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Chapter 46: Problem 20

Rubidium- 87 has a half-life of \(4.9 \times 10^{10}\) years and decays to strontium- 87, which is stable. In an ancient rock, the ratio of \({ }^{87} \mathrm{Sr}\) to \({ }^{87} \mathrm{Rb}\) is \(0.0050 .\) If we assume all the strontium came from rubidium decay, about how old is the rock? Repeat if the ratio is \(0.210\).

Short Answer

Expert verified
The rock is approximately 3.4 million years old with ratio 0.0050 and 64 million years old with ratio 0.210.

Step by step solution

01

Understand the Concept of Half-Life

The half-life of a substance is the time it takes for half of it to decay. For Rubidium-87, the half-life is \(4.9 \times 10^{10}\) years. The decay of Rubidium-87 to Strontium-87 can be described using the half-life formula.
02

Set Up the Decay Equation

The decay of Rubidium-87 over time can be described by the equation: \[N_t = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}\] where \(N_t\) is the amount of Rubidium-87 at time \(t\), \(N_0\) is the initial amount of Rubidium-87, and \(T_{1/2}\) is the half-life. Since the question provides ratios, we rearrange this equation to find \(t\), the age of the rock.
03

Express the Ratio in Terms of the Decay Equation

Given the ratio of \(^{87}\)Sr to \(^{87}\)Rb, let \(x\) be the initial amount of \(^{87}\)Rb. After the decay, the amount of \(^{87}\)Rb is \((1-r)x\) and the amount of \(^{87}\)Sr is \(rx\). Thus, \(\frac{rx}{(1-r)x} = r\). Rearrange to find \(x\) in terms of initial amount \((1-x)\), where \(x = \left(\frac{r}{1-r}\right)(1-x)\).
04

Solve the Decay Equation With the Given Ratios

Substitute the given ratios into the equations. For \(r = 0.0050\), calculate: \[0.0050 = \frac{N_0 - N_t}{N_t}\] \[N_t = \frac{N_0}{1.0050}\]. Thus rearrange: \[ln\left(\frac{N_0}{N_t}\right) = \frac{t \ln(2)}{T_{1/2}}\]. Repeat for \(r = 0.210\).
05

Calculate the Rock's Age Using Both Ratios

Plug the values into the equation from Step 4. For \(r = 0.0050\): \[t = \frac{1}{\ln(2)}(4.9 \times 10^{10}) \cdot \ln(1.0050)\]. Repeat for \(r = 0.210\): \[t = \frac{1}{\ln(2)}(4.9 \times 10^{10}) \cdot \ln(1.210)\]. Calculate each \(t\).

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Key Concepts

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

Half-Life
The concept of half-life is essential to understanding radioactive dating methods like Rubidium-Strontium dating. Half-life refers to the time it takes for half of a quantity of a radioactive isotope to decay into a stable form. For instance, Rubidium-87 has a half-life of approximately \(4.9 \times 10^{10}\) years. This means that over this time span, half of the Rubidium-87 atoms will have decayed to form Strontium-87.

Understanding half-life allows scientists to calculate the age of objects, such as rocks or fossils, by measuring the remaining amount of a radioactive substance compared to its decay products. Here, the initial and remaining amounts of Rubidium-87 are used to tell the age of ancient rocks accurately. It's essential to remember that half-life remains a constant value for any given isotope, making it a reliable metric for dating materials.
Decay Equation
The decay equation is a mathematical representation of how a radioactive substance transforms over time. The decay equation for Rubidium-87 can be written as: \[ N_t = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]Here, \(N_t\) represents the quantity of Rubidium-87 remaining after time \(t\), \(N_0\) is the initial quantity, and \(T_{1/2}\) is the half-life.

This equation allows scientists to calculate the remaining quantity of a substance after a given period. By rearranging the equation, we can determine the time \(t\) 鈥� essentially enabling us to deduce the age of a rock or fossil. For example, by knowing the ratio of remaining Rubidium-87 to the formed Strontium-87, we can input these values into the decay equation and solve for \(t\), providing us with the rock's age. The decay equation is a crucial tool in radiometric dating.
Isotopes
Isotopes are atoms of the same element that differ in the number of neutrons within their nuclei. This deviation usually does not affect the chemical properties significantly, but it can influence an isotope's stability and radioactive properties. In the case of Rubidium-Strontium dating, Rubidium-87 is a radioactive isotope, which means it naturally decays over time into a more stable isotope, Strontium-87.

