Introduction

After discussing an introduction to quantum mechanics in Chapter 29, we have been going inwards, first discussing atomic structure in Chapter 30, and now going further inwards in this chapter to discuss nuclear physics. In the following chapter we'll go yet further inwards, discussing subatomic particles and their internal structure.

Nuclear Structure

We know that protons and neutrons have internal structure; they are made of quarks. But nuclear physicists typically don't focus on this level of detail; they consider atomic nuclei as being made up of protons and neutrons, and do their best to understand the properties of different nuclei based on their differing content of protons and neutrons.

Protons and neutrons are known collectively as nucleons. The number of protons in a particular nucleus is symbolized by Z (also called the atomic number), the number of neutrons is symbolized by N, and the total number of nucleons in the nucleus is symbolized by A, which is also called the atomic mass number of the nucleus. Thus,

A = Z + N

The symbol A is referred to as the atomic mass number because the mass of a proton is very nearly equal to the mass of a neutron (neutrons are more massive than protons by about 0.1%), and the mass of an electron is about 1/2000 times the mass of a

Chapter 31 Nuclear Physics and RadioactivityJanuary-03-15 8:58 PM

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the mass of an electron is about 1/2000 times the mass of a nucleon. Thus, the mass of an atom is nearly equal to the product of A and the mass of a nucleon.

A symbol often used for atoms, and/or their nuclei, is:

AZX

Examples:

Notice that in the symbol for the electron, we use -1 for the number of protons, which seems bizarre. So perhaps it's better to call A the number of "elementary" charge units; then the symbol for the electron makes more sense.

Nuclei that have the same number of protons, but different numbers of neutrons, are called isotopes. (Examples: hydrogen, deuterium, tritium are all isotopes of hydrogen.) Because they have the same number of protons, the chemical properties of Ch31L Page 2

have the same number of protons, the chemical properties of isotopes are the same, although other properties may differ. For example, tritium is radioactive, but "regular" hydrogen and deuterium are not, as we shall see.

Nuclei that have the same number of neutrons, but different numbers of protons, are called isotones. Nuclei that have the same mass number, but different numbers of protons and neutrons are called isobars.

The atomic mass unit, u, which is a convenient unit for doing a number of nuclear physics calculations, is defined to be 1/12th of the mass of a carbon-12 atom. In terms of kilograms, one atomic mass unit is equal to 1.6605 × 10-27 kg. The energy-equivalent of one atomic mass unit is equal to 1.4924 × 10-10 J, which is also equivalent to 931.5 MeV.

Question: Why are the atomic masses listed in the periodic table of the elements not whole numbers?

Experiments show that the radius of a nucleus is given approximately by the following formula:

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where the radius is measured in metres. This assumes that atomic nuclei are approximately spherical.

This makes one wonder about the approximate density of atomic nuclei. Let's calculate:

Now notice that M/A is equal to the mass of a nucleus divided by the number of its nucleons; in other words, it's the mass-per-nucleon, which is the same for all nuclei, and equal to the mass in kg of one nucleon. Thus,

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This is a truly enormous density by the standards of our every-day experience. For example, one cubic millimetre of nuclear matter has a mass of 200,000 tonnes! Clearly, ordinary matter must have a lot of "empty space" in it, otherwise every material object would be a lot more massive than it actually is.____________________________________________________

Strong Nuclear Forces and the Stability of Nuclei

Why do nuclei hold together if protons repel each other? Could it be gravity? Let's see: Using the formula

The radius of a helium atom (2 protons + 2 neutrons = 4 nucleons) is about

Assuming that the two protons in a helium atom are about this far apart, the magnitude of the electrostatic force between the two protons is

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Wow! This is an enormous force considering how small the mass of each proton is. Now for the gravitational force between the two protons:

The gravitational force between two protons is extremely weak compared to the electrostatic force, so much so that it's absolutely negligible. (This explains why we never have to worry about gravitational forces, gravitational potential energy, etc., when we are studying chemical bonding, atomic physics, nuclear physics, etc.) Even when you add in the mutual gravitational attraction of all the nucleons in the helium nucleus, the attractive gravitational forces are absolutely negligible compared to the strong electrostatic repulsion between the protons.

