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 1276: 61

Answer

${\bf 5.94 \times 10^9} \, \rm{years}$

Work Step by Step

To determine the approximate time that has elapsed since a supernova created the isotopes $^{235}\rm{U}$ and $^{238}\rm{U}$, we’ll use their current abundance ratio and their respective half-lives. This allows us to estimate how long ago both isotopes began decaying from a point when they were likely present in equal quantities. Let $ N_{238} $ and $ N_{235} $ represent the number of atoms of $^{238}\rm{U}$ and $^{235}\rm{U}$, respectively. After time $ t $, their ratio now is: $$ \frac{N_{238}}{N_{235}} =\dfrac{0.9928 U}{0.0072U}= \bf 137.9\tag 1 $$ Initially, we assume $ \frac{N_{238}}{N_{235}} = 1 $ right after the supernova. Each isotope decays according to its own half-life, leading to: $$ \frac{N_{238}}{N_{235}} = \frac{\left( N_{238} \right)_0 \left( \frac{1}{2} \right)^{t / t_{1/2}^{238}}}{\left( N_{235} \right)_0 \left( \frac{1}{2} \right)^{t / t_{1/2}^{235}}} $$ - Simplifying this, since $ \frac{\left( N_{238} \right)_0}{\left( N_{235} \right)_0} = 1 $: $$ \frac{N_{238}}{N_{235}} = \frac{\left( \frac{1}{2} \right)^{t / t_{1/2}^{238}}}{\left( \frac{1}{2} \right)^{t / t_{1/2}^{235}}} = \left( \frac{1}{2} \right)^{t \left( \frac{1}{t_{1/2}^{238}} - \frac{1}{t_{1/2}^{235}} \right)} $$ Plug from (1) $$ 137.9 = \left( \frac{1}{2} \right)^{t \left( \frac{1}{t_{1/2}^{238}} - \frac{1}{t_{1/2}^{235}} \right)} $$ Take the Logarithm of Both Sides: $$ \ln(137.9) = -t \left( \frac{\ln(2)}{t_{1/2}^{238}} - \frac{\ln(2)}{t_{1/2}^{235}} \right) $$ $$ t = \frac{-\ln(137.9)}{\left( \frac{\ln(2)}{t_{1/2}^{238}} - \frac{\ln(2)}{t_{1/2}^{235}} \right)} $$ Substitute the known from appendix C, where $ t_{1/2}^{238} = 4.47 \times 10^9 $ years and $ t_{1/2}^{235} = 7.04 \times 10^8 $ years: $$ t = \frac{-\ln(137.9)}{\frac{\ln(2)}{4.47 \times 10^9} - \frac{\ln(2)}{7.04 \times 10^8}} $$ $$ t \approx \color{red}{\bf 5.94 \times 10^9} \, \rm{years} $$ This means that the supernova that produced the heavy elements, including $^{235}\rm{U}$ and $^{238}\rm{U}$, likely occurred approximately 5.94 billion years ago.
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