemistry Online/Nuclear Chemistry

&#160;&#160; Chapter 11. Nuclear Chemistry
'In today’s society, the term radioactivity'' conjures up a variety of images. Nuclear power plants producing hydrocarbon-free energy, but with potentially deadly by-products that are difficult to store safely. Bombs that use nuclear reactions to produce devastating explosions with horrible side effects on the earth as we know it and on the surviving populations that would inhabit it. Medical technology that utilizes nuclear chemistry to peer inside living things to detect disease and the power to irradiate tissues to potentially cure these diseases. Fusion reactors that hold the promise of limitless energy with few toxic side products. Radioactivity has a colorful history and clearly presents a variety of social and scientific dilemmas. In this chapter we will introduce the basic concepts of radioactivity, nuclear equations and the processes involved in nuclear fission and nuclear fusion.'''

&#160;&#160;11.1 Radioactivity
The actual discovery of radioactivity is generally attributed to the French scientist, Henri Becquerel in 1896. As with most discoveries, he was working on something else. In this case it was the nature of phosphorescence; the property of some substances to “glow in the dark” after being exposed to light. In the course of his work, he allowed photographic plates to come in contact with uranium salts, only to find out that the uranium had “fogged” the unexposed plates (Figure 11.1). Further work by Becquerel and others (including Marie Curie) led to the realization that certain elements spontaneously produced a variety of particles, some of which were charged (both positive and negative) and one class that was of higher energy, but appeared to be neutral. The three basic classes of particles were characterized and identified as “alpha”, “beta”, and “gamma” particles (Figure 11.2). Alpha particles were positive, relatively massive and, subsequent work showed that they were identical to the nucleus of the helium atom, containing two protons and two neutrons. Beta particles had a very small mass. They were of higher energy and they carried a negative charge; equivalent in mass and charge to an electron. Gamma particles (actually referred to as gamma rays) were much more energetic, appeared to be neutral and were comparable to a high-energy photon of light. Although it was not apparent immediately, one of the most surprising observations regarding radioactive elements was that as they emitted particles, the identity of the element slowly changed; uranium, for example, slowly became enriched with lead.

When alpha, beta or gamma particles collides with a target, some of the energy in the particle is transferred to the target, typically resulting in the promotion of an electron to an “excited state”. In many “targets”, especially gasses, this results in ionization, and alpha, beta and gamma radiation is broadly referred to as ionizing radiation. A Geiger counter (or Geiger-Müller counter) takes advantage of this in order to detect these particles. In a Geiger tube, the electron produced by ionization of a captive gas travels to the anode and the change in voltage is detected by the attached circuitry. Most counters of this type are designed to emit an audible “click” in response to the change in voltage, and to also show it on a digital or analog meter. A simple schematic of a Geiger counter is shown in Figure 11.3.

Today, we recognize that radioactive decay is actually quite complex, but the basic principles and patterns that were established over 100 years ago still stand. The three basic subatomic particles that occur in radioactive decay are the alpha particle, the beta particle and the gamma ray. The gamma ray is of highest energy (and perhaps the greatest ultimate danger), but from a chemistry standpoint, the alpha and beta particles are of the greatest interest. An alpha particle consists of two protons and two neutrons. It has a mass of four amu and a charge of +2. It is identical with the helium nucleus, and when a radioactive element emits an alpha particle, it loses four amu from its nucleus, including two protons (Figure 11.4). Because the number of protons in a nucleus define the identity of the element, the atomic number of the element decreases by two when it loses an alpha particle; thus uranium ($${}_{92}^{238}U$$) loses an alpha particle and becomes an atom of thorium ($${}_{90}^{234}Th$$); we will discuss this process further in the following section. In order for a beta particle (an electron) to emerge from the nucleus, it must be formed by the decomposition of a neutron (on a very simple scale, think of a neutron as being composed of a positive proton bound to a negative electron). When a neutron decays and emits a beta particle, it leaves behind the newly formed proton. Again, this changes the identity of the element in question.