So what is the attractive force that holds the helium nucleus (and all stable nuclei) together? There must be some other force acting, and experiments show that there is such a force. It is called (rather prosaically) the strong nuclear force, or strong force for short. The strong force acts between any two nucleons; that is, it acts between a proton and a proton,

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nucleons; that is, it acts between a proton and a proton, between a neutron and a neutron, and between a proton and a neutron. (Actually, the strong force acts between quarks, and so it acts between any pair of particles each of which is made up of quarks. Not all particles are made up of quarks; for example, electrons are not made up of quarks, and therefore don't feel the strong force.)

To understand how the strong force varies with distance, recall the potential energy diagram for a simple harmonic oscillator:

Notice that for a harmonic oscillator, the particle is bound, but the greater the total energy, the greater the amplitude of oscillation. Also notice that there is an equilibrium position for the particle, and that a displacement from the equilibrium results in a restoring force. The force on the particle is represented in the figure by the negative of the slope of the potential energy curve.

Now consider the following potential energy curve, which is an approximation to the electric potential energy of a

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an approximation to the electric potential energy of a positive charge near another positive charge:

Notice that the curve is asymptotic to both the vertical and horizontal axes. The curve approaches the horizontal axis slowly enough with increasing r that we say that the electric force is long range. Notice that the positively charged particle in the field created by the central positive charge is not bound.

Now consider the following potential energy curve, which is an approximation to the potential energy of a nucleon in a nucleus:

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What observations can we make about the properties of the strong nuclear force based on this potential energy graph? (The coloured balls and springs in the diagram's upper right allude to the fact that nucleons are made up of quarks. We'll discuss quarks at the end of the course, but nuclear physicists typically strive to understand nuclear structure without going down to the level of quarks.)

Notice that nucleons might or might not be bound in a nucleus, depending on their total energy.

Real nuclei are subject to both strong forces and electric forces, and so the potential energy curve for a nucleon in a real nucleus will be a complex combination of the previous two potential energy curves displayed here. The details of such curves are different for each individual nucleus.

Thus, you can understand nuclear stability by thinking in terms of two opposing influences: the electric repulsion

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terms of two opposing influences: the electric repulsion between the protons and the strong-force attraction between all the nucleons. However, the short range of the strong force and the relatively long range of the electric force means that as the number of protons in a nucleus increases, the number of neutrons must increase faster. Roughly speaking, each proton feels the repulsive force of every other proton in the nucleus, but each nucleon feels the attractive strong force of only the nearest neighbour nucleons. Thus, with an increasing number of protons, one needs yet more neutrons to come between the protons, moving them further apart from each other, so that the electric repulsion does not become too large to break the nucleus apart.

Not all nuclei are stable; some do break apart. Nuclei that spontaneously break apart are called radioactive. The time at which they will break apart is unpredictable, but one can predict the half-life of radioactive nuclei, as we shall discuss later in this chapter.

Every nucleus beyond bismuth is radioactive; consider the following diagram, in which each blue dot represents a stable nucleus.

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"Mass Defect" and Nuclear Binding Energy

For a stable nucleus, each nucleon is bound; that is, each nucleon does not have enough energy to escape the nucleus. The amount of energy that must be supplied to the nucleon so that it can escape is called the binding energy for that nucleon. The total amount of energy that must be supplied to all the nucleons in the nucleus to completely separate all of the nucleons is called the binding energy for the nucleus.

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Nuclei that have greater binding energy per nucleon are more stable; it takes more energy per nucleon to break them apart.

Recalling from our study of special relativity that matter and energy can be converted into one another (E = mc2), it follows that the rest energy of a nucleus is less than the rest energy of its separated constituent nucleons. In other words, the mass of a nucleus is less than the mass of its separated constituent nucleons. This difference in mass is called the mass defect of a nucleus.

In summary: binding energy = (mass defect)c2 = (Δm)c2

The following diagram shows the binding energy per nucleon for selected nuclei.

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Notice that the nuclei with the greatest binding energy per nucleon are those in the vicinity of iron. Under the right conditions, nuclei lighter than iron can fuse to form heavier nuclei, with a net release of energy. This is called nuclear fusion, and it's the process that makes the stars shine, including our own Sun.

