316. HalfLife and ActivityLearning Objectives
Unstable nuclei decay. However, some nuclides decay faster than others. For example, radium and polonium, discovered by the Curies, decay faster than uranium. This means they have shorter lifetimes, producing a greater rate of decay. In this section we explore halflife and activity, the quantitative terms for lifetime and rate of decay. HalfLifeWhy use a term like halflife rather than lifetime? The answer can be found by examining Figure 1, which shows how the number of radioactive nuclei in a sample decreases with time. The time in which half of the original number of nuclei decay is defined as the There is a tremendous range in the halflives of various nuclides, from as short as ${\text{10}}^{\text{23}}$ s for the most unstable, to more than ${\text{10}}^{\text{16}}$ y for the least unstable, or about 46 orders of magnitude. Nuclides with the shortest halflives are those for which the nuclear forces are least attractive, an indication of the extent to which the nuclear force can depend on the particular combination of neutrons and protons. The concept of halflife is applicable to other subatomic particles, as will be discussed in Particle Physics. It is also applicable to the decay of excited states in atoms and nuclei. The following equation gives the quantitative relationship between the original number of nuclei present at time zero (${N}_{0}$) and the number ($N$) at a later time $t$: $$N={N}_{0}{e}^{\mathrm{\lambda t}}\text{,}$$where $e=\text{2.71828}\text{...}$ is the base of the natural logarithm, and $\lambda $ is the To see how the number of nuclei declines to half its original value in one halflife, let $t={t}_{1/2}$ in the exponential in the equation $N={N}_{0}{e}^{\mathrm{\lambda t}}$. This gives $N={N}_{0}{e}^{\mathrm{\lambda t}}={N}_{0}{e}^{\mathrm{0.693}}=0.500{N}_{0}$. For integral numbers of halflives, you can just divide the original number by 2 over and over again, rather than using the exponential relationship. For example, if ten halflives have passed, we divide $N$ by 2 ten times. This reduces it to $N/\text{1024}$. For an arbitrary time, not just a multiple of the halflife, the exponential relationship must be used.
One of the most famous cases of carbon14 dating involves the Shroud of Turin, a long piece of fabric purported to be the burial shroud of Jesus (see Figure 2). This relic was first displayed in Turin in 1354 and was denounced as a fraud at that time by a French bishop. Its remarkable negative imprint of an apparently crucified body resembles the thenaccepted image of Jesus, and so the shroud was never disregarded completely and remained controversial over the centuries. Carbon14 dating was not performed on the shroud until 1988, when the process had been refined to the point where only a small amount of material needed to be destroyed. Samples were tested at three independent laboratories, each being given four pieces of cloth, with only one unidentified piece from the shroud, to avoid prejudice. All three laboratories found samples of the shroud contain 92% of the ${}^{\text{14}}\text{C}$ found in living tissues, allowing the shroud to be dated (see Example 1). Example 1: How Old Is the Shroud of Turin?Calculate the age of the Shroud of Turin given that the amount of ${}^{\text{14}}\text{C}$ found in it is 92% of that in living tissue. Strategy Knowing that 92% of the ${}^{\text{14}}\text{C}$ remains means that $N/{N}_{0}=0\text{.}\text{92}$. Therefore, the equation $N={N}_{0}{e}^{\mathrm{\lambda t}}$ can be used to find $\mathrm{\lambda t}$. We also know that the halflife of ${}^{\text{14}}\text{C}$ is 5730 y, and so once $\mathrm{\lambda t}$ is known, we can use the equation $\lambda =\frac{0\text{.}\text{693}}{{t}_{1/2}}$ to find $\lambda $ and then find $t$ as requested. Here, we postulate that the decrease in ${}^{\text{14}}\text{C}$ is solely due to nuclear decay. Solution Solving the equation $N={N}_{0}{e}^{\mathrm{\lambda t}}$ for $N/{N}_{0}$ gives $$\frac{N}{{N}_{0}}={e}^{\mathrm{\lambda t}}\text{.