&#160;&#160;11.2 The Nuclear Equation: Alpha Particle Emission
In Chapter 1, we described the meaning of the atomic symbol for an element. In the atomic symbol, the atomic number (the number of protons in the nucleus) appears as a subscript preceding the symbol for the element. The mass number appears as a superscript, also preceding the symbol. Thus for uranium (atomic number 92) with a mass of 238, the symbol is $${}_{92}^{238}U$$. To show radioactive decay in a chemical equation, you need to use atomic symbols. Thus, for the loss of an alpha particle from $${}_{92}^{238}U$$, you need to show uranium on the “reactant” side of the equation and thorium and the alpha particle on the “product” side. Just like any other chemical equation, a nuclear equation must balance. The sum of the mass numbers on the reactant side must equal the sum of the mass numbers on the product side. Because we started with uranium-238 and lost four mass units in the alpha particle, the product (or products) of the decay must have a total mass of (238 – 4) = 234. We have also removed two protons from the uranium nucleus, dropping the atomic number by two. The newly formed element is therefore thorium-234.

$${}_{92}^{238}U$$ &rarr; $${}_{2}^{4}He$$ + $${}_{90}^{234}Th$$

In this equation, we have shown the alpha particle using the atomic symbol for helium ($${}_{2}^{4}He$$), but this is often shown using the symbol $${}_{2}^{4}\alpha $$ (Figure 11.5). Compounds that emit alpha particles are very toxic, in spite of the poor penetrating ability of the particle. This is especially true if the emitting element is inhaled or ingested. The toxic dose of the alpha-emitter $${}^{210}Po $$ in a 175-pound person has been estimated to be about one microgram (1 &times; 10-6g).

 Example 11.1 Writing Nuclear Equations

Thorium-230 and polonium-210 both undergo loss of an alpha particle to form different elements. For each of these radioactive decay processes, write the appropriate nuclear equation and show the nature of the elements that are formed.

 Exercise 11.1 Writing Nuclear Equations

Radium-226 and polonium-214 both undergo loss of an alpha particle to form different elements. For each of these radioactive decay processes, write the appropriate nuclear equation and show the nature of the elements that are formed.

&#160;&#160;11.3 Beta Particle Emission
In an element with an “excess” of neutrons, one of these neutrons can break down to form an electron and a proton. In this process, an antinutrino is also produced, but because it has no mass, it is generally ignored in this process. The nuclear equation for the decomposition of a neutron can be written:

$${}_{0}^{1}n $$ &rarr; $${}_{-1}^{0}\beta $$ + $${}_{1}^{1}p $$

where the neutron has the symbol, $${}_{0}^{1}n $$, the proton has the symbol, $${}_{1}^{1}p $$, and the electron that is produced is called a beta particle, with the symbol $${}_{-1}^{0}\beta $$ (Figure 11.6). Because the nuclear equation must balance for mass and atomic numbers, the “atomic number” of the beta particle must be –1. Adding the atomic numbers on the right side of the equation shown above gives {(-1) + (+1) = 0}; identical to the “atomic number” in the neutron ($${}_{0}^{1}n $$); (even though a neutron can break down to produce a proton, there are no actual protons in a neutron, hence its atomic number is zero). Likewise, the “mass number” of the beta particle must be zero because the proton (the product) and the neutron (the reactant) each have a mass of one. Therefore, when a nucleus loses a beta particle, the number of neutrons in the nucleus decreases by one, but the mass number does not change; the neutron is converted into a proton, also having a mass number of one. Because the neutron is converted into a proton, the atomic number of the element increases by one unit, changing the identity of the element to the next highest in the periodic table. For example, thorium-234 undergoes loss of a beta particle to form pennsylvanium-234 by the equation shown below.

$${}_{90}^{234}Th $$ &rarr; $${}_{-1}^{0}\beta $$ + $${}_{91}^{234}Pa $$

Again, with a beta-particle emission, the mass number does not change, but the atomic number increases by one unit.

 Example 11.2 Beta-Particle Emission

Bismuth-210 and lead-214 both undergo loss of a beta particle to form different elements. For each of these radioactive decay processes, write the appropriate nuclear equation and show the nature of the elements that are formed.

 Exercise 11.2 Beta-Particle Emission

Chlorine-39 and strontium-90 both undergo loss of a beta particle to form different elements. For each of these radioactive decay processes, write the appropriate nuclear equation and show the nature of the elements that are formed.

&#160;&#160;11.4 Positron Emission
A positron, also called an antielectron, is an exotic bit of matter, or more correctly, an example of antimatter. A positron is the antimatter equivalent of an electron. It has the mass of an electron, but it has a charge of +1. Positrons are formed when a proton sheds its positive charge and becomes a neutron, as shown below.

$${}_{1}^{1}p $$ &rarr; $${}_{+1}^{0}\beta $$ + $${}_{0}^{1}n $$

Again, in the nuclear equation for positron emission, the sum of protons (atomic numbers) on the right equals the number of protons on the left and the masses all equal one. When an element emits a positron, the identity of the element changes to the one having one fewer protons on the periodic table. An example of a nuclear equation showing positron emission is shown below.