Fusion of light elements takes place in the core of the Sun, where it's extremely hot and dense. The net energy released eventually makes its way to the Sun's surface, where it's radiated into space. Some of this energy reaches Earth, and we depend on it to support our continued lives.

In the early history of the universe, the only elements present were hydrogen and helium, with perhaps trace amounts of lithium. Where did all of the other elements come from? Well, early super-massive stars synthesized some heavier elements via fusion in the later stages of their (relatively short) lives, and synthesized the rest of the heavy elements (past iron) when they experienced supernova explosions. These explosions emit enormous amounts of energy, and for a short time a single super-

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enormous amounts of energy, and for a short time a single super-massive star can emit energy at a greater rate than an entire galaxy. This enormous amount of energy provides the binding energy needed to synthesize heavy elements, in a process called stellar nucleosynthesis. We are truly stardust.____________________________________________________

Problem: The binding energy of a nucleus is 225.0 MeV. Determine the mass defect of the nucleus in atomic mass units.

Solution: The definition of an atomic mass unit is

The mass defect of the nucleus is

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Problem: Determine the total binding energy and the binding energy per nucleon for carbon-12.

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energy per nucleon for carbon-12.

Solution: By definition, the atomic mass of carbon-12 is exactly 12 u. Using Table 31.1 (reproduced earlier in these notes), we can determine the atomic mass of all the constituents of a neutral carbon-12 atom (6 protons, 6 neutrons, and 6 electrons):

Therefore, the mass defect is

The corresponding binding energy is

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The total binding energy for carbon-12 is 92.2 MeV, which amounts to 7.7 MeV per nucleon.__________________________________________________

Problem: Oxygen is produced in stars through the reaction

Determine the amount of energy (in MeV) absorbed or released per reaction. The atomic mass of carbon-12 is exactly 12 u, the atomic mass of helium-4 is 4.002 602 u, and the atomic mass of oxygen-16 is 15.994 914 u.

Solution: The total mass of the reactants is

12 + 4.002 602 = 16.002 602 u

which is greater than the mass of the product of the reaction. The excess mass is converted to energy, and so energy is released by the reaction. The amount of energy released per reaction is

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Radioactivity

Unstable nuclei transform into other nuclei at unpredictable times, through a process known as radioactive decay, which involves the ejection of some particles, known as alpha rays, beta rays, and gamma rays. However, other aspects of such transformations are predictable; for a given sample of a certain type of unstable nucleus, one can predict the proportion of the nuclei that will decay in a given time, as we shall see later in the chapter.

Alpha rays, beta rays, and gamma rays were named before it was known what they were. Alpha, beta, and gamma rays were named according to their penetrating powers; when they were first studied, they were passed through various thicknesses of lead to see if they could pass through. It was found that typically alpha rays were stopped by a sheet of lead about 0.01 mm thick, beta rays were stopped by a sheet of lead about 0.1 mm thick, and gamma rays were stopped by a sheet of lead about 100 mm thick. Nowadays, we know that an alpha ray is a helium-4 nucleus (2 protons and 2 neutrons), a beta ray is an electron, and a gamma ray is a photon of electromagnetic waves. Gamma ray photons have energies that are typically greater than X-ray photons, although in studying radioactive decay the term "gamma ray" is generic and might also apply to X-rays.

Schematic diagram of a particle detector:

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Some history of the discovery of radioactivity:

Wilhelm Röntgen (1895): discovered X-rays

Henri Becquerel (1896): Following up on Röntgen's work, Becquerel was studying phosphorescence, a phenomenon in which certain substances emit light on their own for a short time after being exposed to strong sunlight. However, it was cloudy for a period of days, and so Becquerel prepared his experiment, which included a photographic plate, and placed it in a desk drawer, so that it remained in the dark for some days. To his surprise, the photographic plate was already exposed when he removed it from the desk a few days later. Apparently the substance he was studying (a uranium compound) was emitting some kind of rays, and this was an intrinsic property of the substance; the substance did not need sunlight to stimulate the phenomenon.

Further experimentation confirmed this hypothesis. Marie

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Further experimentation confirmed this hypothesis. Marie Sklodowska-Curie and Pierre Curie, a husband-and-wife team, did intensive research into radioactivity, spending years processing ores to isolate minute quantities of radioactive substances that they could then study.

