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We Are Stardust. In 1952 spectral lines of the element technetium- 99\(\left(^{99} \mathrm{Tc}\right)\) were discovered in a red giant star. Red giants are very old stars, often around 10 billion years old, and near the end of their lives. Technetium has \(no\) stable isotopes, and the half-life of \(^{99} \mathrm{Tc}\) is \(200,000\) years. (a) For how many half-lives has the \(^{99} \mathrm{Tc}\) been in the red-giant star if its age is 10 billion years? (b) What fraction of the original \(^{99} \mathrm{Tc}\) would be left at the end of that time? This discovery was extremely important because it provided convincing evidence for the theory (now essentially known to be true) that most of the atoms heavier than hydrogen and helium were made inside of stars by thermonuclear fusion and other nuclear processes. If the \(^{99} \mathrm{Tc}\) had been part of the star since it was born, the amount remaining after 10 billion years would have been so minute that it would not have been detectable. This knowledge is what led the late astronomer Carl Sagan to proclaim that "we are stardust."

Short Answer

Expert verified
(a) 50,000 half-lives; (b) Practically none remaining (\((1/2)^{50,000}\)).

Step by step solution

01

Determine the Number of Half-Lives

To find the number of half-lives, divide the total time (the age of the star) by the half-life of technetium-99. The age of the star is 10 billion years, and the half-life of technetium-99 is 200,000 years. Thus, the number of half-lives is: \[ n = \frac{10,000,000,000}{200,000} = 50,000. \]
02

Calculate the Remaining Fraction of Technetium

For a given number of half-lives, the remaining fraction of a radioactive isotope is given by \((1/2)^n\), where \(n\) is the number of half-lives. Here, \(n = 50,000\), so the remaining fraction of technetium is: \[ \left(\frac{1}{2}\right)^{50,000}. \]
03

Interpret the Result

The computation \((1/2)^{50,000}\) yields an extremely small number that is practically zero (well below any observable threshold). Thus, any original technetium-99 would have completely decayed over the course of 10 billion years, making it statistically impossible for any of it to remain from the birth of the star.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Technetium-99
Technetium-99 is quite a unique element in the universe. It was first discovered in stars like red giants, which are nearing the end of their life cycle. What makes technetium-99 fascinating is that it doesn't have any stable isotopes. This means all of its isotopes eventually decay into other elements. It's also important to note that technetium-99 is not naturally occurring on Earth, but is observed in stars, which tells us a lot about nuclear processes in stellar environments. The presence of technetium-99 in such old stars is a strong clue about its formation. Stars create heavier elements through complex nuclear reactions. Technetium-99 is particularly interesting because its presence in stars suggests that these elements are continuously formed and not leftover from the beginning of the star's life.
Radioactive Decay
Radioactive decay refers to the process where an unstable atomic nucleus loses energy by emitting radiation. This process is random and spontaneous. Technetium-99 undergoes radioactive decay, which means it transforms over time into lighter elements. This decay happens at a fixed rate characterized by its half-life. During radioactive decay, technetium-99 emits
  • Beta particles,
  • Gamma rays,
  • Other forms of radiation that help it become more stable over time.
Understanding radioactive decay is essential because it explains why technetium-99 couldn’t have existed in its original quantity if it had been in the star for billions of years. Its presence indicates continuous formation through stellar processes.
Thermonuclear Fusion
Thermonuclear fusion is a process where two lighter atomic nuclei fuse to form a heavier nucleus, releasing a tremendous amount of energy. This process occurs at extremely high temperatures and pressures, typically found in the cores of stars. It's the powerhouse behind star energy and is responsible for the creation of elements heavier than hydrogen and helium. In stars like red giants, fusion reactions can produce intermediate-mass elements, including technetium-99. The discovery of technetium-99 in stars offers strong evidence of ongoing fusion processes.
  • This discovery was crucial because it supported the theory that heavier elements originate from stars.
  • It showed that elements beyond hydrogen and helium are not primordial but rather products of stellar activity.
Half-Life
The concept of half-life is a key aspect of understanding radioactive decay. A half-life is the time it takes for half of the radioactive isotope in a sample to decay. For technetium-99, this period is 200,000 years. Calculating the number of half-lives over the star's age (10 billion years) helps determine how much, if any, technetium would remain. Knowing that technetium-99 has such a short half-life helps us understand why we don’t see it leftover in older parts of the universe without newly forming processes.
  • After one half-life, only 50% of the original substance remains.
  • After 50,000 half-lives, as in this case, virtually nothing remains, due to the exponential nature of decay.
This calculation showed that finding technetium-99 in an old star implies it's continuously regenerated within the star.

