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\(\textbf{We Are Stardust.}\) In 1952 spectral lines of the element technetium-99 (\(^9$$^9\)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 \(^9$$^9\)Tc is 200,000 years. (a) For how many halflives has the \(^9$$^9\)Tc been in the red giant star if its age is 10 billion years? (b) What fraction of the original \(^9$$^9\)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 \(^9$$^9\)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) Remaining fraction is nearly zero.

Step by step solution

01

Calculate the number of half-lives

To find out how many half-lives have elapsed over 10 billion years, we divide the total age of the star by the half-life of technetium-99. \[\text{Number of half-lives} = \frac{10,000,000,000 \text{ years}}{200,000 \text{ years/half-life}} \] Calculate this to find the number of half-lives.
02

Calculate the number of half-lives

\[\text{Number of half-lives} = \frac{10,000,000,000}{200,000} = 50,000 \text{ half-lives}\]This indicates that the technetium-99 in the star has undergone 50,000 half-lives.
03

Calculate the remaining fraction of technetium-99

After knowing the number of half-lives, use the formula for exponential decay to determine the remaining fraction of technetium-99: \[\text{Remaining fraction} = \left(\frac{1}{2}\right)^{n} \] where \(n\) is the number of half-lives (50,000).
04

Calculate the small fraction

Insert \(n = 50,000\) into the formula to calculate: \[\text{Remaining fraction} = \left(\frac{1}{2}\right)^{50,000} \approx 0 \]Practically, this fraction is extremely small, close to zero, indicating that almost none of the original technetium-99 remains after 10 billion years.

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Key Concepts

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

Half-life
Understanding the concept of "half-life" is crucial when studying radioactive elements like technetium-99. A "half-life" is the time it takes for half of a radioactive substance to decay. This means that after every half-life, only 50% of the radioisotope's original quantity remains.

For example:
  • If a radioactive substance has a half-life of 200,000 years, like technetium-99, after 200,000 years, you would have only half of the substance left.
  • After 400,000 years, you would have a quarter left, as it would half again.
  • This process continues over successive half-lives, reducing the substance to smaller and smaller fractions.
You can see the power of this concept when applied to the red giant star. Over 10 billion years, technetium-99 in the star would undergo 50,000 half-lives, greatly diminishing its original quantity.
Exponential decay
Exponential decay describes how the quantity of a radioactive element decreases over time. It's characterized by a rapid drop-off followed by a slower decrease, a process best illustrated through the half-life concept.

The mathematical formula for exponential decay is:
  • \[N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}}\]
where:
  • \(N(t)\) is the remaining quantity of the substance at time \(t\).
  • \(N_0\) is the initial quantity of the substance.
  • \(t_{1/2}\) is the half-life of the substance.
For technetium-99 in a red giant, after 50,000 half-lives, the remaining quantity becomes practically zero. This is due to how quickly and exponentially the element decays over multiple half-lives, emphasizing the significant reduction from the original amount.
Red giant star
Red giant stars are fascinating bodies in space, often marking the late stages of stellar evolution. As stars exhaust the hydrogen in their cores, they undergo expansion, cooling the outer layers to form a red giant.

Some key characteristics of red giants include:
  • They are typically over 10 billion years old, representing an advanced age for stellar bodies.
  • The surface appears red due to lower temperatures compared to younger, hotter stars.
  • Red giants play a significant role in the synthesis of elements through nuclear processes.
Discovering elements like technetium in such stars provides evidence for ongoing elemental synthesis, important for understanding stellar life cycles.
Thermonuclear fusion
Thermonuclear fusion is a powerful process occurring in the heart of stars, including red giants. It involves fusing lighter atomic nuclei to form heavier nuclei, releasing immense amounts of energy.

Highlights of thermonuclear fusion in stars include:
  • Hydrogen atoms fuse into helium, which in later stages, can fuse into heavier elements.
  • This process is how stars, including our sun, produce the energy that powers them and emits light and heat.
  • Elements heavier than helium, such as technetium, are formed in stars through fusion and subsequent processes.
Technetium's presence in a red giant proves that elements heavier than hydrogen and helium originate in stars, supporting theories like "we are stardust." This phrase suggests that many elements found in our solar system and planet were created long ago in the fiery cores of ancient stars.

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Most popular questions from this chapter

Measurements indicate that 27.83% of all rubidium atoms currently on the earth are the radioactive \(^8$$^7\)Rb isotope. The rest are the stable \(^8$$^5\)Rb isotope. The half-life of \(^8$$^7\)Rb is 4.75 \(\times\) 10\(^1$$^0\) y. Assuming that no rubidium atoms have been formed since, what percentage of rubidium atoms were \(^8$$^7\)Rb when our solar system was formed 4.6 \(\times\) 10\(^9\) y ago?

\(\textbf{Radioactive Fallout.}\) 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 (\(^9$$^0\)Sr), 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 \(^9$$^0\)Sr decay? (b) What percentage of the original level of \(^9$$^0\)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?

\(\textbf{Comparison of Energy Released per Gram of Fuel.}\) (a) When gasoline is burned, it releases 1.3 \(\times\) 10\(^8\) J of energy per gallon (3.788 L). Given that the density of gasoline is 737 kg/m\(^3\), express the quantity of energy released in J/g of fuel. (b) During fission, when a neutron is absorbed by a \(^2$$^3$$^5\)U nucleus, about 200 MeV of energy is released for each nucleus that undergoes fission. Express this quantity in J/g of fuel. (c) In the proton-proton chain that takes place in stars like our sun, the overall fusion reaction can be summarized as six protons fusing to form one \(^4\)He nucleus with two leftover protons and the liberation of 26.7 MeV of energy. The fuel is the six protons. Express the energy produced here in units of J/g of fuel. Notice the huge difference between the two forms of nuclear energy, on the one hand, and the chemical energy from gasoline, on the other. (d) Our sun produces energy at a measured rate of 3.86 \(\times\) 10\(^2$$^6\) W. If its mass of 1.99 \(\times\) 10\(^3$$^0\) kg were all gasoline, how long could it last before consuming all its fuel? (\(Historical\) \(note\): Before the discovery of nuclear fusion and the vast amounts of energy it releases, scientists were confused. They knew that the earth was at least many millions of years old, but could not explain how the sun could survive that long if its energy came from chemical burning.)

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

\(\textbf{An Oceanographic Tracer.}\) Nuclear weapons tests in the 1950s and 1960s released significant amounts of radioactive tritium (\(^{3}_{1}H\), 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, \(^{3}_{2}He\), to the remaining tritium in the water. For example, if the ratio of \(^{3}_{2}He\) to \(^{3}_{1}H\) 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 \(^{3}_{2}He\) to \(^{3}_{1}H\) is 4.3 to 1.0. How many years ago did this water sink below the surface?

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