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Chapter 25: Problem 60

One method of dating rocks is based on their \(^{87} \mathrm{Sr} /^{87} \mathrm{Rb}\) ratio. \(^{87} \mathrm{Rb}\) is a \(\beta^{-}\) emitter with a half- life of \(5 \times 10^{11}\) years. A certain rock has a mass ratio \(^{87} \mathrm{Sr} /^{87} \mathrm{Rb}\) of \(0.004 / 1.00 .\) What is the age of the rock?

Short Answer

Expert verified
The age of the rock is calculated by substituting the given values into the derived formula. This will give the final age in years.

Step by step solution

01

Identify given information

From the exercise, the half-life of \(^{87}\mathrm{Rb}\) is \(5 \times 10^{11}\) years. Also, the mass ratio \(^{87}\mathrm{Sr} /^{87}\mathrm{Rb}\) of the rock is 0.004/1.00.
02

Use the Decay Equation to find decay constant

The decay constant (\(λ\)) is given by \(\frac{ln 2}{t_{1/2}}\), where \(t_{1/2}\) is the half-life. So, \(λ = \frac{ln 2}{5 \times 10^{11}}\).
03

Use the ratio to compute time

In radioactive decay, the ratio of Sr-87 to Rb-87 is related to time (t) by the equation \(\frac{Nf}{Ni} = e^{-λt}\), where \(Nf\) is the final quantity (Sr-87 in this case), \(Ni\) is the initial quantity (Rb-87 in this case), and \(e\) is Euler's number. Rearranging for t gives \(t = -\frac{ln (\frac{Nf}{Ni})}{λ}\).
04

Substitute values and solve for t

Substitute \(Nf/Ni = 0.004/1.00\) and our previously calculated \(λ\) into the equation to find t. \(t = -\frac{ln (0.004)}{\frac{ln 2}{5 \times 10^{11}}}\)

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Most popular questions from this chapter

Which of the following types of radiation is deflected in a magnetic field? (a) \(X\) ray; (b) \(\gamma\) ray; (c) \(\beta\) ray; (d) neutrons.

Calculate the minimum kinetic energy (in megaelectronvolts) that \(\alpha\) particles must possess to produce the nuclear reaction $$_{2}^{4} \mathrm{He}+^{14}_{7} \mathrm{N} \longrightarrow^{17}_{8} \mathrm{O}+_{1}^{1} \mathrm{H}.$$ The nuclidic masses are \(_{2}^{4} \mathrm{He}=4.00260 \mathrm{u}\); \(_{7}^{14} \mathrm{He}=14.00307\mathrm{u}\);\(_{1}^{1} \mathrm{H}=1.00783 \mathrm{u}\);\(_{8}^{17} \mathrm{H}=16.99913 \mathrm{u}\);

The immediate decay product of element 118 is thought to be element \(116 .\) Write a complete nuclear equation for this reaction.

Of the following nuclides, the highest nuclear binding energy per nucleon is found in (a) \(_{1}^{3} \mathrm{H} ;\) (b) \(_{8}^{16} \mathrm{O} ;\) (c) \(_{26}^{56} \mathrm{Fe}\); (d) \(_{92}^{235} \mathrm{U}\).

Scientists from Dubna, Russia, observed the existence of elements 118 and 116 at the Joint Institute for Nuclear Research U400 cyclotron in 2005. This was the result of bombarding calcium- 48 ions on a californium-249 target. Write a complete nuclear equation for this reaction.
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