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Most popular questions from this chapter

An Oceanographic Tracer. Nuclear weapons tests in the 1950 s and 1960 s released significant amounts of radioactive tritium \((^{3}_{1} \mathrm{H},\) half-life 12.3 years \()\) into the atmosphere. The tritium atoms were quickly bound into water molecules and rained out of the air, most of them ending up in the ocean. For any of this tritium-tagged water that sinks below the surface, the amount of time during which it has been isolated from the surface can be calculated by measuring the ratio of the decay product, \(^{3}_{2} \mathrm{He},\) to the remaining tritium in the water. For example, if the ratio of \(_{2}^{3} \mathrm{He}\) to \(_{1}^{3} \mathrm{H}\) in a sample of water is \(1 : 1,\) the water has been below the surface for one half-life, or approximately 12 years. This method has provided oceanographers with a convenient way to trace the movements of subsurface currents in parts of the ocean. Suppose that in a particular sample of water, the ratio of \(_{2}^{3}\) He to \(_{1}^{3} \mathrm{H}\) is 4.3 to 1.0. How many years ago did this water sink below the surface?

Consider the nuclear reaction $$ _{14}^{28} \mathrm{Si}+\gamma \rightarrow_{12}^{24} \mathrm{Mg}+\mathrm{X} $$ where \(X\) is a nuclide. (a) What are \(Z\) and \(A\) for the nuclide \(X\) ? (b) Ignoring the effects of recoil, what minimum energy must the photon have for this reaction to occur? The mass of a \(_{14}^{28}\) Si atom is 27.976927 \( \mathrm{u},\) and the mass of a \(^{24}_{12} \mathrm{Mg}\) atom is 23.985042 \(\mathrm{u}\)

BIO Radioactive Fallout. One of the problems of in-air testing of nuclear weapons (or, even worse, the use of such weapons!) is the danger of radioactive fallout. One of the most problematic nuclides in such fallout is strontium-90 \(\left(^{90} \mathrm{Sr}\right),\) which breaks down by \(\beta^{-}\) decay with a half-life of 28 years. It is chemically similar to calcium and therefore can be incorporated into bones and teeth, where, due to its rather long half-life, it remains for years as an internal source of radiation. (a) What is the daughter nucleus of the \(^{90}\) Sr decay? (b) What percentage of the original level of \(^{90}\) Sr is left after 56 years? (c) How long would you have to wait for the original level to be reduced to 6.25\(\%\) of its original value?

Industrial Radioactivity. Radioisotopes are used in a variety of manufacturing and testing techniques. Wear measurements can be made using the following method. An automobile engine is produced using piston rings with a total mass of \(100 \mathrm{g},\) which includes 9.4\(\mu \mathrm{Ci}\) of \(^{59} \mathrm{Fe}\) whose half-life is 45 days. The engine is test-run for 1000 hours, after which the oil is drained and its activity is measured. If the activity of the engine oil is 84 decays/s, how much mass worn from the piston rings per hour of operation?

How many protons and how many neutrons are there in a nucleus of the most common isotope of (a) silicon, \(_{14}^{28} \mathrm{Si} ;\) (b) rubidium, \(\frac{85}{37} \mathrm{Rb} ;\) (c) thallium, \(_{81}^{205} \mathrm{Tl} ?\)

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