$${}_{6}^{11}C $$ &rarr; $${}_{+1}^{0}\beta $$ + [$${}_{5}^{11}B $$

Boron has one fewer protons in its nucleus than carbon, but the mass is unchanged because the proton has been replaced by a neutron.

$${}_{9}^{18}F $$ &rarr; $${}_{+1}^{0}\beta $$ + $${}_{8}^{18}O $$ Positron emission from Fluorine-18, as shown above, has become an important medical diagnostic tool; Positron Emission Tomography (a PET scan)(Figure 11.7, Figure 11.8, Figure 11.9). The heart of this technique is based on the fact that positrons undergo instant annihilation when they collide with an electron (an example of matter-antimatter annihilation). When this occurs, two high-energy gamma rays are produced and exit the scene of the annihilation in exactly opposite directions. During a PET scan, a patient is given an injection containing fluorodeoxyglucose (FDG), a sugar analog. The glucose analog is absorbed by metabolically active cells, where the FDG accumulates and undergoes positron decay. After a short waiting period, the patient is scanned using a circular array of gamma-radiation detectors. The fact that the gamma rays are emitted in opposite directions allows the attached computer to “draw a line” through the patient, where the line passes through the point of annihilation. Because this occurs through many directions, the exact location of the emission can be accurately calculated and then imaged as a three-dimensional picture showing the intensity of the emission.

&#160;&#160;11.5 Radioactive Half-Life
Elements such as $${}_{92}^{238}U$$ that emit radioactive particles do so at rates that are constant and unique for each element. The rate at which an radioactive element decays is measured by its half-life ; the time it takes for one half of the radioactive atoms to decay, emitting a particle and forming a new element. Half-lives for elements vary widely, from billions of years to a few microseconds. On a simple, intuitive level, if you begin with 1.00 gram of a radioactive element, after one half-life there will be 0.500 grams remaining; after two half-lives, half of this has decayed, leaving 0.250 grams of the original element; after three half-lives, 0.125 grams would remain, etc. For those that prefer equations, the amount remaining after n half-lives can be calculated as follows:

$$R=I\left( {}^{1}\!\!\diagup\!\!{}_{2}\; \right)^{n}$$

Where I represents the initial mass of the element and R represents the mass remaining. For example, the half-life of Actinium-225 is 10.0 days. If you have a 1.00 gram sample of Actinium-225, how much is remaining after 60.0 days?

The number of half-lives is 6.00 (that is n) and I = 1.00 gram. Substituting:

$$R=\left( 1.00\text{ }gram \right)\left( {}^{1}\!\!\diagup\!\!{}_{2}\; \right)^{6.00}\text{;     }R=\left( 1.00\text{ }gram \right)\left( \text{0}\text{.0156} \right)\text{;     }R=\text{ 0}\text{.0156 }gram$$

 Exercise 11.3 Radioactive Half-Life

The half-life of Antimony-124 is 60.20 days. If you have a 5.00 gram sample of Actinium, how much is remaining after 5.0 half-lives?

One of the interesting uses for half-life calculations involves radiocarbon dating, where the content of carbon-14 in organic (formally living matter) is used to calculate the age of a sample. The process begins in the upper atmosphere, where nitrogen is bombarded constantly by high-energy neutrons from the sun. Occasionally, one of these neutrons collides with a nitrogen nucleus and the isotope that is formed undergoes the following nuclear equation:

$${}_{0}^{1}n$$ + $${}_{7}^{14}N$$ &rarr; $${}_{1}^{1}p$$ + $${}_{6}^{14}C$$

Plants take up atmospheric carbon dioxide by photosynthesis, and are ingested by animals, so every living thing is constantly exchanging carbon-14 with its environment as long as it lives. Once it dies, however, this exchange stops, and the amount of carbon-14 gradually decreases through radioactive decay with a half-life of about 5,730 years, following the nuclear equation shown below:

$${}_{6}^{14}C$$ &rarr; $${}_{-1}^{0}\beta $$ + $${}_{7}^{14}N$$

Thus, by measuring the carbon-14/carbon-12 ratio in a sample and comparing it to the ratio observed in living things, the number of half-lives that have passed since new carbon-14 was absorbed by the object can be calculated.