Isotopes like Rubidium-87 are used in radioactive dating because they provide a natural clock for determining the age of materials. As the isotope decays, it undergoes a predictable transformation allowing for the calculation of elapsed time. Understanding isotopes and their properties helps scientists apply models like the decay equation accurately in dating studies. They form the foundation of the scientific approach to determining the ages of rocks and other geological features.
Rubidium-Strontium Dating
Rubidium-Strontium dating is a type of radiometric dating method that utilizes the decay of Rubidium-87 to Strontium-87 to determine the age of rocks and minerals. This technique is particularly useful in dating ancient geological formations, as its long half-life (\(4.9 \times 10^{10}\) years) allows for dating materials that are billions of years old. It works on the principle that the ratio of Rubidium-87 to Strontium-87 changes over time as Rubidium decays into Strontium.

By measuring the ratio of these isotopes in a rock sample and comparing it to the known decay rate, scientists can calculate the time elapsed since the rock formed. This method assumes that the Strontium-87 detected is solely from the decay of Rubidium-87 and not from external sources. Therefore, when calculating the age of a rock, assumptions and corrections may need to be made to account for any extraneous Strontium not derived from Rubidium decay. The Rubidium-Strontium dating method is integral to understanding the history of our planet and its geological processes.

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Most popular questions from this chapter

A beam of alpha particles passes through flesh and deposits \(0.20 \mathrm{~J}\) of energy in each kilogram of flesh. The \(Q\) for these particles is \(12 \mathrm{~Sv} / \mathrm{Gy}\). Find the dose in \(\mathrm{Gy}\) and \(\mathrm{rd}\), as well as the effective dose in Sv and rem. Recall that \(H=Q D\) where $$ D=\text { Dose }=\frac{\text { Absorbed energy }}{\text { Mass }}=0.20 \mathrm{~J} / \mathrm{kg}=0.20 \mathrm{~Gy}=20 \mathrm{rd} $$ Hence, \(H=\) Effective dose \(=Q(\) dose \()=(12 \mathrm{~Sv} / \mathrm{Gy})(0.20 \mathrm{~Gy})=2.4 \mathrm{~Sv}=2.4 \times 10^{2} \mathrm{rem}\)

A tumor on a person's leg has a mass of \(3.0 \mathrm{~g} .\) What is the minimum activity a radiation source can have if it is to furnish a dose of \(10 \mathrm{~Gy}\) to the tumor in \(14 \mathrm{~min}\) ? Assume each disintegration within the source, on the average, provides an energy \(0.70 \mathrm{MeV}\) to the tumor. A dose of 10 Gy corresponds to \(10 \mathrm{~J}\) of radiation energy being deposited per kilogram. Since the tumor has a mass of \(0.0030 \mathrm{~kg}\), the energy required for a 10 Gy dose is \((0.0030 \mathrm{~kg})(10 \mathrm{~J} / \mathrm{kg})=0.030 \mathrm{~J}\). Each disintegration provides \(0.70 \mathrm{MeV}\), which in joules is $$ \left(0.70 \times 10^{6} \mathrm{eV}\right)\left(1.60 \times 10^{-19} \mathrm{~J} / \mathrm{eV}\right)=1.12 \times 10^{-13} \mathrm{~J} $$ A dose of 10 Gy requires that an energy of \(0.030 \mathrm{~J}\) be delivered. That total energy divided by the energy per disintegration, yields the number of disintegrations: $$ \frac{0.030 \mathrm{~J}}{1.12 \times 10^{-13} \mathrm{~J} / \text { disintegration }}=2.68 \times 10^{11} \text { disintegrations } $$ They are to occur in \(14 \mathrm{~min}\) (or \(840 \mathrm{~s}\) ), and so the disintegration rate is $$ \frac{2.68 \times 10^{11}}{840 \mathrm{~s}} \text { disintegrations }=3.2 \times 10^{8} \text { disintegrations/s. } $$ Hence, the source activity must be at least \(3.2 \times 10^{8} \mathrm{~Bq}\). Since \(1 \mathrm{Ci}=3.70 \times 10^{10} \mathrm{~Bq}\), the source activity must be at least \(8.6 \mathrm{mCi}\).