Rutherford came up with the terms alpha, beta, and gamma rays, and by 1902 determined that all radioactive substances diminish in activity (decay) exponentially with time. In 1907 Rutherford and his student Hans Geiger developed a number of radiation detectors and counters, and later in the same year Rutherford and his student Thomas Royds conducted experiments to show that alpha particles are helium nuclei. In 1913, Rutherford and his student Frederick Soddy discovered the fact that radioactive substances transform into different elements upon radioactive emissions; this was called "transmutation" at the time.__________________________________________________

The most radioactive places on Earth (Derek Muller, Veritasium):

https://www.youtube.com/watch?v=TRL7o2kPqw0

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The discovery and analysis of radioactivity and other particle interactions has led to new conservation laws being postulated. One such new conservation law is called the law of conservation of baryon number (nucleon number), which for the purposes of analyzing radioactivity can be stated as follows: In any radioactive transformation, the number of nucleons remains the same.

Some examples of radioactive decay follow:

Alpha-Decay

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Alpha-Decay

Problem: Radon-222 transforms to polonium-218 by alpha-decay. Determine the energy released in the reaction. (The mass of radon-222 is 222.017 570 u and the mass of polonium-218 is 218.0089 u. The mass of an alpha particle is 4.001 506 u.)

Solution: The mass released is

222.017 570 u − 218.0089 u = 4.00867 u

The alpha particle accounts for 4.001 506 u of the released mass, and the two stray electrons account for

2(0.000 548 580) = 0.001 097 u

The remainder of the released mass,

4.00867 u − 4.001 506 u − 0.001 097 u = 0.006 067 u

is converted to kinetic energy. Thus, the amount of kinetic energy produced by the reaction is

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Smoke detectors

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Quantum tunnelling

The released kinetic energy is carried by all the particles participating in the reaction, which are the two large nuclei, the alpha particle, and the two electrons._____________________________________________________

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Beta-Decay

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Quantum tunnelling is relevant for both alpha-decay and nuclear fusion (the latter will be discussed in the next chapter).

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Problem: Carbon-11 transforms by beta-positive decay. Determine the products of the reaction and the energy released. Ignore the mass of the emitted neutrino. The mass of carbon-11 is 11.011 434 u.

Solution:

The mass of boron-11 is 11.009 305 u and the mass of a positron is 0.000 548 580 u. Thus, ignoring the mass of the neutrino, the mass that has been converted to energy is

11.011 434 u − 11.009 305 u − 0.000 548 580 u = 0.001 580 u

The amount of energy released is therefore

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Gamma-Decay

When a nucleus transforms to a different nucleus via alpha-decay, beta-decay, or some other mode of transformation, it may end up in an excited state. In the same way that electrons in an atom can be in various energy levels, nucleons in a nucleus can also be in various energy levels. If the transformed nucleus is not in its ground state, it will soon reach its ground state by emitting a photon or several photons. This photon emission is called gamma-decay, and it does not transform the nucleus into a different species, but it does transform the nucleus from an excited state to a lower-energy state.

For atoms, the differences in electron energy levels amounts to at most a few eV, so emitted photons have at most a few eV of energy. However, energy transitions for nucleons in an atomic nucleus are typically in the range of 1 keV to 10 MeV, so the resulting photon energies are much greater than for electron transitions. Electron transitions result in the emission of visible

The experimental value is 0.96 MeV, so something is wrong with the calculation. Evidently ignoring the neutrino is the problem, but at this level we'll leave it at that.___________________________________________________

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transitions. Electron transitions result in the emission of visible light, or UV light, or microwaves, but nucleon transitions result in the emission of gamma rays.

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Neutrinos

The neutrino is a ghostly particle that is produced in enormous quantities in the fusion reactions in the cores of stars. Solar neutrinos spread out in all directions, and by the time they reach Earth, about 70 billion of them pass across each square centimetre of cross-sectional area per second. This means that many trillions of neutrinos pass through our bodies every second.

Doesn't this onslaught cause our bodies damage? No. Neutrinos barely interact with matter, so a neutrino could typically pass through a sheet of lead light-years thick with little chance of interacting with any of it! Other species of particles (electrons, protons, neutrons, etc.) flying through our bodies at this rate, with such energies, would utterly destroy us. Neutrinos are odd.

