Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 42 - Nuclear Physics - Exercises and Problems - Page 1275: 53

Answer

$ \approx \bf 207\rm \;million \;years$

Work Step by Step

This problem involves dating a lava rock using the potassium-argon dating method. The isotope $^{40}\text{K}$ has a half-life of 1.28 billion years and decays by two routes: - 89% decays into $^{40}\text{Ca}$ (calcium) by beta-minus decay. - 11% decays into $^{40}\text{Ar}$ (argon) by electron capture. The ratio of $^{40}\text{Ar}$ to $^{40}\text{K}$ in the rock is 0.013. We need to find the age of the rock. - Let $ N_0 $ represent the initial number of $^{40}\text{K}$ atoms at the time of lava solidification. - Let $ N_K $ represent the current number of $^{40}\text{K}$ atoms remaining. - Let $ N_{\text{Ar}} $ represent the number of $^{40}\text{Ar}$ atoms formed by the decay of $^{40}\text{K}$. Since 11% of $^{40}\text{K}$ decays into $^{40}\text{Ar}$, so $$ N_{\text{Ar}} = 0.11 (N_0 - N_K)\tag 1 $$ The ratio of $^{40}\text{Ar}$ to $^{40}\text{K}$ atoms in the rock is given as: $$ \frac{N_{\text{Ar}}}{N_K} = 0.013 $$ So, $$ N_{\text{Ar}}=0.013 N_K\tag 2 $$ Plug into (1); $$ 0.013 N_K= 0.11 (N_0 - N_K) $$ Rearrange to isolate $ \dfrac{N_0}{N_K} $: $$ \frac{0.11 (N_0 - N_K)}{N_K} = 0.013 $$ $$ 0.11 \left( \frac{N_0}{N_K} - 1 \right) = 0.013 $$ $$ \frac{N_0}{N_K} - 1 = \frac{0.013}{0.11} $$ $$ \frac{N_0}{N_K} = 1 + \frac{0.013}{0.11}=\bf 1.11818 $$ Therefore: $$ \frac{N_K}{N_0} = \frac{1}{1.11818} \approx \bf 0.89431\tag 3 $$ This means that 89.4% of the original $^{40}\text{K}$ atoms remain at the time when the $^{40}\text{Ar}/^{40}\text{K}$ ratio is 0.013. Now we need to calculate the Age Using the Decay Formula: $$ N_K = N_0 \left( \frac{1}{2} \right)^{t / t_{1/2}} $$ $$ \frac{N_K}{N_0}= \left( \frac{1}{2} \right)^{t / t_{1/2}} $$ Plug from (4); $$ 0.89431= \left( \frac{1}{2} \right)^{t / 1.28 } $$ Take the Natural Logarithm of Both Sides: $$ \ln(0.89431) = \frac{t}{1.28 } \ln\left( \frac{1}{2} \right) $$ Solve for $ t $: $$ t=\dfrac{ 1.28 \ln(0.89431)}{\ln\left( \frac{1}{2} \right)} $$ $$ t=\bf 0.206276\;\rm bilion\;yesr=\color{red}{\bf 206,276,000}\;yr $$
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