A beam of \(5.0 \mathrm{MeV}\) alpha particles \((q=2 e)\) has a cross-sectional area of \(1.50 \mathrm{~cm}^{2} .\) It is incident on flesh ( \(\rho=950 \mathrm{~kg} / \mathrm{m}^{3}\) ) and penetrates to a depth of \(0.70 \mathrm{~mm} .\) ( \(a\) ) What dose (in Gy) does the beam provide to the flesh in a time of \(3.0 \mathrm{~s}\) ? (b) What effective dose does it provide? Assume the beam to carry a current of \(2.50 \times 10^{-9} \mathrm{~A}\) and to have \(Q=14\). Using the current, find the number of particles deposited in the flesh in \(3.0 \mathrm{~s}\), keeping in mind that for each particle \(q=2 e:\) Number in \(3.0 \mathrm{~s}=\frac{I t}{q}=\frac{\left(2.50 \times 10^{-9} \mathrm{C} / \mathrm{s}\right)(3.0 \mathrm{~s})}{3.2 \times 10^{-19} \mathrm{C}}=2.34 \times 10^{10}\) particles Each \(5.0-\mathrm{MeV}\) alpha particle deposits an energy of \(\left(5.0 \times 10^{6} \mathrm{eV}\right)\left(1.60 \times 10^{-19} \mathrm{~J} / \mathrm{eV}\right)=8.0 \times 10^{-13} \mathrm{~J} .\) In \(3.0\) s a total energy of \(2.34 \times 10^{10}\) particles) \(\left(8.0 \times 10^{-13} \mathrm{~J} /\right.\) particle \()\) is deposited. And it is delivered to a volume of area \(1.50 \mathrm{~cm}^{2}\) and thickness \(0.70 \mathrm{~mm}\). Therefore, $$ \text { Dose }=\frac{\text { Energy }}{\text { Mass }}=\frac{\left(2.34 \times 10^{10}\right)\left(8.0 \times 10^{-13} \mathrm{~J}\right)}{\left(950 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(0.070 \times 1.5 \times 10^{-6} \mathrm{~m}^{3}\right)}=188 \mathrm{~Gy}=1.9 \times 10^{2} \mathrm{~Gy} $$ Effective dose \(=Q(\) dose \()=(14)(188)=2.6 \times 10^{4} \mathrm{~Sv}\)

An alpha-particle beam enters a charge collector and is measured to carry \(2.0 \times 10^{-14} \mathrm{C}\) of charge into the collector each second. The beam has a cross-sectional area of \(150 \mathrm{~mm}^{2}\), and it penetrates human skin to a depth of \(0.14 \mathrm{~mm}\). Each particle has an initial energy of \(4.0 \mathrm{MeV}\). The \(Q\) for such particles is about 15 . What effective dose, in \(\mathrm{Sv}\) and in rem, does a person's skin receive when exposed to this beam for \(20 \mathrm{~s}\) ? Take \(\rho=900 \mathrm{~kg} / \mathrm{m}^{3}\) for skin.

Neutrons produced by fission reactions must be slowed by collisions with moderator nuclei before they are effective in causing further fissions. Suppose an \(800-\mathrm{keV}\) neutron loses 40 percent of its energy on each collision. How many collisions are required to decrease its energy to \(0.040 \mathrm{eV} ?\) (This is the average thermal energy of a gas particle at \(35^{\circ} \mathrm{C}\).) After one collision, the neutron energy is down to \((0.6)(800 \mathrm{keV})\). After two, it is \((0.6)(0.6)(800 \mathrm{keV})\); after three, it is \((0.6)^{3}(800 \mathrm{keV})\). Therefore, after \(n\) collisions, the neutron energy is \((0.6)^{n}(800 \mathrm{keV})\). We want \(n\) large enough so that $$ (0.6)^{n}\left(8 \times 10^{5} \mathrm{eV}\right)=0.040 \mathrm{eV} $$ Taking the logarithms of both sides of this equation yields $$ \begin{array}{c} n \log _{10} 0.6+\log _{10}\left(8 \times 10^{5}\right)=\log _{10} 0.04 \\ (n)(-0.222)+5.903=-1.398 \end{array} $$ from which we find \(n\) to be 32.9. So 33 collisions are required.
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