If neutrinos barely interact with ordinary matter, how can one detect them? As you can imagine, based on the previous few paragraphs, interactions of neutrinos with ordinary matter are extremely rare, so the only hope of detecting them is to place your detector in a place where there are many, many neutrinos flying through. However, such detectors are likely to be sensitive to other particles, so they must be placed in locations where there are not many other types of particles flying about. A major neutrino detector is located deep in an abandoned mine near Sudbury, Ontario, and it's called the Sudbury Neutrino Observatory (SNO, https://www.snolab.ca/ ).

Neutrinos were first hypothesized by Wolfgang Pauli in 1930 in Ch31L Page 25

Neutrinos were first hypothesized by Wolfgang Pauli in 1930 in an attempt to explain energy conservation in beta-decay experiments. The experimental data seemed to indicate that energy was not conserved in such experiments, and Pauli figured that there must be another particle emitted in such reactions that was not being detected.

The hypothetical particle makes it easy to see how momentum and energy could be conserved and still be consistent with the experimental data. However, much more precise experiments would have to be performed to test the hypothesis. Pauli called the hypothetical neutral particles "neutrons," which caused confusion in 1932 when James Chadwick discovered the particles that we now call neutrons. Enrico Fermi worked out the first theory of beta decay at about the same time, and called Pauli's hypothesized particles neutrino, which means "little neutral one," to reflect the fact that neutrinos are much, much less massive than neutrons.

The Cowan-Reines experiment (results first published in 1956) was the first to confirm the existence of neutrinos. Because neutrinos interact so infrequently with matter, their confirmation had to await the development of nuclear reactors, which produce neutrinos at a very high rate.

The question of the mass of a neutrino is currently a hot topic in particle physics. Work along these lines dates back to Arthur Eddington in 1920, who first proposed that nuclear fusion (which we'll study in the next chapter) is responsible for energy production in stars. Further work by many others culminated in a detailed theory created by Hans Bethe in 1939 which made very

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detailed theory created by Hans Bethe in 1939 which made very definite predictions for the flux of neutrinos emitted by the core of the Sun. Measurements of the neutrino flux from the Sun by Roy Davis and John Bahcall (the "Homestake" experiment), which ran from the 1960s until 1994, detected between one-third and one-half of the predicted number of solar neutrinos. These results were confirmed by other detectors around the world, and led to a period of puzzlement over what became known as the solar neutrino problem.

Many ideas were proposed to solve the solar neutrino problem, most involving tinkering with the theory of stellar energy production. Throughout this period it was thought that neutrinos are massless, but the first evidence that neutrinos might not be massless came in 1987 when Supernova 1987A was observed. Measurements of the time difference between observations of a spike in neutrino flux at detectors in Japan and Ohio suggested that the speed of a neutrino might be a little less than the speed of light, which means that neutrinos have mass (only massless particles travel at the speed of light). However, the timers were not precise enough to be definitive. In 1998, the Super-Kamiokande detector in Japan observed strong evidence for one type of neutrino transforming into another type of neutrino, which is understood to be possible only if neutrinos have mass. (This process is called neutrino oscillation.) Starting in 2001, definitive evidence for neutrino oscillations was reported by SNO in Sudbury, and there is widespread agreement now that neutrinos have mass. How much mass (it's thought to be very, very small) is still a matter of intensive investigation.

The SNO observations also definitely resolved the solar neutrino problem; the reason that not as many electron neutrinos are observed coming from the Sun is that some of them are transforming into the other two types of neutrinos. The SNO

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transforming into the other two types of neutrinos. The SNO observations agree precisely with the current state of neutrino oscillation theory.

Investigations into the properties of neutrinos is ongoing. One such experiment sends neutrinos from CERN on the France-Switzerland border through more than 700 km of the interior of the Earth to Gran Sasso in Italy, where they are detected and analyzed. In 2011 there was a brief media frenzy when it was announced that the neutrinos were arriving sooner than expected, and were therefore travelling faster than the light! You can imagine that the experimenters must have been extremely careful before announcing such a surprising result, knowing the scrutiny that such a revolutionary announcement would attract. Virtually instantly a number of problem areas were identified by other workers, and after many months of checking and re-checking, problems were found with the electronics in the experiment, and the results were retracted. Re-analysis of the data suggested that the neutrons were travelling more slowly than the speed of light.