}$$Thus, $$0\text{.}\text{92}={e}^{\mathrm{\lambda t}}\text{.}$$Taking the natural logarithm of both sides of the equation yields $$\text{ln}\phantom{\rule{0.25em}{0ex}}0\text{.}\text{92}=\mathrm{\u2013\lambda t}$$so that $$0\text{.}\text{0834}=\mathrm{\lambda t}\text{.}$$Rearranging to isolate $t$ gives $$t=\frac{0\text{.}\text{0834}}{\lambda}\text{.}$$Now, the equation $\lambda =\frac{0\text{.}\text{693}}{{t}_{1/2}}$ can be used to find $\lambda $ for ${}^{\text{14}}\text{C}$. Solving for $\lambda $ and substituting the known halflife gives $$\lambda =\frac{0\text{.}\text{693}}{{t}_{1/2}}=\frac{0\text{.}\text{693}}{\text{5730 y}}\text{.}$$We enter this value into the previous equation to find $t$: $$t=\frac{0\text{.}\text{0834}}{\frac{0\text{.}\text{693}}{\text{5730 y}}}=\text{690 y.}$$Discussion This dates the material in the shroud to 1988–690 = a.d. 1300. Our calculation is only accurate to two digits, so that the year is rounded to 1300. The values obtained at the three independent laboratories gave a weighted average date of a.d. $\text{1320}\pm \text{60}$. The uncertainty is typical of carbon14 dating and is due to the small amount of ${}^{\text{14}}\text{C}$ in living tissues, the amount of material available, and experimental uncertainties (reduced by having three independent measurements). It is meaningful that the date of the shroud is consistent with the first record of its existence and inconsistent with the period in which Jesus lived. There are other forms of radioactive dating. Rocks, for example, can sometimes be dated based on the decay of ${}^{\text{238}}\text{U}$. The decay series for ${}^{\text{238}}\text{U}$ ends with ${}^{\text{206}}\text{Pb}$, so that the ratio of these nuclides in a rock is an indication of how long it has been since the rock solidified. The original composition of the rock, such as the absence of lead, must be known with some confidence. However, as with carbon14 dating, the technique can be verified by a consistent body of knowledge. Since ${}^{\text{238}}\text{U}$ has a halflife of $4\text{.}5\times {\text{10}}^{9}$ y, it is useful for dating only very old materials, showing, for example, that the oldest rocks on Earth solidified about $3\text{.}5\times {\text{10}}^{9}$ years ago. Activity, the Rate of DecayWhat do we mean when we say a source is highly radioactive? Generally, this means the number of decays per unit time is very high. We define where $\text{\Delta}N$ is the number of decays that occur in time $\text{\Delta}t$. The SI unit for activity is one decay per second and is given the name Activity $R$ is often expressed in other units, such as decays per minute or decays per year. One of the most common units for activity is the or $3\text{.}\text{70}\times {\text{10}}^{\text{10}}$ decays per second. A curie is a large unit of activity, while a becquerel is a relatively small unit. $\text{1 MBq}=\text{100 microcuries}\phantom{\rule{0.25em}{0ex}}(\mu \text{Ci})$. In countries like Australia and New Zealand that adhere more to SI units, most radioactive sources, such as those used in medical diagnostics or in physics laboratories, are labeled in Bq or megabecquerel (MBq). Intuitively, you would expect the activity of a source to depend on two things: the amount of the radioactive substance present, and its halflife. The greater the number of radioactive nuclei present in the sample, the more will decay per unit of time. The shorter the halflife, the more decays per unit time, for a given number of nuclei. So activity $R$ should be proportional to the number of radioactive nuclei, $N$, and inversely proportional to their halflife, ${t}_{1/2}$. In fact, your intuition is correct. It can be shown that the activity of a source is $$R=\frac{0\text{.}\text{693}N}{{t}_{1/2}}$$where $N$ is the number of radioactive nuclei present, having halflife ${t}_{1/2}$. This relationship is useful in a variety of calculations, as the next two examples illustrate. Example 2: How Great Is the ${}^{\text{14}}\text{C}$ Activity in Living Tissue?Calculate the activity due to ${}^{\text{14}}\text{C}$ in 1.00 kg of carbon found in a living organism. Express the activity in units of Bq and Ci. Strategy To find the activity $R$ using the equation $R=\frac{0\text{.