&#160;&#160;11.6 Nuclear Fission
The process by which nitrogen is converted to carbon-14 is an example of neutron capture, in which particles are absorbed by the nucleus of another atom to form a new element. These types of reactions are actually quite common in nuclear chemistry. In Figure 11.10, a uranium-235 nucleus is shown capturing a “slow-moving” neutron, just like nitrogen captures a neutron, leading to the formation of carbon-14. Initially, uranium-236 is formed, but this nucleus has a neutron-to-proton ratio that makes it exceptionally unstable. The unstable nucleus instantaneously breaks apart (undergoes fission) to form lighter elements and to release additional free neutrons. As the nucleus breaks apart, a significant amount of energy is also released. A nuclear equation showing a typical fission of uranium-235 is shown below:

$${}_{92}^{235}\text{U}$$+ $${}_{0}^{1}\text{n}$$ &rarr; $${}_{56}^{141}\text{Ba}$$+ $${}_{36}^{92}\text{Kr}$$ + 3 $${}_{0}^{1}\text{n}$$

The three neutrons that are released are now speeding through the mass of uranium. If these are captured by another nucleus, the process happens again and three more neutrons are released. This represents a chain reaction, and in order to sustain a chain reaction like this, the mass or uranium must be large enough so that the probability of every released neutron being captured by another uranium is high. The mass of uranium (or other fissile element) that is required in order to sustain a chain reaction is called the critical mass.

The process of nuclear fission is best known within the context of fission bombs and as the process that operates within nuclear power plants. Designing a workable fission bomb presents many technical challenges. A mass of fissile material that exceeds the critical mass is unstable, so you must begin with a smaller, non-critical mass and somehow create one within a few microseconds. In the original design, this was accomplished by taking two non-critical pieces and forcing them together (very rapidly). This is typically referred to as a “gun assembly”, in which one piece of fissile uranium is fired at a fissile uranium target at the end of the weapon, similar to firing a bullet down a gun barrel (Figure 11.11A)

Each of the uranium fragments are less than a critical mass, but when they collide, they form a mass capable of sustaining the nuclear chain reaction. The assembly stays together for a few microseconds before the energy released from the fission blows it to pieces. The trick is designing nuclear devices like this is to keep them together long enough so that enough energy is released. Neutron reflectors and “boosters” are generally used to accomplish this, nonetheless, this basic type of weapon is inefficient, although easy to design and incredibly deadly. A critical mass of uranium-235 is a sphere that is slightly less than 7 inches in diameter.

A much more efficient fission bomb is based on achieving a critical mass of fissile material, not by combining smaller fragments, but by increasing the density of a sub-critical mass to the point that the rate of neutron capture sustains the chain reaction. This design is called the “implosion” bomb and it basically consists of a sphere of fissile material surrounded by shaped explosives that must be detonated simultaneously. The resulting shock wave compresses the fissile material, allowing the chain reaction to occur. This type of design requires much less fissile material, but is technically challenging. Modern devices, such as that shown in Figure 11.11B, have neutron reflectors, “neutron initiators”, etc., and sophisticated bombs can be efficient, have a high yield and a relatively small physical size.

In a nuclear reactor designed to heat water, produce steam and electrical power, the chemistry is the same, but control rods are introduced between the pieces of fissile material to absorb some of the neutrons that are produced so that a critical mass is never achieved and the chain reaction can be controlled (Figure 11.12). In this set-up, as the control rods are withdrawn, the chain reaction speeds up, and as they are inserted, the reaction slows down. Even under “meltdown” conditions, where control rods fail, the critical mass of fissile material would be formed slowly. The resulting explosion would be a bad thing, but would not compare with the energy released from a well-designed fission weapon.

&#160;&#160;11.7 Nuclear Fusion
As we saw in the preceding section, when the nuclei of heavy atoms split, energy is released. For light atoms, the opposite is true; when these nuclei combine (fuse together), energy is released. This is the process of nuclear fusion. Fusion of light elements, mostly hydrogen, is the force that powers energy release in the sun and in sun-like stars. Imagine the sun as a huge sphere of hydrogen. Because a star is so massive, the gravitation pull on the hydrogen atoms is sufficient to overcome the repulsion between the two nuclei to force them together to form an unstable $${}_{2}^{2}\text{He}$$ nucleus. This immediately ejects a positron, leaving deuterium, $${}_{1}^{2}\text{H}$$, and releasing a significant amount of energy. In the cascade of reactions shown in Figure 11.13, deuterium fuses with another hydrogen to give $${}_{2}^{3}\text{He}$$, and two of these combine to form helium, ejecting two high-energy protons in the process.

In stars that are larger and heavier than our sun, the “triple alpha process” is the dominant nuclear reaction. In this, helium nuclei fuse to eventually form carbon, releasing significant energy in the process (Figure 11.14).