This is one of the strengths of science: it is self-correcting. We humans are very good at fooling ourselves, but collectively we help each other by correcting each other.

Remember the slogan, due to Carl Sagan, that extraordinary claims require extraordinary evidence. Scientific claims are checked and double-checked, and the most extraordinary claims are given special scrutiny. In science it's rare that a single experiment is considered definitive; what carries the most weight is a diverse collection of experiments or observations from different perspectives with consistent results.

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Radioactive Decay and Activity

The transformation ("decay") of a radioactive nucleus is unpredictable. However, what can be predicted, and very precisely, is the proportion of nuclei in a sample that decays in a certain time. It is found experimentally that the rate at which transformations ("decays," "disintegrations") occur (called the activity) is proportional to the number N of nuclei in the sample:

Where the positive number λ is called the decay constant, and depends on the species of nuclei. The negative sign is present because the number of nuclei of the given species decreases as they transform into another species.

Using calculus, you calculus lovers can derive a formula for the number of radioactive nuclei as a function of time:

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One way of comparing different radioactive nuclei is by comparing the decay constants; the larger the decay constant, the faster the rate at which radioactive nuclei decay. You might call λ the relative activity, since it's a measure of the proportion of a sample of nuclei that decays per unit time. However, most people find it easier to understand half-life, which is the time needed for half of a particular sample of radioactive nuclei to transform. The longer the half-life, the slower the rate at which radoactive nuclei decay. Half-life T is related to the decay constant as follows:

Thus, the larger the decay constant the smaller the half-life, and Ch31L Page 30

Thus, the larger the decay constant the smaller the half-life, and vice versa. Note that the textbook uses a subscript 1/2 on the symbol "T" to indicate half-life. The following graph may help you understand the concept of half-life.

The following table lists half-lives for some radioactive nuclei:

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Some people prefer to write the formula for N in terms of the half-life instead of the decay constant. Observe:

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Thus, we have two equivalent formulas that can be used to solve problems of radioactive decay, depending on whether we prefer to work with the decay constant λ or the half-life T:

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Problem: In 11.4 days, the number of radioactive nuclei in a sample decays to 1/8th of the original number. Determine the half-life of this species of nucleus.

Solution: Let t = 11.4 days, and therefore:

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Does this make sense? When the number of radioactive nuclei has decreased by a factor of 1/8, the number of half-lives that has passed is 3, because

(1/2)*(1/2)*(1/2) = 1/8.

Because there are 3 half-lives in an 11.4 day time period, each half-life lasts for 11.4/3 days.

Alternative Solution: First determine the decay constant; once again, let t = 11.4 days, and therefore:

For the half-life,

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Problem: Two radioactive waste products from nuclear reactors are strontium-90, with a half-life of 29.1 years, and cesium-134, with a half-life of 2.06 years. These two species are present initially in a ratio N0, Sr /N0, Cs = 7.80 × 10-3. Determine the ratio NSr /NCs 14 years later.

The two solutions are equivalent, and the results are the same, as they must be. Which solution do you prefer?

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Determine the ratio NSr /NCs 14 years later.

Solution: Because we are given information about half-lives, it's probably easier to use the form of the decay equation that is expressed in terms of half-lives.

Thus,

The sample starts off almost entirely cesium-134 and ends up 15 years later being predominantly strontium-90._____________________________________________________

Problem: A certain radioisotope has a half-life of 8.10 days. What percentage of an initial sample of this isotope remains after 26 days?

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days?

Solution:

Thus, about 11% of the original amount of the radioisotope remains after 26 days.__________________________________________________

Problem: If the activity of a radioactive substance is initially 483 disintegrations/min and two days later it is 365 disintegrations/min, determine the activity four days later still, which means six days after the start. Quote your result in disintegrations/min.