}\text{693}N}{{t}_{1/2}}$, we must know $N$ and ${t}_{1/2}$. The halflife of ${}^{\text{14}}\text{C}$ can be found in Appendix B, and was stated above as 5730 y. To find $N$, we first find the number of ${}^{\text{12}}\text{C}$ nuclei in 1.00 kg of carbon using the concept of a mole. As indicated, we then multiply by $1\text{.}3\times {\text{10}}^{\text{12}}$ (the abundance of ${}^{\text{14}}\text{C}$ in a carbon sample from a living organism) to get the number of ${}^{\text{14}}\text{C}$ nuclei in a living organism. Solution One mole of carbon has a mass of 12.0 g, since it is nearly pure ${}^{\text{12}}\text{C}$. (A mole has a mass in grams equal in magnitude to $A$ found in the periodic table.) Thus the number of carbon nuclei in a kilogram is $$N({}^{12}\text{C})=\frac{6.02\times {\text{10}}^{\text{23}}\phantom{\rule{0.25em}{0ex}}{\text{mol}}^{\mathrm{\u20131}}}{\mathrm{12.0\; g/mol}}\times \text{(1000 g)}=\text{5.02}\times {\text{10}}^{\text{25}}\text{.}$$So the number of ${}^{\text{14}}\text{C}$ nuclei in 1 kg of carbon is $$N({}^{\text{14}}\text{C})=(5.02\times {\text{10}}^{\text{25}})(1.3\times {\text{10}}^{\text{\u221212}})=6.52\times {\text{10}}^{\text{13}}\text{.}$$Now the activity $R$ is found using the equation $R=\frac{0\text{.}\text{693}N}{{t}_{1/2}}$. Entering known values gives $$R=\frac{0\text{.}\text{693}(6\text{.}\text{52}\times {\text{10}}^{\text{13}})}{\text{5730 y}}=7\text{.}\text{89}\times {\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}{\text{y}}^{\mathrm{\u20131}},$$or $7\text{.}\text{89}\times {\text{10}}^{9}$ decays per year. To convert this to the unit Bq, we simply convert years to seconds. Thus, $$R=(\text{7.89}\times {\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}{\text{y}}^{\mathrm{\u20131}})\frac{\mathrm{1.00\; y}}{3\text{.}\text{16}\times {\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{s}}=\text{250 Bq,}$$or 250 decays per second. To express $R$ in curies, we use the definition of a curie, $$R=\frac{\text{250 Bq}}{3.7\times {\text{10}}^{\text{10}}\phantom{\rule{0.25em}{0ex}}\text{Bq/Ci}}=6.76\times {\text{10}}^{9}\phantom{\rule{0.25em}{0ex}}\text{Ci.}$$Thus, $$R=6.76\phantom{\rule{0.25em}{0ex}}\text{nCi.}$$Discussion Our own bodies contain kilograms of carbon, and it is intriguing to think there are hundreds of ${}^{\text{14}}\text{C}$ decays per second taking place in us. Carbon14 and other naturally occurring radioactive substances in our bodies contribute to the background radiation we receive. The small number of decays per second found for a kilogram of carbon in this example gives you some idea of how difficult it is to detect ${}^{\text{14}}\text{C}$ in a small sample of material. If there are 250 decays per second in a kilogram, then there are 0.25 decays per second in a gram of carbon in living tissue. To observe this, you must be able to distinguish decays from other forms of radiation, in order to reduce background noise. This becomes more difficult with an old tissue sample, since it contains less ${}^{\text{14}}\text{C}$, and for samples more than 50 thousand years old, it is impossible. Humanmade (or artificial) radioactivity has been produced for decades and has many uses. Some of these include medical therapy for cancer, medical imaging and diagnostics, and food preservation by irradiation. Many applications as well as the biological effects of radiation are explored in Medical Applications of Nuclear Physics, but it is clear that radiation is hazardous. A number of tragic examples of this exist, one of the most disastrous being the meltdown and fire at the Chernobyl reactor complex in the Ukraine (see Figure 3). Several radioactive isotopes were released in huge quantities, contaminating many thousands of square kilometers and directly affecting hundreds of thousands of people. The most