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Chapter 43: Problem 61

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?

Short Answer

Expert verified
When the solar system was formed, approximately 29.76% of rubidium atoms were \(^{87}\text{Rb}\).

Step by step solution

01

Understand the Current Ratio

At present, 27.83% of rubidium atoms are \(^{87}\text{Rb}\);thus, \(100\% - 27.83\% = 72.17\%\) are \(^{85}\text{Rb}\). This establishes our current ratio between the isotopes.
02

Use the Half-Life Formula

To find out the initial percentage of \(^{87}\text{Rb}\), we need to use the half-life formula:\[N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}}\]where\(N(t)\)is the current amount,\(N_0\)is the initial amount,\(t\)is the elapsed time, and\(t_{1/2}\)is the half-life.
03

Calculate Remaining Fraction After Elapsed Time

Using the half-life formula, calculate the fraction of \(^{87}\text{Rb}\) remaining after \(4.6 \times 10^9\, \text{years}\):\[N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{4.6\times10^9}{4.75\times10^{10}}}\]\[N(t) = N_0 \left(\frac{1}{2}\right)^{\approx 0.0968}\]
04

Determine Remaining Fraction

Compute the exponent to find the remaining fraction:\[N(t) \approx N_0 \left(0.935\right)\]Thus, about \(93.5\%\)of the initial amount remains.
05

Calculate Initial Percentage of \(^{87}\text{Rb}\)

If the current remaining \(^{87}\text{Rb}\)is 27.83% of the total, and it's \(93.5\%\)of the original,then:\[0.935N_0 = 27.83\%\]Solving for \(N_0\):\[N_0 = \frac{27.83}{0.935}\]\[N_0 \approx 29.76\%\]
06

Conclusion about the Initial State

Approximately\(29.76\%\)of all rubidium atoms were \(^{87}\text{Rb}\)when the solar system was formed.

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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 radioactive decay, as it refers to the time required for half of a given amount of a radioactive substance to decay. This means that after one half-life, only 50% of the original radioactive material remains. After two half-lives, only 25% remains, continuing this pattern.
To solve problems involving half-life, we typically use the half-life formula: \[N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}}\]where:
  • \(N(t)\) is the quantity that remains after time \(t\).
  • \(N_0\) is the initial quantity.
  • \(t\) is the elapsed time.
  • \(t_{1/2}\) is the half-life of the substance.
In the problem given, the half-life of \(^{87}\text{Rb}\) is 4.75 \(\times\) 10\(^{10}\) years. By using this formula, we can calculate how much of the \(^{87}\text{Rb}\) remains after a period of 4.6 \(\times\) 10\(^{9}\) years, which is crucial to determining the past composition of a substance when analyzed today.
Isotope Ratio
The ratio of different isotopes of the same element is a valuable tool in understanding both the chemistry and the history of the universe. Isotopes are atoms with the same number of protons but different numbers of neutrons.
In our specific exercise, we deal with two isotopes of rubidium: \(^{87}\text{Rb}\) and \(^{85}\text{Rb}\). Each has a different ratio in current geological samples:
  • \(27.83\%\) of rubidium atoms are \(^{87}\text{Rb}\).
  • \(72.17\%\) are \(^{85}\text{Rb}\).
Determining the ratio of these isotopes requires understanding the historical context and the changes over time due to the decay of \(^{87}\text{Rb}\). By knowing these ratios, geologists can backtrack and determine the initial composition of elements when our solar system was formed. This can lead to insights about the age and the processes that have occurred over billions of years.
Rubidium Isotopes
Rubidium is a fascinating element with isotopes that have applications in various scientific fields, such as geology and pharmacology. Two important isotopes of rubidium are:
  • \(^{87}\text{Rb}\): A radioactive isotope with a half-life of 4.75 \(\times\) 10\(^{10}\) years.
  • \(^{85}\text{Rb}\): A stable isotope.
Rubidium-87 decays into strontium-87, which is used in rubidium-strontium dating to estimate the age of rocks and minerals. This decay makes \(^{87}\text{Rb}\) significant in studies involving long-term geological changes. The presence of \(^{85}\text{Rb}\), being stable, acts as a baseline to compare against the decreasing amounts of \(^{87}\text{Rb}\).
Understanding these rubidium isotopes not only helps in dating terrestrial formations but also opens up wider applications in studying astronomical samples and providing information on planetary formations.