Solution: As discussed earlier in these lecture notes, the activity is proportional to the number of radioactive nuclei present. Thus, if

then it follows that

and therefore

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Radioactive Dating

Radioactive decay provides a tool for dating objects of archeological interest. The basic idea is that if you know the amount of some radioactive element at two different times, then you can calculate the time interval. This works because the

Thus,

Could you determine the half-life of this radioactive species of nucleus if you had to?__________________________________________________

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you can calculate the time interval. This works because the formula

is a function (in the mathematical sense) with a very specific graph. If two N(t) values are specified, then there is a definite corresponding time interval on the graph:

We can determine the amount N of radioactive material that is present now by measuring its activity. But how do we know how much radioactive material there was initially, which is the value of N0? Typically some assumptions must be made, which renders the whole method a bit uncertain. For this reason, having several different methods for estimating the ages of artifacts is beneficial, so that if all methods agree, one has greater confidence in the result.

For carbon-14 dating, the basic assumption one makes (as a first approximation) is that the proportion of carbon-14 (as compared

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approximation) is that the proportion of carbon-14 (as compared to all isotopes of carbon) in the environment has been nearly constant over time. This is not quite true, and there is other evidence that helps us to better estimate the proportion of carbon-14 in the environment in the past. (See the footnote on page 969 of the textbook.)

For a living creature, the proportion of carbon-14 in its body is equal to the proportion of carbon-14 in the environment, because the creature ingests carbon-14 continuously. However, once the creature dies, the proportion of carbon-14 in the creature's body decreases as the carbon-14 in the body decays. Thus, one can use the proportion of carbon-14 in the environment as an estimate of the initial proportion of carbon-14 in the body. The current proportion of carbon-14 in the body can be measured.

The following problems show how it works in practice using this simplified assumption. (More accurate calculations correct for the fact that the proportion of carbon-14 in the environment was different in the past.)___________________________________________________Problem:

A sample has a carbon-14 activity of 0.0087 Bq per gram of carbon. The half-life of carbon-14 is 5730 years.(a) Find the age of the sample, assuming that the activity per gram of carbon in a living organism has been constant at a value of 0.23 Bq. (b) Evidence suggests that the value of 0.23 Bq might have been as much as 45% larger. Repeat part (a), taking into account this 45% increase.

Solution:

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Solution:

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There is clearly some uncertainty in the method if we don't know precisely the proportion of carbon-14 in the past (which is reflected in its activity), but nevertheless, we have a range of ages for the sample that may be quite useful, especially if we have other independent means of estimating its age.___________________________________________________

Problem: The half-life of uranium-238 is 4.47 billion years. Determine the age of a rock sample that contains 70% of its original number of uranium-238 atoms.

Solution:

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Because dating techniques have inherent uncertainties, based on assumptions to determine the initial amounts of radioactive substances, it's important to have several dating methods to increase confidence and accuracy in the results. This is typical in science; typically, there is not one definitive method or experiment, but by having overlapping experiments and methods, ideally of quite different types and from different perspectives, one accumulates evidence, and it is the weight of evidence that tells.

One of the historically very important questions that drastically changed our perspective on who we are and what is our place in the universe is about the age of the Earth. William Thomson, Lord Kelvin, one of the most prominent physicists and engineers of the 1800s, performed a number of calculations to estimate the age of the Earth, and obtained estimates on the order of 100 million years. Our current estimates are about 4.5 billion years, so Kelvin was off by nearly a factor of 50. Nevertheless, his calculations were influential for a number of reasons. It was unclear in the 1800s whether the Earth was finite or not! Some prominent scientists believed that the Earth had existed forever. Geologists were generally skeptical that physics was relevant for resolving questions of geology. Kelvin's calculations made it reasonable to consider the Earth of finite age, and many geologists were convinced that it was reasonable to apply physics to geology.

The rock sample is about 2.3 billion years old.

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convinced that it was reasonable to apply physics to geology.

Kelvin's calculations were so far off primarily because he neglected convection currents in the Earth's mantle, but it must also be said that radioactivity was unknown in Kelvin's time, and radioactivity plays a role in the calculations.

Kelvin supported his calculations of the age of the Earth by estimating the age of the Sun. He got this wrong as well; nuclear physics was unknown at the time, and the mechanisms by which stars produce energy were therefore also unknown. Using incorrect assumptions about how the Sun produces energy (conversion of kinetic energy and gravitational potential energy from in-falling material), Kelvin estimated an age for the Sun of about 100 million years.