significant releases were of ${}^{\text{131}}\text{I}$, ${}^{\text{90}}\text{Sr}$, ${}^{\text{137}}\text{Cs}$, ${}^{\text{239}}\text{Pu}$, ${}^{\text{238}}\text{U}$, and ${}^{\text{235}}\text{U}$. Estimates are that the total amount of radiation released was about 100 million curies. Human and Medical ApplicationsExample 3: What Mass of ${}^{\text{137}}\text{Cs}$ Escaped Chernobyl?It is estimated that the Chernobyl disaster released 6.0 MCi of ${}^{\text{137}}\text{Cs}$ into the environment. Calculate the mass of ${}^{\text{137}}\text{Cs}$ released. Strategy We can calculate the mass released using Avogadro’s number and the concept of a mole if we can first find the number of nuclei $N$ released. Since the activity $R$ is given, and the halflife of ${}^{\text{137}}\text{Cs}$ is found in Appendix B to be 30.2 y, we can use the equation $R=\frac{0\text{.}\text{693}N}{{t}_{1/2}}$ to find $N$. Solution Solving the equation $R=\frac{0\text{.}\text{693}N}{{t}_{1/2}}$ for $N$ gives $$N=\frac{{\text{Rt}}_{\mathrm{1/2}}}{\text{0.693}}\text{.}$$Entering the given values yields $$N=\frac{(\mathrm{6.0\; MCi})(\text{30}\text{.}\mathrm{2\; y})}{0\text{.}\text{693}}\text{.}$$Converting curies to becquerels and years to seconds, we get $$\begin{array}{lll}N& =& \frac{(6\text{.}0\times {\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{Ci})(3\text{.}7\times {\text{10}}^{\text{10}}\phantom{\rule{0.25em}{0ex}}\text{Bq/Ci})(\text{30.2 y})(3\text{.}\text{16}\times {\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{s/y})}{\text{0.693}}\\ & =& \text{}3\text{.}1\times {\text{10}}^{\text{26}}\text{.}\end{array}$$One mole of a nuclide ${}^{A}X$ has a mass of $A$ grams, so that one mole of ${}^{\text{137}}\text{Cs}$ has a mass of 137 g. A mole has $6\text{.}\text{02}\times {\text{10}}^{\text{23}}$ nuclei. Thus the mass of ${}^{\text{137}}\text{Cs}$ released was $$\begin{array}{lll}m& =& \left(\frac{\text{137 g}}{\text{6.02}\times {\text{10}}^{\text{23}}}\right)(3\text{.}1\times {\text{10}}^{\text{26}})=\text{70}\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{g}\\ & =& \text{}\text{70 kg}\text{.}\end{array}$$Discussion While 70 kg of material may not be a very large mass compared to the amount of fuel in a power plant, it is extremely radioactive, since it only has a 30year halflife. Six megacuries (6.0 MCi) is an extraordinary amount of activity but is only a fraction of what is produced in nuclear reactors. Similar amounts of the other isotopes were also released at Chernobyl. Although the chances of such a disaster may have seemed small, the consequences were extremely severe, requiring greater caution than was used. More will be said about safe reactor design in the next chapter, but it should be noted that Western reactors have a fundamentally safer design. Activity $R$ decreases in time, going to half its original value in one halflife, then to onefourth its original value in the next halflife, and so on. Since $R=\frac{0\text{.}\text{693}N}{{t}_{1/2}}$, the activity decreases as the number of radioactive nuclei decreases. The equation for $R$ as a function of time is found by combining the equations $N={N}_{0}{e}^{\mathrm{\lambda t}}$ and $R=\frac{0\text{.}\text{693}N}{{t}_{1/2}}$, yielding $$R={R}_{0}{e}^{\mathrm{\lambda t}}\text{,}$$where ${R}_{0}$ is the activity at $t=0$. This equation shows exponential decay of radioactive nuclei. For example, if a source originally has a 1.00mCi activity, it declines to 0.500 mCi in one halflife, to 0.250 mCi in two halflives, to 0.125 mCi in three halflives, and so on. For times other than whole halflives, the equation $R={R}_{0}{e}^{\mathrm{\lambda t}}$ must be used to find $R$. PhET Explorations: Alpha DecayWatch alpha particles escape from a polonium nucleus, causing radioactive alpha decay. See how random decay times relate to the half life. Section Summary
