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43.65. We Are Stardust. In 1952 spectral lines of the element technetium- 99 ( \(^{99} \mathrm{Tc} )\) 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 \(^{99} \mathrm{Te}\) is \(200,000\) years. (a) For how many half-lives has the \(^{99}\) Te been in the red-giant star if its age is 10 billion years? (b) What fraction of the original "Te 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 \(^{99} \mathrm{Tchadbeen}\) 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 no technetium-99 remains.

Step by step solution

01

Determine the Half-Life Duration in Number of Years

Since the element technetium-99 has a half-life of 200,000 years, we use this value as a basis for calculating how many half-lives fit into 10 billion years.
02

Calculate the Number of Half-Lives over 10 Billion Years

Divide the total time of 10 billion years by the half-life of technetium-99 to find the number of half-lives: \[ \text{Number of half-lives} = \frac{10 \text{ billion years}}{200,000 \text{ years/half-life}} = \frac{10^{{10}}}{2 \times 10^5} = 5 \times 10^4 \text{ half-lives} \]
03

Determine the Remaining Fraction of Original Technetium-99

For each half-life, the remaining fraction of a substance is halved, so after \( n \) half-lives, the fraction remaining is given by \( \left(\frac{1}{2}\right)^n \). \[ \text{Remaining fraction} = \left(\frac{1}{2}\right)^{5 \times 10^4} \] This result is an extremely small number, essentially zero, indicating that after 50,000 half-lives, there would be virtually none left.

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Key Concepts

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

Nuclear Fusion in Stars
Stars shine and produce energy through a process known as nuclear fusion. In the intense heat and pressure at the core of a star, lighter elements like hydrogen are fused together to form heavier elements such as helium. This process releases an enormous amount of energy.

Nuclear fusion in stars is fundamental because it's the source of the light and heat emitted by stars, including our Sun. These energy-producing reactions push back against the force of gravity pulling the star inward, maintaining the star's stability over billions of years.

One key aspect of nuclear fusion is that it operates like a chain reaction. As heavier elements are produced, further fusion reactions can take place, forming elements up to iron. Elements heavier than iron require different processes, like those occurring in supernovae explosions, to form.
  • Fusion mainly occurs in stars' cores, where temperatures reach millions of degrees.
  • It creates elements crucial for life, including carbon, oxygen, and others.
  • The energy created balances the gravitational forces in the star.
Half-Life Calculations
Half-life is a concept used to understand the time it takes for half of a radioactive substance to decay. When calculating how much of a radioactive element remains over time, the concept of half-life allows us to determine the decay process.

For technetium-99, with a half-life of 200,000 years, the half-life formula is essential. The formula for the remaining quantity of a substance after a certain time is:\[\text{Remaining fraction} = \left(\frac{1}{2}\right)^{n}\]
Here, \(n\) is the number of half-lives elapsed, found by dividing the total time by the half-life duration.
  • The amount of technetium-99 left is minuscule after 10 billion years, showing why it was a significant discovery.
  • Knowing how to perform half-life calculations helps in understanding decay processes in various scientific fields.
Stellar Nucleosynthesis
Stellar nucleosynthesis refers to the process by which stars generate new elements in their cores. This process begins with nuclear fusion, but it doesn't stop there. As stars evolve, they create heavier and heavier elements.

In young stars, light elements like hydrogen and helium fuse to form elements up to iron. However, in more massive stars, or in their explosive ends as supernovae, even heavier elements are formed.

The discovery of technetium in a star was a breakthrough that confirmed much of our scientific understanding of stellar nucleosynthesis. Since technetium has no stable isotopes and decays relatively quickly on a cosmic timescale, finding it in a star showed it must have been produced there—not through ancient processes, but in the contemporary life of the star.
  • Processes like the s-process and r-process describe how stars can create these elements.
  • Most elements in the universe were formed inside stars, supporting the idea that "we are stardust."
  • This process is central to our understanding of the chemical composition of the universe.

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Most popular questions from this chapter

43.80. 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 was worn from the piston rings per hour of operation?

43.24. Radioactive Tracers. Radioactive isotopes are often introduced into the body through the bloodstream. Their spread through the body can then be monitored by detecting the appearance of radiation in different organs. 131 \(\mathrm{I}\) , a \(B^{-}\) emitter with a half-life of 8.0 \(\mathrm{d}\) is one such tracer. Suppose a scientist introduces a sample with an activity of 375 \(\mathrm{Bq}\) and watches it spread to the organs. (a) Assuming that the sample all went to the thyroid gland, what will be the decay rate in that gland 24 \(\mathrm{d}\) (about 2\(\frac{1}{2}\) weeks) later? (b) If the decay rate in the thyroid 24 d later is actually measured to be 17.0 \(\mathrm{Bq}\) , what percentage of the tracer went to that gland? (c) What isotope remains after the \(\mathrm{I}-131\) decays?

43.72. An Oceanographic Tracer. Nuclear weapons tests in the 1950 s and 1960 s released significant amounts of radioactive tritium \((3 \mathrm{H}, \text { half-life } 12.3 \text { years) into the atmosphere. The tritium }\) 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, \(\frac{3}{2} \mathrm{He},\) to the remaining tritium in the water. For example, if the ratio of \(\frac{3}{1}\) He to \(\frac{3}{1} \mathrm{H}\) 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 \(\frac{3}{2} \mathrm{He}\) to \(\frac{3}{1} \mathrm{H}\) is 4.3 to 1.0 . How many years ago did this water sink below the surface?

43.55. The atomic mass of \(\frac{25}{12} \mathrm{Mg}\) is 24.985837 \(\mathrm{u}\) , and the atomic mass of 13 \(\mathrm{Al}\) is 24.990429 \(\mathrm{u}\) . (a) Which of these nuclei will decay into the other? (b) What type of decay will occur? Explain how you determined this. (c) How much energy (in MeV) is released in the decay?

43.13. What nuclide is produced in the following radioactive decays? (a) \(\alpha\) decay of \(^{239}_{4} \mathrm{Pu}\) (b) \(\beta^{-}\) decay of \(^{24} \mathrm{Na}\) (c) \(\beta^{+}\) decay of \(\frac{15}{8} 0\)

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