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Chapter 25: Problem 31

A lunar rock was analyzed for argon by mass spectrometry and for potassium by atomic absorption. The results of these analyses showed that the sample contained \(3.02 \times 10^{-5} \mathrm{mL} \mathrm{g}^{-1}\) of argon and \(0.083 \%\) of potassium. The half-life of potassium- 40 is \(1.248 \times\) \(10^{9} \mathrm{y} \cdot\) Calculate the age of the lunar rock.

Short Answer

Expert verified
The age of the lunar rock is approximately \(3.5 \times 10^{9}\) years.

Step by step solution

01

Calculate the number of moles of argon

First calculate the number of moles of argon in the lunar rock. Recall that 1 mole of any gas at standard temperature and pressure (STP) occupies 22.4 liters or \(2.24 \times 10^{4} \mathrm{mL} \). Therefore, if the lunar rock contains \(3.02 \times 10^{-5} \mathrm{mL} \mathrm{g}^{-1}\) of argon, the number of moles of argon (\(n_{Ar}\)) per gram of lunar rock can be calculated as \(n_{Ar} = \frac{(3.02 \times 10^{-5} \mathrm{mL} \mathrm{g}^{-1})}{(2.24 \times 10^{4} \mathrm{mL / mol})} = 1.35 \times 10^{-9} \mathrm{mol} \mathrm{g}^{-1} \)
02

Calculate the number of moles of potassium-40

Next calculate the number of moles of potassium-40 in the lunar rock. The total potassium content is given as \(0.083 \%\). Given natural isotopic abundance of potassium-40 is 0.0117\%, the amount of potassium-40 (\(n_{K}\)) per gram of lunar rock can be calculated as \(n_{K} = 0.083 \times 0.0117 \times \frac{1}{100} = 9.71 \times 10^{-5} \mathrm{mol} \mathrm{g}^{-1} \)
03

Calculate the age of the rock

Use the ratio of the number of moles of argon to the number of moles of potassium-40 to calculate the age of the rock. The ratio of argon-40 to potassium-40 is given by (\(1.35 \times 10^{-9}\) / \(9.71 \times 10^{-5}\)) = \(1.39 \times 10^{-5}\). Use the equation for radioactive decay which is \(N_t = N_0 \times \frac{1}{2}^{\left(\frac{t}{t_{1/2}}\right)}\). Rearranging gives \(t = t_{1/2} \times \frac{log(\frac{N_0}{N_t})}{log(2)} = 1.248 \times 10^{9} years \times \frac{log(1 / 1.39 \times 10^{-5})}{log(2)} = 3.5 \times 10^{9} years\)

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Most popular questions from this chapter

Hydrogen gas is spiked with tritium to the extent of \(5.00 \%\) by mass. What is the activity in curies of a \(4.65 \mathrm{L}\) sample of this gas at \(25.0^{\circ} \mathrm{C}\) and 1.05 atm pressure? [Hint: Use 3.02 u as the atomic mass of tritium and data from elsewhere in the text, as necessary.]

Supply the missing information in each of the following nuclear equations representing a radioactive decay process.(a) \(160_?\mathrm{W} \longrightarrow\\{\mathrm{Hf}+?\) (b) \(38_? \mathrm{Cl} \longrightarrow_{?}^{?} \mathrm{Ar}+?\) (c) \(^{214} ? \longrightarrow_{?}^{?} \mathrm{Po}+_{-1}^{0} \boldsymbol{\beta}\) (d) \(_{17}^{32} \mathrm{Cl} \longrightarrow_{1}^{?} ?+?\)

Both \(\beta^{-}\) and \(\beta^{+}\) emissions are observed for artificially produced radioisotopes of low atomic numbers, but only \(\beta^{-}\) emission is observed with naturally occurring radioisotopes of high atomic number. Why do you suppose this is so?

The following reactions are carried out with HCl(aq) containing some tritium ( \(_{1}^{3} \mathrm{H}\) ) as a tracer. Would you expect any of the tritium radioactivity to appear in the \(\mathrm{NH}_{3}(\mathrm{g}) ?\) In the \(\mathrm{H}_{2} \mathrm{O} ?\) Explain. $$\begin{array}{c} \mathrm{NH}_{3}(\mathrm{aq})+\mathrm{HCl}(\mathrm{aq}) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq}) \\ \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{aq})+\mathrm{NaOH}(\mathrm{aq}) \longrightarrow\\\\\mathrm{NaCl}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(1)+\mathrm{NH}_{3}(\mathrm{g})\end{array}$$

One member each of the following pairs of radioisotopes decays by \(\beta^{-}\) emission, and the other by positron \(\left(\beta^{+}\right)\) emission: \((\mathrm{a})_{15}^{29} \mathrm{P}\) and \(_{15}^{33} \mathrm{P} ;(\mathrm{b}) \stackrel{120}{53} \mathrm{I}\) and \(_{53}^{134} \mathrm{I} .\) Which is which? Explain your reasoning.
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