Conceptual QuestionsExercise 1In a $3\times {\text{10}}^{9}$yearold rock that originally contained some ${}^{\text{238}}\text{U}$, which has a halflife of $4.5\times {\text{10}}^{9}$ years, we expect to find some ${}^{\text{238}}\text{U}$ remaining in it. Why are ${}^{\text{226}}\text{Ra}$, ${}^{\text{222}}\text{Rn}$, and ${}^{\text{210}}\text{Po}$ also found in such a rock, even though they have much shorter halflives (1600 years, 3.8 days, and 138 days, respectively)? Exercise 2Does the number of radioactive nuclei in a sample decrease to exactly half its original value in one halflife? Explain in terms of the statistical nature of radioactive decay. Exercise 3Radioactivity depends on the nucleus and not the atom or its chemical state. Why, then, is one kilogram of uranium more radioactive than one kilogram of uranium hexafluoride? Exercise 4Explain how a bound system can have less mass than its components. Why is this not observed classically, say for a building made of bricks? Exercise 5Spontaneous radioactive decay occurs only when the decay products have less mass than the parent, and it tends to produce a daughter that is more stable than the parent. Explain how this is related to the fact that more tightly bound nuclei are more stable. (Consider the binding energy per nucleon.) Exercise 6To obtain the most precise value of BE from the equation $\text{BE=}\left[\text{ZM}\left({}^{1}\text{H}\right)+{\text{Nm}}_{n}\right]{c}^{2}m\left({}^{A}\mathrm{X}\right){c}^{2}$, we should take into account the binding energy of the electrons in the neutral atoms. Will doing this produce a larger or smaller value for BE? Why is this effect usually negligible? Exercise 7How does the finite range of the nuclear force relate to the fact that $\text{BE}/A$ is greatest for $A$ near 60? Problems & ExercisesData from the appendices and the periodic table may be needed for these problems. Exercise 1An old campfire is uncovered during an archaeological dig. Its charcoal is found to contain less than 1/1000 the normal amount of ${}^{\text{14}}\text{C}$. Estimate the minimum age of the charcoal, noting that ${2}^{\text{10}}=\text{1024}$. Show/Hide Solution Solution57,300 y Exercise 2A ${}^{\text{60}}\text{Co}$ source is labeled 4.00 mCi, but its present activity is found to be $1\text{.}\text{85}\times {\text{10}}^{7}$ Bq. (a) What is the present activity in mCi? (b) How long ago did it actually have a 4.00mCi activity? Exercise 3(a) Calculate the activity $R$ in curies of 1.00 g of ${}^{\text{226}}\text{Ra}$. (b) Discuss why your answer is not exactly 1.00 Ci, given that the curie was originally supposed to be exactly the activity of a gram of radium. Show/Hide Solution Solution(a) 0.988 Ci (b) The halflife of ${}^{\text{226}}\text{Ra}$ is now better known. Exercise 4Show that the activity of the ${}^{\text{14}}\text{C}$ in 1.00 g of ${}^{\text{12}}\text{C}$ found in living tissue is 0.250 Bq. Exercise 5Mantles for gas lanterns contain thorium, because it forms an oxide that can survive being heated to incandescence for long periods of time. Natural thorium is almost 100% ${}^{\text{232}}\text{Th}$, with a halflife of $1\text{.}\text{405}\times {\text{10}}^{\text{10}}\phantom{\rule{0.25em}{0ex}}\text{y}$. If an average lantern mantle contains 300 mg of thorium, what is its activity? Show/Hide Solution Solution$1.22\times {\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{Bq}$ Exercise 6Cow’s milk produced near nuclear reactors can be tested for as little as 1.00 pCi of ${}^{\text{131}}\text{I}$ per liter, to check for possible reactor leakage. What mass of ${}^{\text{131}}\text{I}$ has this activity? Exercise 7(a) Natural potassium contains ${}^{\text{40}}\text{K}$, which has a halflife of $1\text{.