One of the great challenges in physics and engineering today is to replicate fusion of this sort under controlled conditions, harvesting the energy released and converting it, indirectly, into electrical power. The extremely high temperatures and pressures that are required to initiate and sustain fusion reactions thwarted, thus far, attempts to build a fusion reactor that is “break even” in terms of the energy released relative to the energy required to produce the fusion events. Uncontrolled fusion is certainly possible, and fusion bombs exist, but these typically use an advanced fission bomb to create the temperatures and pressures necessary to promote the fusion of the lighter elements. Clearly, this approach does not work in the laboratory! Work on fusion reactors continues at a fast pace and includes novel approaches such as aneutronic fusion reactions that utilize proton-boron fusion to produce charged particles rather than a barrage of neutrons. The advantage here is that few neutrons are produced, reducing the need for shielding, and the charged particles formed can potentially be captured directly as electricity.

&#160;&#160;Study Points

 * In most atoms, a nucleus containing an “excess” of neutrons (more neutrons than protons) is unstable and the nucleus will decompose by radioactive decay, in which particles are emitted until a stable nucleus is achieved. Common particles emitted during radioactive decay include:
 * Alpha particles, consist of two protons and two neutrons. This is equivalent to a helium nucleus and an alpha particle has a charge of 2+. Because it is positive, it will be attracted towards a negative charge in an electric field. The atomic symbol for an alpha particle is $${}_{2}^{4}He $$, or sometimes $${}_{2}^{4}\alpha $$. Alpha particles are slow-moving and are easily absorbed by air or a thin sheet of paper. When an element ejects an alpha particle, the identity of the element changes to the element with an atomic number that is two less than the original element. The mass number of the element decreases by four units.
 * Beta particles are electrons, are considered to have negligible mass and have a single negative charge. They will be attracted towards a positive charge in an electric field. The atomic symbol for a beta particle is $${}_{-1}^{0}\beta $$, or sometimes $${}_{-1}^{0}e $$. Beta particles have “intermediate” energy and typically require thin sheets of metal for shielding. A beta particle is formed in the nucleus when a neutron “ejects” its negative charge (the beta particle) leaving a proton behind. When an element ejects a beta particle, the identity of the element changes to the next higher atomic number, but the mass number does not change.
 * Gamma particles (gamma rays) are high-energy photons. They have no mass and can be quite energetic, requiring thick shielding.
 * Positrons are anti-electrons, are considered to have negligible mass and have a single positive charge. They will be attracted towards a negative charge in an electric field. The atomic symbol for a positron is symbol $${}_{+1}^{0}\beta $$. Positrons have “intermediate” energy and typically require thin sheets of metal for shielding. A positron is formed in the nucleus when a proton “ejects” its positive charge (the positron) leaving a neutron behind. When an element ejects a positron, the identity of the element changes to the next lower atomic number, but the mass number does not change.
 * In a nuclear equation, elements and sub-atomic particles are shown linked by a reaction arrow. When you balance a nuclear equation, the sums of the mass numbers and the atomic numbers on each side must be the same.
 * Radioactive elements decay at rates that are constant and unique for each element. The rate at which an radioactive element decays is measured by its half-life ; the time it takes for one half of the radioactive atoms to decay, emitting a particle and forming a new element. The amount of an original element remaining after n half-lives can be calculated using the equation:

$$R=I\left( {}^{1}\!\!\diagup\!\!{}_{2}\; \right)^{n}$$

Where I represents the initial mass of the element and R represents the mass remaining.


 * In nuclear fission, a nucleus captures a neutron to form an unstable intermediate nucleus, which then splits (undergoes fission) to give nuclei corresponding to lighter elements. Typically, neutrons are also ejected in the process. For heavy isotopes, the process of fission also releases a significant amount of energy. A nuclear equation for a classical fission reaction is shown below:

$${}_{92}^{235}\text{U}$$+ $${}_{0}^{1}\text{n}$$ &rarr; $${}_{56}^{141}\text{Ba}$$+ $${}_{36}^{92}\text{Kr}$$ + 3 $${}_{0}^{1}\text{n}$$


 * In nuclear fusion, nuclei combine to form a new element. For light isotopes, the process of fusion also releases a significant amount of energy. A nuclear equation for the fusion cascade that typically occurs in stars the size of our sun is shown below:

2 $${}_{1}^{1}p $$ &rarr; $${}_{+1}^{0}\beta $$ + $${}_{2}^{3}He $$

2 $${}_{2}^{3}He $$ &rarr; 2 $${}_{1}^{1}p $$ + $${}_{2}^{4}He $$

&#160;&#160;Supplementary Problems
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