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Measurements indicate that 27.83\(\%\) of all rubidium atoms currently on the earth are the radioactive \(^{87} \mathrm{Rb}\) isotope. The rest are the stable \(^{87} \mathrm{Rb}\) isotope. The half-life of \(^{87} \mathrm{Rb}\) is \(4.75 \times 10^{10} \mathrm{y}\) . Assuming that no rubidium atoms have been formed since, what percentage of rubidium atoms were \(^{87} \mathrm{Rb}\) when our solar system was formed \(4.6 \times 10^{9} \mathrm{y}\) ago?

Short Answer

Expert verified
Approximately 31.97\% of rubidium atoms were \(^ {87} \mathrm{Rb} \) when the solar system formed.

Step by step solution

01

Understand the Given and Required

We know that 27.83% of the rubidium atoms currently are the radioactive isotope \(^ {87} \mathrm{Rb} \). We need to find the percentage of \(^ {87} \mathrm{Rb} \) atoms when the solar system formed 4.6 billion years ago. The half-life of \(^ {87} \mathrm{Rb} \) is given as \(4.75 \times 10^{10} \) years.
02

Write the Decay Formula

Use the decay formula \( N = N_0 \times \left( \frac{1}{2} \right)^{t / t_{1/2}} \), where \( N \) is the current number of \(^{87} \mathrm{Rb} \) atoms, \( N_0 \) is the initial number, \( t \) is the time, and \( t_{1/2} \) is the half-life.
03

Express Current and Initial Percentages

Let the original percentage of \(^{87} \mathrm{Rb} \) be \( P_0 \) and the current percentage be \( P \). We have \( P = P_0 \times \left( \frac{1}{2} \right)^{4.6 \times 10^9 / 4.75 \times 10^{10}} \).
04

Calculate Current Ratio of Decay

Compute the decay factor for the 4.6 billion years. \(\left( \frac{1}{2} \right)^{4.6 \times 10^9 / 4.75 \times 10^{10}} \approx 0.8707\).
05

Solve for Initial Percentage

Since \( P = 27.83\% \), solve the equation \( 27.83 = P_0 \times 0.8707 \) to find \( P_0 \). Calculate \( P_0 = \frac{27.83}{0.8707} \approx 31.97\%\).
06

Conclusion

The percentage of \(^{87} \mathrm{Rb} \) when the solar system was formed was approximately \(31.97\%\).

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Key Concepts

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

Half-life calculation
The concept of half-life is crucial in understanding radioactive decay. It refers to the amount of time it takes for half of a given amount of a radioactive isotope to decay. For example, in the case of the radioactive isotope Rubidium-87, its half-life is given as \(4.75 \times 10^{10}\) years. This means that every \(4.75 \times 10^{10}\) years, half of the Rubidium-87 atoms will have decayed into another element.

When calculating how much of a radioactive isotope remains after a certain period, we use the formula:
  • \( N = N_0 \times \left( \frac{1}{2} \right)^{t / t_{1/2}} \)
Where:
  • \(N\) is the current amount of the isotope.
  • \(N_0\) is the initial amount of the isotope.
  • \(t\) is the elapsed time.
  • \(t_{1/2}\) is the half-life of the isotope.
This formula helps in determining how much of a radioactive substance remains after a specific amount of time, given its half-life. In the provided problem, we calculated the decay over \(4.6 \times 10^9\) years to find out how much Rubidium-87 remained from the initial composition.
Isotope composition
Isotopes are different forms of an element's atoms, which have the same number of protons but a different number of neutrons. This difference in neutron number leads to variations in the atomic mass and can alter the stability of the atom.

In naturally occurring elements, isotopes are often found in specific and stable ratios, but some isotopes are radioactive. Radioactive isotopes, like Rubidium-87, decay over time at a predictable rate, quantified by their half-lives. Knowing the isotope composition, including both stable and radioactive isotopes, enables scientists to date materials and understand historical geological and astronomical events.
  • The original problem involves understanding that only \(27.83\%\) of present rubidium atoms are Rubidium-87.
  • Back in the solar system's formation, a higher percentage of natural Rubidium would have been Rubidium-87, due to the lapse of radioactive decay over billions of years.
Using the decay calculations, the initial isotope composition is estimated to determine how the ratios have shifted over time, giving insights into the processes that have occurred since the solar system's formation.
Rubidium
Rubidium is a soft, silvery-white metallic element in the alkali metal group, with the symbol \(Rb\) and atomic number 37. It's naturally occurring, mostly as the stable isotope Rubidium-85, which accounts for 72.17\% of naturally occurring rubidium. However, Earth also contains a minor percentage of the radioactive isotope Rubidium-87, as discussed in our exercise.

Rubidium-87 has been useful in dating geological materials due to its long half-life. The isotopic composition of rubidium is crucial for scientists using radiometric dating techniques, particularly the rubidium-strontium method. This method relies on the decay of Rubidium-87 into Strontium-87 over time, a decay process that can help date ancient rocks and minerals.
  • Rubidium was not synthesized but is found naturally as the isotopes \(^ {85} \text{Rb}\) and \(^ {87} \text{Rb}\).
  • The decay process and isotope ratios help study the Earth's history.
  • Major uses of rubidium include deep scientific research and various industrial applications like specialty glasses.
Understanding its isotopic nature aids in interpreting geological data and advances knowledge in fields that study the Earth's processes and history.

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Most popular questions from this chapter

Hydrogen atoms are placed in an external \(1.65-\mathrm{T}\) magnetic field. (a) The protons can make transitions between states where the nuclear spin component is parallel and antiparallel to the field by absorbing or emitting a photon. Which state has lower energy: the state with the nuclear spin component parallel or antiparallel to the field? What are the frequency and wavelength of the photon? In which region of the electromagnetic spectrum does it lie? (b) The electrons can make transitions between states where the electron spin component is parallel and antiparallel to the field by absorbing or emitting a photon. Which state has lower energy: the state with the electron spin component parallel or antiparallel to the field? What are the frequency and wavelength of the photon? In which region of the electromagnetic spectrum does it lie?

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?

BIO 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 \(\beta^{-}\) 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 d (about 3\(\frac{1}{2}\) weeks) later? (b) If the decay rate in the thyroid 24 d later is actually measured to be 17.0 Bq, what percentage of the tracer went to that gland? (c) What isotope remains after the I-131 decays?

BIO To Scan or Not to Scan? It has become popular for some people to have yearly whole-body scans (CT scans, formerly called CAT scans) using x rays, just to see if they detect anything suspicious. A number of medical people have recently questioned the advisability of such scans, due in part to the radiation they impart. Typically, one such scan gives a dose of 12 \(\mathrm{mSv}\) , applied to the whole body. By contrast, a chest \(x\) ray typically administers 0.20 mSv to only 5.0 kg of tissue. How many chest \(x\) rays would deliver the same total amount of energy to the body of a 75 -kg person as one whole-body scan?

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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