}\text{277}\times {\text{10}}^{9}$ y. What mass of ${}^{\text{40}}\text{K}$ in a person would have a decay rate of 4140 Bq? (b) What is the fraction of ${}^{\text{40}}\text{K}$ in natural potassium, given that the person has 140 g in his body? (These numbers are typical for a 70kg adult.) Show/Hide Solution Solution(a) 16.0 mg (b) 0.0114% Exercise 8There is more than one isotope of natural uranium. If a researcher isolates 1.00 mg of the relatively scarce ${}^{\text{235}}\text{U}$ and finds this mass to have an activity of 80.0 Bq, what is its halflife in years? Exercise 9${}^{\text{50}}\text{V}$ has one of the longest known radioactive halflives. In a difficult experiment, a researcher found that the activity of 1.00 kg of ${}^{\text{50}}\text{V}$ is 1.75 Bq. What is the halflife in years? Show/Hide Solution Solution$1.48\times {\text{10}}^{\text{17}}\phantom{\rule{0.25em}{0ex}}\text{y}$ Exercise 10You can sometimes find deep red crystal vases in antique stores, called uranium glass because their color was produced by doping the glass with uranium. Look up the natural isotopes of uranium and their halflives, and calculate the activity of such a vase assuming it has 2.00 g of uranium in it. Neglect the activity of any daughter nuclides. Exercise 11A tree falls in a forest. How many years must pass before the ${}^{\text{14}}\text{C}$ activity in 1.00 g of the tree’s carbon drops to 1.00 decay per hour? Show/Hide Solution Solution$5.6\times {\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\mathrm{y}$ Exercise 12What fraction of the ${}^{\text{40}}\text{K}$ that was on Earth when it formed $4\text{.}5\times {\text{10}}^{9}$ years ago is left today? Exercise 13A 5000Ci ${}^{\text{60}}\text{Co}$ source used for cancer therapy is considered too weak to be useful when its activity falls to 3500 Ci. How long after its manufacture does this happen? Show/Hide Solution Solution2.71 y Exercise 14Natural uranium is 0.7200% ${}^{\text{235}}\text{U}$ and 99.27% ${}^{\text{238}}\text{U}$. What were the percentages of ${}^{\text{235}}\text{U}$ and ${}^{\text{238}}\text{U}$ in natural uranium when Earth formed $4\text{.}5\times {\text{10}}^{9}$ years ago? Exercise 15The ${\beta}^{}$ particles emitted in the decay of ${}^{3}\text{H}$ (tritium) interact with matter to create light in a glowinthedark exit sign. At the time of manufacture, such a sign contains 15.0 Ci of ${}^{3}\text{H}$. (a) What is the mass of the tritium? (b) What is its activity 5.00 y after manufacture? Show/Hide Solution Solution(a) 1.56 mg (b) 11.3 Ci Exercise 16World War II aircraft had instruments with glowing radiumpainted dials (see Section 312 Figure 1). The activity of one such instrument was $1.0\times {\text{10}}^{5}$ Bq when new. (a) What mass of ${}^{\text{226}}\text{Ra}$ was present? (b) After some years, the phosphors on the dials deteriorated chemically, but the radium did not escape. What is the activity of this instrument 57.0 years after it was made? Exercise 17(a) The ${}^{\text{210}}\text{Po}$ source used in a physics laboratory is labeled as having an activity of $1.0\phantom{\rule{0.25em}{0ex}}\mu \text{Ci}$ on the date it was prepared. A student measures the radioactivity of this source with a Geiger counter and observes 1500 counts per minute. She notices that the source was prepared 120 days before her lab. What fraction of the decays is she observing with her apparatus? (b) Identify some of the reasons that only a fraction of the $\alpha $ s emitted are observed by the detector. Show/Hide Solution Solution(a) $\text{1.23}\times {\text{10}}^{3}$ (b) Only part of the emitted radiation goes in the direction of the detector. Only a fraction of that causes a response in the detector. Some of the emitted radiation (mostly $\alpha $ particles) is observed within the source. Some is absorbed within the source, some is absorbed by the detector, and some does not penetrate the detector. Exercise 18Armorpiercing shells with depleted uranium cores are fired by aircraft at tanks. (The high density of the uranium makes them effective.) The uranium is called depleted because it has had its ${}^{\text{235}}\text{U}$ removed for reactor use and is nearly pure ${}^{\text{238}}\text{U}$. Depleted uranium has been erroneously called nonradioactive. To demonstrate that this is wrong: (a) Calculate the activity of 60.0 g of pure ${}^{\text{238}}\text{U}$. (b) Calculate the activity of 60.0 g of natural uranium, neglecting the ${}^{\text{234}}\text{U}$ and all daughter nuclides. Exercise 19The