The rough agreement between Kelvin's estimates for the ages of the Sun and Earth gave him confidence in both results. However, geologists figured the Earth was much older than Kelvin's estimate, and Darwin's recently created theory of evolution was consistent with a very old Earth, as Darwin figured it would take a very long time for modern species to evolve from simpler organisms. This inconsistency is a hallmark of young theories; it typically takes time and lots of observations, experiments, and reflection to polish and sharpen scientific theories.

In the 20th century, it was eventually realized that both of Kelvin's estimate were wrong, and the conflict between biological, geological, and physical theories were removed. With the rise of radioactive dating, it became possible to estimate more carefully the age of the Earth, which is about 4.5 billion years old. Developments in nuclear physics and particle physics helped us to understand how energy is produced in the Sun, and helped us to estimate the age of the Sun, which is also about 4.5

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helped us to estimate the age of the Sun, which is also about 4.5 billion years old. These ages are consistent with current theories of the development and evolution of life on Earth.

This illustrates one of the strengths of science: It is self-correcting. Hypotheses and theories are not accepted just because the people proposing them are famous; they are continually tested by experimentation and observation, and modified, updated, or discarded depending on the weight of new evidence brought to bear.

For more information on the history of how the age of the Earth was discovered, see the following very interesting article:

http://www.americanscientist.org/issues/page2/kelvin-perry-and-the-age-of-the-earth

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Radioactive Decay Series

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Radiation Detectors

Geiger-Muller counter:

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Scintillation counter:

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Appendix: Review of Logarithm Functions and their Properties

Logarithm functions are useful for many reasons, but the main use we'll make of them in this chapter is to help us solve exponential equations (that is, equations where there are unknown quantities in one or more exponents).

Review of exponential functions:

Logarithm and exponential functions are inverses of each other:

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The graphs of y = 2x and y = log2 x are inverses of each other; graphically, each is the reflection of the other in the line y = x. Numerically, if you interchange the columns of the table of values for an exponential function you obtain the table of values for the corresponding logarithmic function, and vice versa. Algebraically, you can obtain the formula for one function from the formula for the other function by interchanging x and y. Thus, starting with the exponential function

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Interchange x and y to obtain a formula for the corresponding logarithm function:

An equivalent way to write the logarithm function is:

Carefully study the previous three displayed formulas, and you'll understand the slogan "a logarithm is an exponent:"

This basic equivalence between exponential and logarithmic perspectives can be expressed as follows:

You can consider exponential functions with various bases, and so you can also consider logarithm functions with various bases; the same basic equivalence between exponential and logarithmic perspectives holds for other bases as well:

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For example:

The most commonly used bases are 10 and e; the latter is an irrational number that is approximately equal to 2.7. Logarithm to base e is called "natural logarithm" and is symbolized by ln instead of log, at least in most books. Some advanced books use log for natural logarithm, so watch out for this if you read advanced books. Logarithm to base 10 is often symbolized by log without specifying a base; base 10 is understood.

Most calculators have logarithms to both base 10 and base e. Natural logarithms are most popular in applications involving calculus, because the derivative formula for the natural logarithm function is simplest of all the logarithm functions.

Properties of exponents, which lead to properties of logarithms

Let M = Br and let N = Bs. Consider the following properties of exponents:

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exponents:

In equation (1), notice that the exponent of the right side is the sum of the exponents of the factors on the left side. Thus,

Similarly, from equation (2) above, the corresponding relation for logarithms is

Similarly, from equation (3) above, the corresponding relation for logarithms is

Including the following two basic "inverse" identities results in a good list of useful properties of logarithms:

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Solving problems using logarithms

To solve an equation involving multiplication, we use the inverse operation, division:

To solve an equation involving addition, we use the inverse operation, subtraction:

Similarly, because exponential and logarithmic functions are inverses of each other, to solve exponential equations we use logarithms. The following examples illustrate this process.___________________________________________________

Example: Solve the equation

Solution: Take the logarithm of both sides of the equation to any convenient base and simplify:

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Example: Calculate

Solution 1:

Solution 2:

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Note that the following calculation is incorrect:

This is a common error; don't do this! The correct calculation is:

The exponential functions used to model radioactive decay are decreasing functions. You can obtain the graph of a decreasing exponential function from the graph of an increasing exponential function by reflection in the vertical axis. For example, see the following graph:

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The exponential functions used in modeling radioactive decay have graphs that resemble the decreasing exponential function in the graph above:

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