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Most popular questions from this chapter

\(\textbf{We Are Stardust.}\) 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".

Consider the nuclear reaction \(^{4}_{2}He\) + \(^{7}_{3}Li\) \(\rightarrow\) X + \(^{1}_{0}n\) where X is a nuclide. (a) What are Z and A for the nuclide X? (b) Is energy absorbed or liberated? How much?

\(\textbf{An Oceanographic Tracer.}\) 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?

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 g, which includes \(9.4 \mu Ci\) of \(^{59}Fe\) 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?

Many radioactive decays occur within a sequence of decays for example, \(^{234}_{92}U\) \(\rightarrow\) \(^{230}_{88}Th\) \(\rightarrow\) \(^{226}_{84}Ra\). The half-life for the \(^{234}_{92}U\) \(\rightarrow\) \(^{230}_{88}Th\) decay is \(2.46 \times 10^{5}\) y, and the half-life for the \(^{230}_{88}Th\) \(\rightarrow\) \(^{226}_{84}Ra\) decay is \(7.54 \times 10^{4}\) y. Let 1 refer to \(^{234}_{92}U\), 2 to \(^{230}_{88}Th\), and 3 to \(^{226}_{84}Ra\); let \(\lambda\)1 be the decay constant for the \(^{234}_{92}U\) \(\rightarrow\) \(^{230}_{88}Th\) decay and \(\lambda\)2 be the decay constant for the \(^{230}_{88}Th\) \(\rightarrow\) \(^{226}_{84}Ra\) decay. The amount of \(^{230}_{88}Th\) present at any time depends on the rate at which it is produced by the decay of \(^{234}_{92}U\) and the rate by which it is depleted by its decay to \(^{226}_{84}Ra\). Therefore, \(d$$N_{2}\)(\(t\))/\(d$$t\) = \(\lambda$$_1$$N$$_1\)(\(t\)) - \(\lambda$$_2$$N$$_2\)(\(t\)). If we start with a sample that contains \(N_{10}\) nuclei of \(^{234}_{92}U\) and nothing else, then \(N{(t)}\) = \(N_{01}\)e\(^{-\lambda_{1}t}\). Thus \(dN_{2}(t)/dt\) = \(\lambda\)1\(N_{01}\)e\(^{-\lambda_{1}t}\)- \(\lambda2N_{2}(t)\). This differential equation for \(N_{2}(t)\) can be solved as follows. Assume a trial solution of the form \(N_{2}(t)\) = \(N_{10} [h_{1}e^{-\lambda_1t}\) + \(h_{2}e^{-\lambda_2t}\)] , where \(h_{1}\) and \(h_{2}\) are constants. (a) Since \(N_{2}(0)\) = 0, what must be the relationship between \(h_{1}\) and \(h_{2}\)? (b) Use the trial solution to calculate \(dN_{2}(t)/dt\), and substitute that into the differential equation for \(N_{2}(t)\). Collect the coefficients of \(e^{-\lambda_{1}t}\) and \(e^{-\lambda_{2}t}\). Since the equation must hold at all t, each of these coefficients must be zero. Use this requirement to solve for \(h_{1}\) and thereby complete the determination of \(N_{2}(t)\). (c) At time \(t = 0\), you have a pure sample containing 30.0 g of \(^{234}_{92}U\) and nothing else. What mass of \(^{230}_{88}Th\) is present at time \(t = 2.46 \times 10^{5}\) y, the half-life for the \(^{234}_{92}U\) decay?
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