ceramic glaze on a redorange Fiestaware plate is ${\text{U}}_{2}{\text{O}}_{3}$ and contains 50.0 grams of ${}^{238}\text{U}$ , but very little ${}^{235}\text{U}$. (a) What is the activity of the plate? (b) Calculate the total energy that will be released by the ${}^{238}\text{U}$ decay. (c) If energy is worth 12.0 cents per $\text{kW}\cdot \text{h}$, what is the monetary value of the energy emitted? (These plates went out of production some 30 years ago, but are still available as collectibles.) Show/Hide Solution Solution(a) $1.68\times {10}^{\u20135}\phantom{\rule{0.25em}{0ex}}\text{Ci}$ (b) $8.65\times {10}^{10}\phantom{\rule{0.25em}{0ex}}\text{J}$ (c) $\$2.9\times {10}^{3}$ Exercise 20Large amounts of depleted uranium (${}^{238}\text{U}$) are available as a byproduct of uranium processing for reactor fuel and weapons. Uranium is very dense and makes good counter weights for aircraft. Suppose you have a 4000kg block of ${}^{238}\text{U}$. (a) Find its activity. (b) How many calories per day are generated by thermalization of the decay energy? (c) Do you think you could detect this as heat? Explain. Exercise 21The Galileo space probe was launched on its long journey past several planets in 1989, with an ultimate goal of Jupiter. Its power source is 11.0 kg of ${}^{238}\text{Pu}$, a byproduct of nuclear weapons plutonium production. Electrical energy is generated thermoelectrically from the heat produced when the 5.59MeV $\text{\alpha}$ particles emitted in each decay crash to a halt inside the plutonium and its shielding. The halflife of ${}^{238}\text{Pu}$ is 87.7 years. (a) What was the original activity of the ${}^{238}\text{Pu}$ in becquerel? (b) What power was emitted in kilowatts? (c) What power was emitted 12.0 y after launch? You may neglect any extra energy from daughter nuclides and any losses from escaping $\text{\gamma}$ rays. Show/Hide Solution Solution(a) $6.97\times {10}^{15}\phantom{\rule{0.25em}{0ex}}\text{Bq}$ (b) 6.24 kW (c) 5.67 kW Exercise 22Construct Your Own Problem Consider the generation of electricity by a radioactive isotope in a space probe, such as described in Exercise 21. Construct a problem in which you calculate the mass of a radioactive isotope you need in order to supply power for a long space flight. Among the things to consider are the isotope chosen, its halflife and decay energy, the power needs of the probe and the length of the flight. Exercise 23Unreasonable Results A nuclear physicist finds $1.0\phantom{\rule{0.25em}{0ex}}\mu \text{g}$ of ${}^{236}\text{U}$ in a piece of uranium ore and assumes it is primordial since its halflife is $2.3\times {10}^{7}\phantom{\rule{0.25em}{0ex}}\text{y}$. (a) Calculate the amount of ${}^{236}\text{U}$that would had to have been on Earth when it formed $4.5\times {10}^{9}\phantom{\rule{0.25em}{0ex}}\text{y}$ ago for $1.0\phantom{\rule{0.25em}{0ex}}\mu \text{g}$ to be left today. (b) What is unreasonable about this result? (c) What assumption is responsible? Exercise 24Unreasonable Results (a) Repeat Exercise 14 but include the 0.0055% natural abundance of ${}^{234}\text{U}$ with its $2.45\times {10}^{5}\phantom{\rule{0.25em}{0ex}}\text{y}$ halflife. (b) What is unreasonable about this result? (c) What assumption is responsible? (d) Where does the ${}^{234}\text{U}$ come from if it is not primordial? Exercise 25Unreasonable Results The manufacturer of a smoke alarm decides that the smallest current of $\text{\alpha}$ radiation he can detect is $1.00\phantom{\rule{0.25em}{0ex}}\mu \text{A}$. (a) Find the activity in curies of an $\text{\alpha}$ emitter that produces a $1.00\phantom{\rule{0.25em}{0ex}}\mu \text{A}$ current of $\text{\alpha}$ particles. (b) What is unreasonable about this result? (c) What assumption is responsible? Show/Hide Solution Solution(a) 84.5 Ci (b) An extremely large activity, many orders of magnitude greater than permitted for home use. (c) The assumption of $1.00\phantom{\rule{0.25em}{0ex}}\text{\mu A}$ is unreasonably large. Other methods can detect much smaller decay rates.
