Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

The isotope ${}^{40}K$ can decay to either ${}^{40}C a$ or ${}^{40}A r$ ; assume both decays have a half-life of $1.26 \times 10^{9} y$ . The ratio of the Ca produced to Ar the produced is 8.54/1 = 8.54. A sample originally had only ${}^{40}K$ . It now has equal amounts of ${}^{40}K$ and ${}^{40}A r$ ; that is, the ratio of K to Ar **is 1/1 = 1. How old is the sample? (Hint: Work this like other radioactive-dating problems, except that this decay has two products.)**

The sample is $4.24 \times 10^{9} y$ old.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

Half-life of $^{40} Ca$ and $^{40} A r$ , $T_{1 / 2} = 1.26 \times 10^{9} y$

Ratio of Ca to Ar is $N_{c a} / N_{k} = 8.54 / 1$ .

Ratio of K to Ar is $N_{A r} / N_{k} = 1$ .

Step 2: Understanding the concept of radioactive dating

The concept of radioactive dating is used to know the age of the fossil using the given decay rates. Here, the problem is of a parent nucleus that can decay into two products. Being the products of the decay, the number of atoms of calcium and argon were initially potassium atoms before the decay. Thus, the total number of initial potassium atoms can be given as the sum of all the three atoms present in the rock after the decay at the given time.

Formulae:

The disintegration constant, $λ = \frac{\ln 2}{T_{1 / 2}} \dots . . . . (1)$

where, $T_{1 / 2}$ is the half-life of the substance.

The undecayed sample remaining after a given time, $N = N_{0} e^{- λ t} . . . . . . . . . . . . (2)$

Step 3: Calculation of the age of the sample

We note that every calcium-40 atom and krypton-40 atom found now in the sample was once one of the original numbers of potassium atoms.

Thus, the total number of initial potassium-40 atoms can be given as:

$N_{k o} = N_{k} + N_{A r} + N_{c a}$

Thus, substituting equation (1) in (2) with the above data, we can get the age of the sample as follows:

$\ln (\frac{N_{k}}{N_{k o}}) = - (\frac{\ln 2}{T_{1 / 2}}) t \ln (\frac{N_{k}}{N_{k} + N_{A r} + N_{c a}}) = - (\frac{\ln 2}{1.26 \times 10^{9} y}) t \ln (\frac{1}{1 + 1 + 8.54}) = - (0.55501 \times 10^{- 9} / y) t \ln \frac{1}{10.54} = (0.55501 \times 10^{- 9} / y) t t = \frac{2.3552}{0.55501 \times 10^{- 9} / y t} = 4.24 \times 10^{9} y$

Hence, the age of the sample is $4.24 \times 10^{9} y$ .

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Fundamentals Of Physics

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Short Answer

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Step 1: Given data

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Step 3: Calculation of the age of the sample

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